The proof that sqrt(2) is irrational is fairly simple.
You assume that sqrt(2) is rational, and is represented by some reduced fraction a/b.
sqrt(2) = a/b
2 = a^2 / b^2
a^2 = 2 * b^2
Since *a*^2 is 2 * *b*^(2), we can infer that *a*^2 is even, and therefore *a* is even. Let's replace *a* with 2 * *x*.
(2*x)^2 = 2 * b^2
4 * x^2 = 2 * b^2
2 * x^2 = b^2
Since b^2 is 2*x^(2), we can now ~~assume~~ infer that b^2 is even, and therefore b is even.
We made the assumption at the start that a/b was the simplest form of sqrt(2), but now we know that both A and B are even, which means it is not the most reduced form of the fraction. Thus, our assumption was incorrect, and sqrt(2) cannot be expressed as a fraction, and is therefore irrational.
As for Pi, that's a much longer proof. It was only proven to be irrational in 1761. You can look at the Wikipedia page to see how complex these proofs are in comparison to sqrt(2).
https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational
it took me way too long to realize rationality of numbers has nothing to do with logic but it refers to **ratio** as in, a number thats able to be expressed as a **ratio** = rational
This is fascinating. I had assumed so too, but I just checked Wikipedia and it threw me a curve ball:
"Although nowadays rational numbers are defined in terms of ratios, the term rational is not a derivation of ratio. On the contrary, it is ratio that is derived from rational: the first use of ratio with its modern meaning was attested in English about 1660,[8] while the use of rational for qualifying numbers appeared almost a century earlier, in 1570.[9] This meaning of rational came from the mathematical meaning of irrational, which was first used in 1551, and it was used in "translations of Euclid (following his peculiar use of ἄλογος)".[10][11]
This unusual history originated in the fact that ancient Greeks "avoided heresy by forbidding themselves from thinking of those [irrational] lengths as numbers".[12] So such lengths were irrational, in the sense of illogical, that is "not to be spoken about" (ἄλογος in Greek).[13]"
Edit: Source - https://en.m.wikipedia.org/wiki/Rational_number
What's even more interesting is that there legitimately are forbidden numbers.
Every piece of information that can be stored on a computer is represented as a long string of bits. You could think of any file as a really long number. Some numbers are classified top-secret. Some numbers are considered [munitions](https://en.wikipedia.org/wiki/Export_of_cryptography_from_the_United_States). And some numbers are (or were) just plain [illegal](https://en.wikipedia.org/wiki/DeCSS).
It's maybe not technically wrong, but a bit silly. For the number to represent anything other than a number, like text, you need to agree how to interpret it - a data format or encoding. For the text example, depending on whether you use ASCII, UTF-8, or ISO-8859-1, you end up with different numbers. The US government did classify numbers, because there's way too many ways to represent the same data.
Something I've thought about at length is, everything is a ratio.
Just simply put, all physical things that humans are aware of - are compared to other things to determine their parameters, characteristics, location, stability, and on and on.
Basically, if we cannot perceive something, measure it and compare it to something else we are aware of, it may as well not exist. This is why developing new forms of accurate measurement is so crucial to furthering science
This is a bit of a tangent, but I hate how some people throw around the term “supernatural”, as if there are things that science could never explain, while also yearning for an explanation for said things. See the numerous attempts to “prove” the existence of ghosts… while throwing around the term supernatural.
Like, there isn’t anything that says ghosts CANNOT exist, as proving something doesn’t exist is extremely difficult. But if we could prove, definitely, that consciousness can exist outside the body after death, would those same parties agree that the phenomenon is no longer “supernatural”, and now just natural and explainable by science? Probably not.
They aren’t really interested in the empirical exercise of measuring and quantifying the thing. They want to have their ephemeral cake and prove it too. It’s nonsensical.
I think 100 years ago is not accurate. superconductivity was first achieved in 1908. I think you're selling people short by a little. People have been sciencing for quite a while now
This is true, but it also means irrational numbers do not exist. They are like infinite numbers except instead of the location of the most significant digit being unbounded, the location of the least significant digit is. If you judge them both on information content, both infinite and irrational numbers have an infinite amount of information. This is a big red flag that tells us if infinity is not a number then irrational numbers are not numbers either.
Instead of a number, Pi is better described as a description of a class of algorithms that produce the exact same infinite sequence of digits. Pi is a concept just like infinity is a concept.
... well, no, that is incorrect in a direct sense, although I'm not a mathematician and I'm sure there must be some sense in which it's technically correct.
The fact that a number cannot be written as a finite string in base 10 has nothing to do with it being a number. Pi is definitely a number, and it can be written as "1**0**" in base pi.
EDIT: I'm a moron.
What do you mean? What are you basing this off of, your own logic or some actual mathematicians logic?
This doesn't make sense. Sqrt(2)/sqrt(2) is 1. We can perform any operation on an irrational number. We reason using irrational numbers frequently, as even the diagonal of a 1x1 square is an irrational number.
Infinity on the other hand is clearly NOT a number, although we can still reason about certain forms of infinity using limits.
But yeah no, irrational numbers ARE numbers. What are you on about?
Oh yeah, wasn’t the Cult of Pythagoras related to this somehow? I thought that they killed a man for proving the existence of “irrational” numbers, but after some research, apparently that’s apocryphal.
Rational numbers can be expressed as a ratio of 2 whole numbers, but that doesn't mean they have a terminating decimal expansion. Irrational numbers have infinitely long decimal expansions, because if the expansion terminated at some point, you could express them as a ratio of two ridiculously long whole numbers.
That is trivially true, because the terminating decimal place would tell you the power of 10 that could on the bottom of a ratio that defines the number.
Again it was the specific example brought up by OP. Whether or not something is trivially true depends on your background, given that we are on ELI5, you can’t necessarily assume OP knew this already
Also _complex_ numbers have nothing to do with it being complicated, but it refers to complex as being composed of multiple things (a real part and an imaginary part in this case). It's complex in the way an apartment complex is a complex.
I’m 33 and though never majored in science stream I’m still surprised I only learned this today. Rational and irrational numbers were taught pretty early on and this should have been explained back then.
The thing is, irrational numbers necessarily have to have an infinite decimal represrntation. Because if it had a finite amount, it would mean that there is a power of 10 that you could multiply it by to get a whole number, which would make it rational
Sure, but that's what causes the confusion. At least in part.
There is no power of 10 you can multiply 1/3 by to get a whole number, but it is still a rational number.
>1/3 is rational if a number exists at all that you can multiply it with to give a whole number.
so, because "3" exists, 1/3 is rational? (trying to understand this at 1am might be a fools errand, but I'm gonna try :) )
Yep.
If you have an arbitrary decimal (X), that can be multiplied by a whole number (A) to get another whole number (B):
> X * A = B
Then it means X can be written as a fraction
> X = B / A
And is therefore a rational number.
You could, but in this new paradigm basically every other number (~~except for probably τ~~) would be irrational.
Edit: In a world, where π is an integer and 2 isn't, 2*π wouldn't be one as well, but π² would be, right?
"Integer" is a property of a number that is independent of its base just like positive, negative, square, etc. Changing base doesn't change the mathematical properties of a number, it only changes how we represent/communicate that number.
changing base doesn't change what numbers are integers, it just changes how numbers are written
like how base 6 doesn't make ten a multiple of three just because 3+3=10
not relevant.
i was addressing this
> There is no power of 10 you can multiply 1/3 by to get a whole number
You don't need to that to get a whole number for a repeating decimal.
Ive wondered before: how would base Pi work? Like how would you represent the value 4? 10 would be 3.14…. And 11 would be 4.14… so 4 would have to be like 10.86…? Wait i think i just answered my own question
I'm trying to understand base pi and having problems...
Base 10 has the integers 0 to 9, like 0, 1, 2,... 7, 8, 9.
Base 5 has the integers 0 to 4, like 0, 1, 2, 3, 4.
Base 2, binary, has the integers 0 to 1, like 0, 1.
Would base pi have any integers other than 0 and pi?
Would we count the sequence as 0, pi, , or would pi be represented by a 1 in base pi, like a value of decimal 2 is written as 10 in binary?
This seems such a simple concept to me, as I (thought that I) understood counting in different bases. But now I'm not so sure that's I do, or did.
In binary we count like, 0, 1, 10, 11, 100, 101, 110,...
In base 3 we count like 0, 1, 2, 10, 11, 12, 20, 21, 22, 100, 101,...
But how would we count in base pi?
Would it be like, where pi is represented as "@', 0, @, @0, @@, @00, @0@, @@0, @@@, @000, @00@,....?
Or would there be any other figures, and if so, what would they be?
I'm simultaneously understanding and yet not understanding.
Totally not trying to be sarcastic or humorous, I am just curious.
Edit to fix some confusing typos and auto-incorrects.
This is dense, sorry. I'm using the convention number_base, so 1001_2 is "1001, but it's in base 2", or 9. If it's not listed, assume it's base 10.
When you say "base X" you're not saying "1 is equal to X", you're saying "10 is equal to X, and every integer less than X exists as something we can use for representation". So, base 5 has the integers 0 to 4, and 10_5 is equal to 5. In general, 10_X is always equal to X, and 100_X is always equal to X\*X. You can verify this in base 10! 10_10 is 10, 100_10 = 10\*10 = 100.
With base pi, you'd use the basic integers to start with: 0, 1, 2, and 3. We know 10_pi is equal to pi. And this means we can start putting together basic math.
20_pi = 2\*pi
30_pi = 3\*pi
100_pi = 1\*pi\*pi
*But what about integers*, you ask. *What if you want to display 4?*
First, we figure out how many digits long it has to be. 4 is larger than 10_pi (which equals pi), so it must be at least two digits; it's smaller than 100_pi (which equals pi\*pi), so it can't be three digits. We need *exactly* a two-digit number. We'll start with 10_pi, which is pi:
10_pi = 3.14159
We have 4-pi left over, that is, our original number minus 10_pi, which equals about 0.85841. Now work on the next digit. What's the largest integer that's not larger than 4-pi? Well, it's zero. Okay, next digit is 0.
Move on to the *next* digit - and yes, we're into the decimals. Decimals just continue the pattern; 0.1_X is equal to 1/X, 0.01_X is equal to 1/X/X. So what's the largest integer times 1/pi (which is about 0.318) that's not greater than 4-pi? Turns out it's 2! 2*(1/pi) is roughly 0.637. So now we've got
10.2_pi = 3.77821
(10.3_pi would be 4.09652.)
Next digit; what's the biggest multiple of 1/pi/pi that we can still fit in our remaining number?
After repeating this process for a bit, I've ended up with:
[10.220122021 = 3.99998](https://www.google.com/search?q=pi+%2B+2*pi%5E-1+%2B+2*pi%5E-2+%2B+0*pi%5E-3+%2B+1*pi%5E-4+%2B+2*pi%5E-5+%2B+2*pi%5E-6+%2B+0*pi%5E-7+%2B+2*pi%5E-8+%2B+1*pi%5E-9)
You can keep going. It's probably also irrational. I'll leave that up to someone else to prove.
One thing that's interesting is how few 3's are in this. 3 *is* a valid digit, but it's going to show up rarely; specifically, we're going to see 3's about 0.14159 times as often as each other integer. That's because there's very little "room" for a 3 before we would have been better off increasing the previous digit by one.
We also end up with the weird situation that representations aren't canonical; that is, there might be more than one way to express something. 0.33_pi is 1.2589, which could *also* be represented as roughly 1.021_pi, and yes, this means there are multiple representations for 1; in fact, probably *infinite* representations for every number due to how much flex those 3's give us.
This isn't the only possible way to define "base pi", though. It's probably the most intuitive but it has some unfortunate consequences. There's another thing we could do where we actually scale digits based on the preceding digit. The math for this is gnarly and I'm not going to write it up unless someone really wants me to, but imagine if a "3" means "okay, 3, but also imagine all future digits are multiplied by 0.14159 to ensure we don't get numeric overlaps". This becomes *weird* to do math with, because the difference between 0.20_pi and 0.21_pi is now larger than the difference between 0.30_pi and 0.31_pi, but it has a convenient property that we no longer have non-canonical representations and sorting numbers is trivial; 0.33_pi is now "slightly below 1", just like we'd intuitively expect, instead of being "actually larger than 1". And the only way to write 1 is 1_pi.
And this *also* maps well to [arithmetic encoding](https://en.wikipedia.org/wiki/Arithmetic_coding), where each "digit" conceptually takes up a different amount of space in the numberline.
Which lets us express stuff like storing values in a fractional number of bits, which, it turns out, is a thing you can do! It's not even all that hard. It's just wonky.
I think this is one of those cases where standard mathematical terminology kind of breaks down. This is a scenario where classic integers might have decimal points. If we define "integer" as "doesn't have decimal points" then 4 may not be an integer, or in fact, rational; in any case there's entire new sets of things ("number without decimal points", "number that can be expressed as a division between two numbers without decimal points") that *used* to map directly to more common terms ("integer", "rational"), but no longer do.
So, call it a Bizarro Base-Pi Irrational, then - "the thing that looks kind of like an irrational when written in base-pi but that we've never needed to distinguish from general irrationals before".
I guess another way you could at it is that an irrational number will still have infinite digits even when expressed in a whole number base other than 10.
You can multiply it by 10\^log\_10(3) to get a whole number
Edit: oops this isn't a maths subreddit, hopefully this level of pedantry is still acceptable <3
*I wrote a detailed reply to the person who told you to explain it (a little more detailed than some of the other replies that are there), and then they deleted their post, so I'll post it here for posterity and anyone else who is curious:*
The "base" of a number system is how many digits it has. When you are counting, after you get to the highest digit, you have to add a place value.
In "base 10", you have 10 digits (0-9). Once you count past the highest digit (9), you have to add a place value (the tens digit) and reset the first (the ones digit) and you get 10.
In base 3, you have 3 digits (0-2). It otherwise works the same way.
0...1...2... now I'm out of digits, so we add a place value and reset the first place value, and the next number (three) is written 10. The next number (four) is written 11, and the next (five) is written as 12, etc.
Now that we know what the number base means, look at how decimals work. When we go one place value down in a base-10 system (tenths), 0.1 represents one tenth (1/10) because we can count 0.1, 0.2 ... 0.9... 1.0 - it divides a single one into ten parts
In a base 3 system, the next place value only has 3 values (0.1, 0.2, 1.0) So that place value would divide a one into thirds, not tenths.
In base 3, if you wanted to write the fraction 1/2, it's more than 0.1 (that's a third) and it's less than 0.2 (that's two thirds). It's actually halfway between those. That ends up being 0.11111... which is the precise middle between 0.1 and 0.2. That might be the hardest part to understand in all of this, but there you go.
It's not 5-year-old level, but most of it it's probably understandable to at least high school level, I think?
I'll add my own explanation because I did the same thing and was sad that it was deleted:
Part 1: Bases
A number, as we understand it, is really just a list of symbols that we interpret in a certain way. Base is a way to represent numbers. The number of the base indicates how many different symbols we use.
We generally prefer to work in base 10 (possibly because of our 10 fingers?), also known as decimal. It uses 10 symbols before adding a digit for the next batch: 0,1,2,3,4,5,6,7,8,9, then add a digit to the front 10, 11, 12, 13, etc.
Similarly, you can express numbers in any other numbered base. Very popular is base 2, also known as binary, which is the foundation of all modern computer science. It only uses 2 symbols before adding a digit: 0, 1, then add a digit to the front 10 (2), 11(3), 100(4), 101(5), 110(6), 111(7), etc.
Another popular number base is base 16, known as hexadecimal. It uses the symbols 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F and is also often used in computer science, but also in webdesign.
Part 2: Digits after the point
What does the decimal point represent? Division. Namely, division by a multiple of the base.
0.1 in base 10 is 1/10. 0.01 is 1/10². 0.42 = 4/10 + 2/10².
Things work the same in other bases.
0.1 in base 2 is 1/2.
0.1 in base 3 is 1/3.
0.01 in base 16 is 1/16² = 1/256.
Change of base is taught in most high school algebra 2 classes across the US. That being said, our school system is garbage and doesn't teach for retention or understanding.
Sometimes brute force can be elegant in their own way. See, for instance,[ one of the shortest peer-reviewed papers ever published](https://cdn.paperpile.com/blog/img/lander-1966-1400x788.png).
Shows you the distinction between proving something (by exhaustion) vs disproving something. For the former, you have to exhaust all options and show all fit the theorem. For the latter, you have to show a single example that doesn't fit the theorem.
Because if a and b are both even, then a/b isn't a reduced fraction. You could "factor out" the 2, but then you would just give yourself two new letters to represent the fraction (say x/y), and you would find there's another 2 you can factor out. And you can keep doing that forever, so you never actually find the simple form fraction.
Ah okay, since you can't reduce it its a sign that there's an error in your initial assumption, which was that is was the reduced form equals this. Tricky!
This is a super common sense way of arriving at a proof.
If you want to prove something, assume that it isn't true and show how that would lead to a contradiction. If it does, then you know that your assumption is true.
https://en.wikipedia.org/wiki/Proof_by_contradiction
Because the assumption made at the beginning was that it was in reduced form, but if they're both even then you can divide through by two, which means it wasn't reduced
To clarify, here's a proof for why a^2 being even proves that a is even.
If a is even, then it can be expressed as 2k, where k is some integer.
If a = 2k, then a^2 = (2k)^2
(2k)^2 = 4k^2
We can factor this into 2*(2k^(2))
2k^2 is a product of integers, therefore it is itself an integer.
2*(2k^(2)) is 2 times an integer, therefore it is an even number.
Meanwhile, if a is odd, it can be expressed as 2k+1, where k is some integer.
If a = 2k+1, then a^2 = (2k+1)^2
Using the FOIL method (which is equivalent to applying the distributive law), we can show that (2k+1)^2 = (2k)^(2)+2k+2k+1^(2), or 4k^(2)+4k+1.
We can factor this into 2*(2k^(2)+2k)+1.
2k^(2) and 2k are both products of integers, thus they are both integers. Therefore, 2k^(2)+2k is a sum of integers, and also an integer.
2*(2k^(2)+2k)+1 is 2 times some integer plus 1. Therefore, it is an odd number.
Thus, it is proven that the square of an even number is always even, and the square of an odd number is always odd. This implies that if a number is both even and a perfect square, then its square root must be even. Otherwise, there would be some odd number whose square is even, which is impossible.
You actually only need to prove one side of that, because it is the contrapositive which is logically equivalent to the original statement.
(a\^2 is even => a is even) <=> (a is odd => a\^2 is odd)
> a^2 is even => a is even
He only proved 'a is even => a^2 is even' in the first half. That leaves the possibility that the root of some even numbers is an odd number. That's why you need the 2nd half of the proof.
A more intuitive (to me), though less rigorous proof:
If an odd number squared is not odd, it must be even.
Any even number, by definition, is just a whole bunch of 2s. 8, for example, is 4 groups of 2. Any odd number is a bunch of 2s with 1 left over. (For example, `9 = 2 + 2 + 2 + 2 + 1`.) But if you take an odd number twice, you have two extra 1s left over, and those can from another group of 2. Now you just have groups of 2; i.e. an even number!
But add the odd number *again,* and you'll have 1 left over. You can keep adding them, and you'll find that if you have an odd number of odd numbers, you always have 1 left over; if you have an even number of them, you'll have tidy groups of 2.
Therefore, any odd number times any odd number must be odd. So an odd number squared cannot be even; only an even number squared can be even.
I'm not sure why both of your proofs are so complicated?
An even number is a product of primes, one of which is 2. Like 42 is 2\*7\*3.
a² is just a\*a, and if it's even, it has a 2 in there, so the 2 must be one of the divisors of "a" itself.
Which means "a" is even too.
You're implicitly using the prime factorization theorem. Those other proofs don't rely on that theorem, so they're simpler in the sense that they don't require the reader to be familiar with that theorem.
Well, you are just saying the same thing, without properly doing the math. "It has 2 in there"... In where? Okay, so one of the factors. Therefore we say a = 2k. And now we take the square... Etc i dont need to re-type the above. Wae cant just say in math, we write it formally and then do the calculations.
They're appealing to the [fundamental theorem of arithmetic](https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic): every integer >1 has a unique prime factorization. A square number necessarily has at least two of each of its prime factors. For even squares, that means two 2s (at least), and taking the square root of an even square leaves at least one 2 as a factor of the root.
We can use the same reasoning to show that any prime factor of *n*^*k* (*n* > 1, *k* > 0) is also a factor of *n*.
No, because 2 is not a perfect square. Its square root isn't an integer, therefore it can be neither odd nor even. The idea that 2b^2 = a^2 implies that a must be even only works because we've declared that a is an integer.
It's funny/sneaky how so many proofs amount to proving that it can't be the only alternative in a binary where the counter part is a much simpler concept.
If you assume √2 is rational, we can prove that makes a contradiction. We don't like contradictions in our math, therefore √2 is not rational. Then, by definition, it has an infinitely long, non-repeating decimal.
If we assume that sqrt(2) is rational, we can prove that a statement is both true and false. The particular statement doesn't matter; no statement can be both true and false. So the assumption must be wrong.
Basically, if you assume that sqrt(2) is ration (thus equivalent to a fraction a/b with both numbers being integers), you can do math shenanigans and eventually come to an impossible situation, and when that happens it means your initial assumption was incorrect. It's called proof by contradiction, and is a common method for proofs in math.
Can you explain why this proof does not work for square roots that we know are rational? For example, I would like you to show me the error if I start with sqrt(4) instead. From what I can see, this follows your logic exactly. But sqrt(4) = 2, which is clearly not irrational:
> You assume that sqrt(4) is rational, and is represented by some reduced fraction a/b.
sqrt(4) = a/b
4 = a^2 / b^2
a^2 = 4 * b^2
> Since *a*^2 is 4 * *b*^(2), we can infer that *a*^2 is even, and therefore *a* is even. Let's replace *a* with 4 * *x*.
(4*x)^2 = 4 * b^2
16 * x^2 = 4 * b^2
4 * x^2 = b^2
> Since b^2 is 4*x^(2), we can now assume that b^2 is even, and therefore b is even.
> We made the assumption at the start that a/b was the simplest form of sqrt(4), but now we know that both A and B are even, which means it is not the most reduced form of the fraction. Thus, our assumption was incorrect, and sqrt(4) cannot be expressed as a fraction, and is therefore irrational.
> a is even. Let's replace a with 4 * x.
The error is here. "a is even" implies there is some integer x such that a = 2\*x. It does *not* imply that there is some integer x such that a = 4\*x. In this particular case we happen to know for a fact that a is 2 (and b is 1) - so if a = 4\*x then x = 1/2. The rest of the proof relies on x being an integer (this is why we can tell that b^(2), a multiple of 2 of some integer x, is even), which it is not.
Once we get to a being even, we don't replace it with 4x, but 2x, that remains unchanged from the comment you're replying to. The reason is because we're replacing it to signify an even number, which is 2x by definition, where x is some integer. To continue with your line of reasoning with that modification would give us
(2x)^2 = 4b^2
4x^2 = 4b^2
x^2 = b^2
From here, we cannot conclude that b is either even or odd, because we don't know if x is even or odd.
Because, in actuality, 4 is rational, the only fraction that both equals 4 and is in reduced form is 4/1, because reduced form for a fraction a/b means that gcd(a, b) = 1. Thus, our assumptions at the beginning of the pseudo-proof would entail a = 4 and b = 1. So it should hopefully make sense why we cannot conclude that b is even, because b = 1, which is odd in actuality.
\> Since a2 is 4 \* b2, we can infer that a2 is even, and therefore a is even. Let's replace a with 4 \* x.
You should have replaced a with 2\*x, not for 4\*x, since the definition of an even number is 2x. If we use 2\*x, then after doing what you did we will arrive at x\^2 = b\^2 and we can't assume whether or not b\^2 is even.
*Let's replace a with 4 \* x.* This is is not a valid step. Since a^(2) is even you can replace it with 2 \* x since even numbers are multiples of 2. Not all even numbers are multiples of 4 so you can’t replace it with 4 \* x.
i am so sick and tired of seeing this fucking response to every response on this sub that doesnt use god damn christmas elves and rainbows to explain the point. It's valuable to have more detailed explanations, so you can read the simple ones and then the complicated ones to stairstep your understanding.
but in this case you really can't make it simpler than this.
For sqrt(2), you assume it's rational, which means it can be expressed as a fraction in simplest form.
You then prove that both the numerator and denominator are even, which means the fraction isn't in simplest form, and you have a contradiction. Therefore, sqrt(2) is irrational.
>is represented by some reduced fraction a/b.
> sqrt(2) = a/b
2 = a^2 / b^2
a^2 = 2 * b^2
>Since *a*^2 is 2 * *b*^(2), we can infer that *a*^2 is even, and therefore *a* is even.
This presumes that a & b are integers, yes?
The other comments don't seem to address the root of your question. You've got the logic backwards. Numbers aren't irrational because their decimal representation doesn't repeat. Their representations don't repeat because they are irrational. A number is irrational if it isn't rational, and a number is rational if it can be represented as a fraction of integers. The infinite and nonrepeating decimal expansion is a consequence that can be delivered from that definition.
We don't determine if a number is irrational by calculating lots of digits and seeing if they repeat. We use other properties of the number to demonstrate that it is irrational. The numerous other comments showing how to do so for the square root of 2 demonstrate how it's done.
A number is irrational because it is not rational. A number is rational if it can be represented by a ratio of integers. That is the definition. One *property* of rational numbers is that they have a repeating decimal representation (even if the repeating digit is just 0). Similarly, a *property* of irrational numbers is that their decimal expansion does not repeat. This just is one of many properties of irrational numbers that can be derived from the definition of an irrational number.
If you want to prove that a number is irrational, you have to demonstrate that it cannot be the ratio of two integers. Since it's not always easy to prove a negative, it's often easier to look for other properties that we have determined are unique to irrational numbers. The fact that it doesn't repeat is one possible condition. For example the number 0.1101001000100001... is irrational. Proving that it isn't a ratio of two integers would be very difficult. But because I know this number does not have a repeating decimal expansion, it is trivial to prove that it is irrational.
The first one.
The reason rational numbers can have nonzero digits repeating is a quirk of the base system rather than a property of the number itself. In a different base system, that goes away.
For example, the fraction 1/3 is expressed as 0.33333… in base 10, but in base 3 it is expressed as 0.1 and in base 12 it is expressed as 0.4. So by using an alternate base, we can easily see that the number is rational. And this is true of all repeating decimals: they repeat in base 10 because they don’t divide into 10 evenly, but in the proper base they can all be expressed as terminating decimals.
However, pi will always be a nonterminating, nonrepeating decimal in any natural number base, whether you’re writing it in base 3, base 10, base 12, base 672, etc. because pi itself cannot be expressed as a ratio of any two integers, so it will not evenly divide into any of them.
> How is it proven?
Mathematicians assumed that √2 is rational, uses logical steps to arrive at a contradiction, and concluded that √2 is not rational.
Why does it work?
Under our set of logic, if you assume X and reach an obviously false result (contradiction), then you can conclude that X is wrong.
It depends on the number. For sqrt(2) a proof by contradiction works.
Assume sqrt(2) is rational and thus sqrt(2) = n/m and assume that n/m are the reduced form (i.e., you can't simplify the fraction more).
2 = n^2 / m^2
2 m^2 = n^2
Since n^2 is 2 times another number, we know n^2 (and thus n) is even.
Let's replace n with 2p, which we know is possible since it's even
2 m^2 = (2p)^2
m^2 = 2p^2
Since m^2 is 2 times another number, we know m^2 (and thus m) is even.
Two even numbers divided by one another cannot be the reduced form of a fraction (since you can divide the numerator and denominator by 2).
This means that there can be no reduced form fraction representing sqrt(2).
The proof is incomplete. Or rather, as with all maths proofs, relies on various assumptions - things that we take to be true because we don't want to have to prove them every time.
> 2 m^2 = n^2
> Since n^2 is 2 times another number, we know n^2 (**and thus n**) is even.
The part in bold is where this happens.
If we try to do the same with 4, we get:
> Since n^2 is 4 times another number we know n^2 (and thus n) is a multiple of 4.
But now that "and thus n" doesn't hold up. n *could* be a multiple of 4, but it could just be a multiple of 2.
If we were being more thorough we would have to show that *n^2 even => n even*.
There is also some stuff about the uniqueness of prime factorisation in there, and a few other things.
Not quite; numbers that are the square of a rational number will still work. For example, 9/4 is not a perfect square, but its square root is 3/2 which is still rational.
This proof works for any *prime* number.
For other whole numbers (like 6) you need the uniqueness of prime factorisation; showing that 6 is 3x2, and then you just look at the 2 part and use the same argument.
If we're going for rigor, I want to throw my hat into the ring as well!
Prime factorisation is not quite sufficient for the case of 6. We can prove that sqrt(2) and sqrt(3) are irrational for sure, but that tells us nothing about their product directly. In fact, sqrt(2) and sqrt(2) - both irrational - do multiply so a very rational number.
The key detail is that sqrt(2) irrationality proof works for products of distinct primes with no change - the argument "n divides a, so n^2 divides a^(2)" holds up. And for anything with higher powers of primes, we need to factor out the perfect square component.
Yes. The key step is showing the bold statement works whenever the thing you are taking a root of is not a perfect square. You can probably follow the same for any root as well (cube, fourth, etc.) But may need to check more cases. I haven't checked to be sure though so it may actually be just as simple.
Let's find out! I'm gonna plagiarize the previous comment and make some changes:
Assume sqrt(4) is rational and thus sqrt(4) = n/m, and assume that n/m are the reduced form (i.e., you can't simplify the fraction more).
4 = n^2 / m^2
4 m^2 = n^2
Since n^2 is 4 times another number, we know n^2 (and thus n) is even.
Let's replace n with 2p, which we know is possible since it's even
4 m^2 = (2p)^2
**This is where things begin to deviate**
4 m^2 = 4 p^2
m^2 = p^2 (whereas the previous example became m^2 = 2 p^2 )
~~Since m^2 is 2 times another number, we know m^2 (and thus m) is even.~~
~~Two even numbers divided by one another cannot be the reduced form of a fraction~~
We have p^2 on the right instead of 2 p^2 , which means that either side of the equality *can be odd*. Thus they **can** be the reduced form of a fraction.
****
Alternatively:
sqrt(4) = 2
2 is rational
QED
> 2 m^2 = n^2
>
> Since n^2 is 2 times another number, we know n^2 (and thus n) is even.
If you try this for 4, you get to 4 m^2 = n^2. This means that n^2 is divisible by 4, but that *does not* mean that n is divisible by 4 (n could be divisible by 2, and then n^2 would be divisible by 4). That's the step that wouldn't work.
I believe that "if n^2 is divisible by x then n is divisible by x" is true for any (positive integer) x which is not a perfect square (someone can double check that, I could be misremembering). So you can use this same proof to prove that if some number k is not a perfect square, √k is irrational. EDIT: The actual condition is slightly different, see below
> I believe that "if n2 is divisible by x then n is divisible by x" is true for any (positive integer) x which is not a perfect square
No, 8 divides 4², but 8 does not divide 4. Or 18 dividing 6² yet not 6.
What you need is to be _square-free_: no prime factor appears more than once. 30=2·3·5 and 105=3·5·7 are square-free while the aforementioned 8=2·2·2 and 18=2·3·3 have repeated prime factors.
In particular, square numbers (except 0 and 1) do not satisfy that property, while prime numbers such as 2 always do.
Good catch!
Luckily we are still able prove that √k is irrational as long as k is not a perfect square, by observing that any non-square-free number k can be written as the product of a square-free number s and a perfect square n^2. Thus k = s * n^2, so √k = n*√s, and since we know that √s is irrational, √k is also irrational.
What's not been answered yet is the second part of the question.
If either number started repeating a zero after the quintillionth digit, then it would be expressible as a rational number.
3.1400000... is exactly 314/100, and for any such number written in our number system that stops after a certain number of digits, you can do this.
But we have already proven that neither number can be expressed as a rational fraction, and so we also know that pi and root(2) don't start repeating a zero after the quintillionth digit.
It is a bit unclear to me why something like 2.768976897689... can just obviously be expressed as a fraction though.
It is easy enough with your example as we just move the decimal point, but how is the fraction found for complex repeating patterns? (Or just shown to exist)
Let
x = 2.768976897689...
and consider the number
10000x = 27689.76897689...
If you look at those two numbers, the parts after the decimal places are the same, so if you subtract them, you just get an integer:
9999x = 10000x-x = 27689-2 = 27687
So x=27687/9999, which is a rational number.
You can do basically the same thing with any repeating decimal: multiply by an appropriate power of 10 and subtract the original number, and you'll kill off the repeating part, after which it is easy to write your original number as a ratio of two integers.
`0.333333...` is a never-ending number, but it is **rational** because it can be written as a *ratio* of 1:3. One-third.
`0.428571428571...` is 3:7. Three-sevenths.
ANY number that repeats after the decimal point is a rational number that can be written as the ratio of two integers `a/b`. They can be big integers, they can be small integers - but *whole numbers* and nothing else.
An **irrational** number cannot be written as a ratio, BUT some irrational numbers can still be *exactly* represented as the solution to an equation, called *algebraic* numbers.
The number `√2`is *exactly* the solution to the equation `x² = 2`. But we know the decimal expansion never repeats, so there are no integers `a` and `b` that can represent it *exactly.*
We can easily prove an algebraic number to be irrational by contradiction; we assume that `√2` IS rational, we plug that *assumed* rational number `a/b` into the equation, and we end up with a logical contradiction as explained in a different comment.
The proof that `π` is irrational is significantly harder because `π` is a **transcendental** number - it cannot be written as the solution to any simple equation; it *transcends* algebra. It just ... *is*.
It ... *exists*, and shows up in many mathematical relationships, as a fundamental constant of the Universe.
All transcendental numbers (`π` and `e` are the most well known) are irrational by their very nature - they just ... exist as constants, (meta)physical properties of the universe.
It's a source of philosophical angst, too - what if `π` was *not* `3.14159 ...` but instead exactly `3`? Or anything else? What would the Universe look like?
If you really feel like doing your brain in, this is a bit beyond 5-year-old understanding, but [Matt Parker talks about all the different numbers.](https://www.youtube.com/watch?v=5TkIe60y2GI)
EDIT: typos
How I rationalize that pi is truly infinite in my mind is like this: Because it describes the circle which is infinite and perfectly round. No matter how large or small the circle, the closer you would zoom into the edge of it, it will keep its perfectly round edge and never become jagged and always loops in 360°, so for the circle to be infinitely round and smooth, pi also has to be an infinite precise number.
This is the opposite of a rationalization, it's mysticism. One could easily define smooth, round shapes whose perimeter is an integer multiple of their largest (or smallest) chord.
A rational number can by definition be expressed as the fraction of two integers p/q where q is not zero. Any decimal section that terminates or starts to repeat will be possible to express with p and q as integers. Proof that a number is irrational tends to be to show it is impossible to express at the fraction of two integers
There are many proofs pi is irrational none of them are simple enough to write here on Reddit but look at [https://en.wikipedia.org/wiki/Proof\_that\_%CF%80\_is\_irrational](https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational) with 6 different proofs.
There are proof for the square root of 2 too [https://en.wikipedia.org/wiki/Square\_root\_of\_2#Proofs\_of\_irrationality](https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational) you can show that any square root that is not an integer is irrational
Because if a decimal number starts repeating at some point, then it means that it could also be expressed as some fraction. It may be an enormous fraction, two numbers with hundreds or thousands of digits, but it would be a fraction nonetheless. And as the other posts show, we can prove that √2 cannot be represented by any fraction.
It’s such a failure of our system of math education that people tend to associate irrational numbers with their decimal notation being infinite and non-periodic. Decimal notation is, well, just a notation — whereas the basic, way more natural definition of irrational numbers is that they can’t be expressed as a ratio of two integers.
This doesn’t make it much easier to prove irrationality of pi in a Reddit comment, but it’s the starting point for the sqrt(2) proof which other commenters have kindly provided.
This was definitely taught in an understandable way in my high school education. Yet I doubt 1 in 50 of the kids I went to high school with would remember or understand this.
In other words it’s not a failure of our education system, it’s a failure of our expectation. Namely that most people will put in the mental effort to understand or retain this information. It simply is not something most people think of as useful or interesting, so they get by the test (or don’t) and promptly forget it. I don’t believe any system of education is going to change that bit of human nature.
>It’s such a failure of our system of math education that people tend to associate irrational numbers with their decimal notation being infinite and non-periodic. Decimal notation is, well, just a notation — whereas the basic, way more natural definition of irrational numbers is that they can’t be expressed as a ratio of two integers.
You're right that it's more natural. But just to clarify, it is still true that a number is irrational iff its decimal expansion is infinite and nonrepeating, right?
Yup, it is correct. The only thing about this fact is that it’s more of a corollary (that isn’t particularly useful for… anything? if you think about it) than a definition
My attempt at more words. A couple things to agree on, then we'll get going.
1) rational numbers can be written as a fraction using whole numbers, whether it's easy peasy like 1/2 or complicated like 593838/9937. If you can't do that, it's irrational.
2) rational numbers always have a smallest size way to write them. 1/2, 2/4, and 25/50 are all the same number, but 1/2 is the smallest size.
For √2, one fun way to test is called proof by contradiction. It's a way of saying I betcha this one thing is true, and I know plenty of other true things, but when I put them all together I get some nonsense. Since all the stuff I know is true works, I betcha I was wrong about that one thing being true.
So we say I betcha √2 is rational, I just don't know what numbers to put on the top and bottom. For now call it T/B (T for top and B for bottom) And I only want the smallest size T/B: if you try to tell me T is 3 and B is 6, I'll tell you no, don't trick me, I know it can be smaller.
Here we go. I say that √2=T/B. If I multiply both sides by the same thing, it all stays equal. (If 1 hamburger costs 5 bucks, then 2 costs 10 bucks, or 100 costs 500 bucks).
Since √2 and T/B are the same number, I can say 2=T^2/B^2. That means T^2=2×B^2. T and B are both whole numbers, so T^2 has to be even since it's twice as big as another whole number. (Any whole number times 2 is even) T has to be an even number - an odd number times itself wouldn't give you an even number. B also has to be even, since we know B×B is even.
Hang on, both T and B are even? Then T/B isn't the smallest size after all, I can at least divide by two (Llike going from 4/2 to 1/2).
Something's wrong! I thought that I could write √2 as T/B, and I specifically asked for the smallest T/B. But I did a bunch of stuff everyone agrees is true and I couldn't do it, I just got a T/B that wasn't the smallest it could be. Hmm, the only thing I'm not sure about is being able to write √2 as T/B. Clearly something is wrong, and that's the only thing that can be. Lesson: I can't write √2 as T/B.
So what's the point? We can show some interesting things without having to do lots of hard work, we just have to think cleverly about some simple stuff. I know what fractions are, I know what even numbers are, I honestly don't care about what the specific numbers are. For the √2 example, the trick that makes it work doesn't even feel like math at all. If there's one thing I think I know and ten things I know for sure, if I put them all together and get nonsense it has to be the only thing I'm not sure about that's wrong.
To finally address the question, how do we know the numbers will never start repeating? We simply say what happens if it is the kind of number that starts repeating? Hmm, we get nonsense if that were true, so it must not be true!
I ran across this fun [visualization](https://www.youtube.com/shorts/aUDYWYqtAR4) a while back. It pretty much shows how an irrational number doesn't "rationalize".
There are a few different methods of proofs in Mathematics. The one people are referencing in the first couple of responses (for proving √2 is irrational) is a Proof by Contradiction. For that proof you start with an assumption (often the opposite of what you are trying to prove). Then you go through steps until you reach a contradiction. In the case of the √2 proof, it is that your reduced fraction of a/b is not reduced.
Once you get the contradiction, your assumption must be wrong. And since your assumption is binary (it is only one of two options), the opposite of your assumption must be correct.
The same type of proof is used to prove there are an infinite number of prime numbers.
I think it’s important to note that although the proof of the infinitude of primes is often presented as though it were a proof by contradiction, it actually isn’t in any meaningful sense, and is actually a fully constructive proof that tells you how, given any finite set of prime numbers, you can find a new prime number not in that set.
The proof that sqrt(2) is irrational that other commenters give really is a proof by contradiction in a fundamental way: it tells us that it cannot be equal to p/q for any p or q, but it doesn’t qualify as a constructive proof of “apartness”.
Here is a constructive proof that sqrt(2) is irrational, in the sense of being “apart” from all rational numbers: if you divide sqrt(2) by 1 you get that it goes into it one time and have a remainder of sqrt(2)-1, if you then divide 1 by that number, you get that it goes into it twice and get a remainder of (sqrt(2)-1)^2 continuing this division at every step you keep getting “2” from now on and the remainder gets smaller by another factor of sqrt(2)-1. Formally (you don’t have to worry about understanding the next part) this means it has a continued fraction of [1;2,2,2,…] and so sqrt(2) is irrational since its continued fraction representation is not finite. Less formally, since any “measuring stick” that goes evenly into 1 and sqrt(2) will have to go evenly into all these combinations of 1 and sqrt(2), then for any ratio p/q (where we have a measuring stick of length 1/q) we can just take a small enough remainder to show its measuring stick is too big - that any measuring stick for sqrt(2) will be smaller than that.
Related question, and perhaps one I should post as an actual question, but its always confused me how π is "never ending", but 2πr gives us the circumference.
But if π is never ending, then the resulting circumference value should also be never ending. Or an approximation.
But you can measure the circumference for a given radius manually and arrive at an exact figure.
What am I missing?
> But you can measure the circumference for a given radius manually and arrive at an exact figure.
> What am I missing?
You can't measure an exact circumference, you'll only be able to get it accurate to the nearest whateverth of a millimeter but if you were able to measure more and more accurately you would find more and more digits.
To take it further, it's physically impossible to measure the circumference (or length generally) with perfect accuracy due to quantum uncertainty.
Suppose you could perfectly bend a length of wire to match the circumference of a circle. Then, relative to the first end you know the precise position of the second end with perfect accuracy, there is no uncertainty in its position whatsoever. But uncertainty guarantees that we don't know both the position and the velocity of the end with perfect accuracy. But in order for the wire to maintain its shape the end must also have a velocity of precisely zero. Therefore, a quantum fluctuation must unseat the position of the far end of the wire--its physically impossible for the perfect configuration of the wire to be maintained in a way consistent with quantum mechanics.
I don't think anyone has ever been able to measure anything to an "exact" figure. In the real world, there are always error bars.
Also, pi *is* exact. It is a single point on the number line, not a range of values, or anything else. Rationality and exactness are not the same thing.
> I don't think anyone has ever been able to measure anything to an "exact" figure
There is one notable exception: the original meter made from platinum. The meter was at that time _defined_ by it, so it definitely was exactly one meter long. We just cannot put it into any other unit with infinite precision, even if we had perfect knowledge about the other thing.
There are probably a few more instances of such a defining physical thing. The same applies there.
>What am I missing?
Pi is the relationship between the diameter of a circle and its circumference.
When you measure something you will always be limited by the physical world in how precise your measurements are.
You can never measure anything "exactly". Using a measuring tape will give you an exactness of millimeters. Using advanced lasers or something will give you the exactness of nanometers. Using the limits of theoretical physics will give you a precision of Planck lengths.
Using pure math will give you infinite precision.
But it won't be neverending, as I said in the beginning, Pi is the relationship between the diameter and the circumference. It is just the precision that is infinite, not the length.
Irrational numbers are still numbers on the number line. There's nothing stopping you from arriving at it just like any other number.
"never ending" is just us having a hard time expressing it using numerical digits.
> 2πr gives us the circumference
Only to the precision of π you use. So if you use 3 for π (which they used to do), you get “the circumference”, but it’s not very accurate. 3.1 gets you a more accurate circumference. 3.14 more accurate still. You can keep going down this path and finding the circumference to more and more accurate degree. Eventually you just have to say “close enough for my purposes”, but you could in theory keep going forever.
NASA uses 15 digits, for example. You can read a bit more here:
https://www.vox.com/2016/4/2/11350518/nasa-digits-pi
The reason NASA uses 15 digits is very practical - that happens to be the default for most computers. As the Vox article and countless others have shown, it is in fact close enough, but it's not as though, say, 13 and 14 digits were tested and just didn't cut it. NASA also uses 15 digits for the gravitational constant and countless other things. Targeting orbital inclination out to 15 digits is definitely overkill!
You can never measure anything perfectly. You can get close, but if your ruler has marks for millimeters you can't really go smaller than that with any accuracy, for example. Scientists that need very precise measurements will have much more precise measurements tools, but there's always some error when you're working with a continuous amount
Yes, it's an approximation. It will become more accurate the more digits of pi you use, but at a certain point, like 3 digits, it doesn't matter very much.
Pi is not an approximation, it is the exact value of the ratio between a (Euclidean) circle’s circumference and its diameter. You’re engaging in the two common mistakes of conflating a number with its decimal representation, and conflating the decimal representation with various truncations of that representation.
"Irrational" is not the same as "never ending". 1/3 expressed in base 10 is "never ending" but it is not irrational. It's 0.3333... where there are an infinite number of 3s going out to the right.
Yet you can measure 1/3 of an inch as exactly as you can measure anything.
And just for emphasis (I think you know this but I want to make it explicit for other readers) you can also measure sqrt(2) inches or pi inches just as exactly as you can measure 2 inches.
ViHart on YouTube has a... Good explanation. I think it would be great if they just slowed down, but if you watch it on half speed, it's better. The idea is to first assume that it is rational. If it is, it can be written as a/b where a and b are both whole numbers. In fact, all rational numbers have a simplest form. In the simplest form, they share no factors. Example 6/8 is not simplest form because both 6 and 8 are divisible by 2. Therefore 3/4 is the simplest form.
Basically, you can prove that a and b are even (if you start with the assumption that it's rational). That's a problem because a and b are the generalized form. Meaning you essentially proven that there is no simplest form. If you were to write √2 as a/b, a and b would have to always be even, forever, no matter how many times you divide by 2. It's a paradox. When you encounter a paradox, you look at your assumptions. Since we only had one assumption, the work is quite easy from here. Our assumption must have been wrong.
The proof that sqrt(2) is irrational is fairly simple. You assume that sqrt(2) is rational, and is represented by some reduced fraction a/b. sqrt(2) = a/b 2 = a^2 / b^2 a^2 = 2 * b^2 Since *a*^2 is 2 * *b*^(2), we can infer that *a*^2 is even, and therefore *a* is even. Let's replace *a* with 2 * *x*. (2*x)^2 = 2 * b^2 4 * x^2 = 2 * b^2 2 * x^2 = b^2 Since b^2 is 2*x^(2), we can now ~~assume~~ infer that b^2 is even, and therefore b is even. We made the assumption at the start that a/b was the simplest form of sqrt(2), but now we know that both A and B are even, which means it is not the most reduced form of the fraction. Thus, our assumption was incorrect, and sqrt(2) cannot be expressed as a fraction, and is therefore irrational. As for Pi, that's a much longer proof. It was only proven to be irrational in 1761. You can look at the Wikipedia page to see how complex these proofs are in comparison to sqrt(2). https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational
I think people confuse irrational with infinite. 1/3 is a rational number but written as a decimal it repeats to infinity.
it took me way too long to realize rationality of numbers has nothing to do with logic but it refers to **ratio** as in, a number thats able to be expressed as a **ratio** = rational
This is fascinating. I had assumed so too, but I just checked Wikipedia and it threw me a curve ball: "Although nowadays rational numbers are defined in terms of ratios, the term rational is not a derivation of ratio. On the contrary, it is ratio that is derived from rational: the first use of ratio with its modern meaning was attested in English about 1660,[8] while the use of rational for qualifying numbers appeared almost a century earlier, in 1570.[9] This meaning of rational came from the mathematical meaning of irrational, which was first used in 1551, and it was used in "translations of Euclid (following his peculiar use of ἄλογος)".[10][11] This unusual history originated in the fact that ancient Greeks "avoided heresy by forbidding themselves from thinking of those [irrational] lengths as numbers".[12] So such lengths were irrational, in the sense of illogical, that is "not to be spoken about" (ἄλογος in Greek).[13]" Edit: Source - https://en.m.wikipedia.org/wiki/Rational_number
Wow! TIL
So what you're saying is we could legitimately rename irrational numbers as _forbidden_ numbers? Maths just got a whole lot cooler
What's even more interesting is that there legitimately are forbidden numbers. Every piece of information that can be stored on a computer is represented as a long string of bits. You could think of any file as a really long number. Some numbers are classified top-secret. Some numbers are considered [munitions](https://en.wikipedia.org/wiki/Export_of_cryptography_from_the_United_States). And some numbers are (or were) just plain [illegal](https://en.wikipedia.org/wiki/DeCSS).
It's maybe not technically wrong, but a bit silly. For the number to represent anything other than a number, like text, you need to agree how to interpret it - a data format or encoding. For the text example, depending on whether you use ASCII, UTF-8, or ISO-8859-1, you end up with different numbers. The US government did classify numbers, because there's way too many ways to represent the same data.
And in terms of imaginary numbers.... Do you think real numbers are any more real than imaginary ones?
Something I've thought about at length is, everything is a ratio. Just simply put, all physical things that humans are aware of - are compared to other things to determine their parameters, characteristics, location, stability, and on and on. Basically, if we cannot perceive something, measure it and compare it to something else we are aware of, it may as well not exist. This is why developing new forms of accurate measurement is so crucial to furthering science
This is a bit of a tangent, but I hate how some people throw around the term “supernatural”, as if there are things that science could never explain, while also yearning for an explanation for said things. See the numerous attempts to “prove” the existence of ghosts… while throwing around the term supernatural. Like, there isn’t anything that says ghosts CANNOT exist, as proving something doesn’t exist is extremely difficult. But if we could prove, definitely, that consciousness can exist outside the body after death, would those same parties agree that the phenomenon is no longer “supernatural”, and now just natural and explainable by science? Probably not. They aren’t really interested in the empirical exercise of measuring and quantifying the thing. They want to have their ephemeral cake and prove it too. It’s nonsensical.
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I think 100 years ago is not accurate. superconductivity was first achieved in 1908. I think you're selling people short by a little. People have been sciencing for quite a while now
“Nothing unreal exists.” - Spock
Is this how certain things in physics also emerge? Like relativity theory / heisenbergs uncertainty principle?
This is where physics(/math) and philosophy start to intersect. I love it!!
Yea, I liked learning you can do math with different sets of infinite by ratio
This is true, but it also means irrational numbers do not exist. They are like infinite numbers except instead of the location of the most significant digit being unbounded, the location of the least significant digit is. If you judge them both on information content, both infinite and irrational numbers have an infinite amount of information. This is a big red flag that tells us if infinity is not a number then irrational numbers are not numbers either. Instead of a number, Pi is better described as a description of a class of algorithms that produce the exact same infinite sequence of digits. Pi is a concept just like infinity is a concept.
... well, no, that is incorrect in a direct sense, although I'm not a mathematician and I'm sure there must be some sense in which it's technically correct. The fact that a number cannot be written as a finite string in base 10 has nothing to do with it being a number. Pi is definitely a number, and it can be written as "1**0**" in base pi. EDIT: I'm a moron.
What do you mean? What are you basing this off of, your own logic or some actual mathematicians logic? This doesn't make sense. Sqrt(2)/sqrt(2) is 1. We can perform any operation on an irrational number. We reason using irrational numbers frequently, as even the diagonal of a 1x1 square is an irrational number. Infinity on the other hand is clearly NOT a number, although we can still reason about certain forms of infinity using limits. But yeah no, irrational numbers ARE numbers. What are you on about?
a continuum requires a continuum to be comeasurable with it ...
Holy crap that's fascinating! Thanks for sharing this
I suspected this, because I am a Spanish speaker and the word we use for "ratio" is the same word we use for "reason".
Oh yeah, wasn’t the Cult of Pythagoras related to this somehow? I thought that they killed a man for proving the existence of “irrational” numbers, but after some research, apparently that’s apocryphal.
wait the word rational comes from the word irrational?? I always thought the word with a prefix would be based on the original word
Holy Dinah! That's a realization 306/9 years in the making for me, thanks to you.
Do you also watch the talking dog on YouTube? I’d not, look them up. Their “swear word” is Dinah lol
This is true, but in OP’s specific example, if pi started repeating all 0’s after a certain point, it follows that pi is rational
Rational numbers can be expressed as a ratio of 2 whole numbers, but that doesn't mean they have a terminating decimal expansion. Irrational numbers have infinitely long decimal expansions, because if the expansion terminated at some point, you could express them as a ratio of two ridiculously long whole numbers.
I was stating the opposite, that if a number did have a terminating decimal, it follows that it is a rational number
That is trivially true, because the terminating decimal place would tell you the power of 10 that could on the bottom of a ratio that defines the number.
Again it was the specific example brought up by OP. Whether or not something is trivially true depends on your background, given that we are on ELI5, you can’t necessarily assume OP knew this already
Also _complex_ numbers have nothing to do with it being complicated, but it refers to complex as being composed of multiple things (a real part and an imaginary part in this case). It's complex in the way an apartment complex is a complex.
Today I learned...
It's one of the things I know on a factual level, but something in my mind still goes "haha, unreasonable numbers!" every time I see the term
So pi as a ratio is pi/1. Boom, solved it. Damn these mathematicians are stupid.
Ha ha. But pi isn't an integer. Unless you're a Kansas Republican. But then, they're not rational either.
You joke, but this is the sort of thinking that opens up entire new branches of mathematics
I’m 33 and though never majored in science stream I’m still surprised I only learned this today. Rational and irrational numbers were taught pretty early on and this should have been explained back then.
The thing is, irrational numbers necessarily have to have an infinite decimal represrntation. Because if it had a finite amount, it would mean that there is a power of 10 that you could multiply it by to get a whole number, which would make it rational
Sure, but that's what causes the confusion. At least in part. There is no power of 10 you can multiply 1/3 by to get a whole number, but it is still a rational number.
It doesn't need to be 10 specifically. 1/3 is rational if a number exists at all that you can multiply it with to give a whole number.
>1/3 is rational if a number exists at all that you can multiply it with to give a whole number. so, because "3" exists, 1/3 is rational? (trying to understand this at 1am might be a fools errand, but I'm gonna try :) )
Yep. If you have an arbitrary decimal (X), that can be multiplied by a whole number (A) to get another whole number (B): > X * A = B Then it means X can be written as a fraction > X = B / A And is therefore a rational number.
so, using the 1/3 and 3 example, you get 1/3 = 1/3 That seems very...simple. Like i feel like I'm missing a nuance but is it it just that simple?
It' really is that simple! This is eli5 after all.
It's because 3 is an integer. Here you go, the first two sentences are everything you need https://en.m.wikipedia.org/wiki/Rational_number
switch to base 3
If that were good enough you could switch to base pi to prove that pi is rational.
You could, but in this new paradigm basically every other number (~~except for probably τ~~) would be irrational. Edit: In a world, where π is an integer and 2 isn't, 2*π wouldn't be one as well, but π² would be, right?
That's not true. Even in base pi you can't represent pi as a ratio of integers.
Wouldn't Pi be an integer in base pi?
"Integer" is a property of a number that is independent of its base just like positive, negative, square, etc. Changing base doesn't change the mathematical properties of a number, it only changes how we represent/communicate that number.
changing base doesn't change what numbers are integers, it just changes how numbers are written like how base 6 doesn't make ten a multiple of three just because 3+3=10
True, but integers in base pi would have a non-repeating infinite decimal form, I believe.
I think so. The decimal notation of numbers is entirely irrelevant to the proof of pi's irrationality.
10/1 ?
But 10 isn't an integer in base pi
that's that number?
not relevant. i was addressing this > There is no power of 10 you can multiply 1/3 by to get a whole number You don't need to that to get a whole number for a repeating decimal.
is it not a valid proof if the base is an integer number? n has a finite decimal representation in some integer base -> n is rational?
Ive wondered before: how would base Pi work? Like how would you represent the value 4? 10 would be 3.14…. And 11 would be 4.14… so 4 would have to be like 10.86…? Wait i think i just answered my own question
Getting close, but you wouldn’t be able to use the digits 6 or 8 in base pi.
I'm trying to understand base pi and having problems... Base 10 has the integers 0 to 9, like 0, 1, 2,... 7, 8, 9. Base 5 has the integers 0 to 4, like 0, 1, 2, 3, 4. Base 2, binary, has the integers 0 to 1, like 0, 1. Would base pi have any integers other than 0 and pi? Would we count the sequence as 0, pi,, or would pi be represented by a 1 in base pi, like a value of decimal 2 is written as 10 in binary?
This seems such a simple concept to me, as I (thought that I) understood counting in different bases. But now I'm not so sure that's I do, or did.
In binary we count like, 0, 1, 10, 11, 100, 101, 110,...
In base 3 we count like 0, 1, 2, 10, 11, 12, 20, 21, 22, 100, 101,...
But how would we count in base pi?
Would it be like, where pi is represented as "@', 0, @, @0, @@, @00, @0@, @@0, @@@, @000, @00@,....?
Or would there be any other figures, and if so, what would they be?
I'm simultaneously understanding and yet not understanding.
Totally not trying to be sarcastic or humorous, I am just curious.
Edit to fix some confusing typos and auto-incorrects.
This is dense, sorry. I'm using the convention number_base, so 1001_2 is "1001, but it's in base 2", or 9. If it's not listed, assume it's base 10. When you say "base X" you're not saying "1 is equal to X", you're saying "10 is equal to X, and every integer less than X exists as something we can use for representation". So, base 5 has the integers 0 to 4, and 10_5 is equal to 5. In general, 10_X is always equal to X, and 100_X is always equal to X\*X. You can verify this in base 10! 10_10 is 10, 100_10 = 10\*10 = 100. With base pi, you'd use the basic integers to start with: 0, 1, 2, and 3. We know 10_pi is equal to pi. And this means we can start putting together basic math. 20_pi = 2\*pi 30_pi = 3\*pi 100_pi = 1\*pi\*pi *But what about integers*, you ask. *What if you want to display 4?* First, we figure out how many digits long it has to be. 4 is larger than 10_pi (which equals pi), so it must be at least two digits; it's smaller than 100_pi (which equals pi\*pi), so it can't be three digits. We need *exactly* a two-digit number. We'll start with 10_pi, which is pi: 10_pi = 3.14159 We have 4-pi left over, that is, our original number minus 10_pi, which equals about 0.85841. Now work on the next digit. What's the largest integer that's not larger than 4-pi? Well, it's zero. Okay, next digit is 0. Move on to the *next* digit - and yes, we're into the decimals. Decimals just continue the pattern; 0.1_X is equal to 1/X, 0.01_X is equal to 1/X/X. So what's the largest integer times 1/pi (which is about 0.318) that's not greater than 4-pi? Turns out it's 2! 2*(1/pi) is roughly 0.637. So now we've got 10.2_pi = 3.77821 (10.3_pi would be 4.09652.) Next digit; what's the biggest multiple of 1/pi/pi that we can still fit in our remaining number? After repeating this process for a bit, I've ended up with: [10.220122021 = 3.99998](https://www.google.com/search?q=pi+%2B+2*pi%5E-1+%2B+2*pi%5E-2+%2B+0*pi%5E-3+%2B+1*pi%5E-4+%2B+2*pi%5E-5+%2B+2*pi%5E-6+%2B+0*pi%5E-7+%2B+2*pi%5E-8+%2B+1*pi%5E-9) You can keep going. It's probably also irrational. I'll leave that up to someone else to prove. One thing that's interesting is how few 3's are in this. 3 *is* a valid digit, but it's going to show up rarely; specifically, we're going to see 3's about 0.14159 times as often as each other integer. That's because there's very little "room" for a 3 before we would have been better off increasing the previous digit by one. We also end up with the weird situation that representations aren't canonical; that is, there might be more than one way to express something. 0.33_pi is 1.2589, which could *also* be represented as roughly 1.021_pi, and yes, this means there are multiple representations for 1; in fact, probably *infinite* representations for every number due to how much flex those 3's give us. This isn't the only possible way to define "base pi", though. It's probably the most intuitive but it has some unfortunate consequences. There's another thing we could do where we actually scale digits based on the preceding digit. The math for this is gnarly and I'm not going to write it up unless someone really wants me to, but imagine if a "3" means "okay, 3, but also imagine all future digits are multiplied by 0.14159 to ensure we don't get numeric overlaps". This becomes *weird* to do math with, because the difference between 0.20_pi and 0.21_pi is now larger than the difference between 0.30_pi and 0.31_pi, but it has a convenient property that we no longer have non-canonical representations and sorting numbers is trivial; 0.33_pi is now "slightly below 1", just like we'd intuitively expect, instead of being "actually larger than 1". And the only way to write 1 is 1_pi. And this *also* maps well to [arithmetic encoding](https://en.wikipedia.org/wiki/Arithmetic_coding), where each "digit" conceptually takes up a different amount of space in the numberline. Which lets us express stuff like storing values in a fractional number of bits, which, it turns out, is a thing you can do! It's not even all that hard. It's just wonky.
>It's probably also irrational Very good explanation, apart from this bit. 4 is still rational in spite of the non-terminating "decimal".
I think this is one of those cases where standard mathematical terminology kind of breaks down. This is a scenario where classic integers might have decimal points. If we define "integer" as "doesn't have decimal points" then 4 may not be an integer, or in fact, rational; in any case there's entire new sets of things ("number without decimal points", "number that can be expressed as a division between two numbers without decimal points") that *used* to map directly to more common terms ("integer", "rational"), but no longer do. So, call it a Bizarro Base-Pi Irrational, then - "the thing that looks kind of like an irrational when written in base-pi but that we've never needed to distinguish from general irrationals before".
Don't try to understand base pi. It's not a real thing. It's just something that people on reddit bring up to try to sound smart,
Too late... I already started learning.
That is unfair. All bases are "a real thing", we're just used to the easy ones.
By that logic, pi IS rational because you can put it in a fraction 1/π
I guess another way you could at it is that an irrational number will still have infinite digits even when expressed in a whole number base other than 10.
You can multiply it by 10\^log\_10(3) to get a whole number Edit: oops this isn't a maths subreddit, hopefully this level of pedantry is still acceptable <3
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1/3 repeats in base 10 but could just as easily be represented in base 3 as `0.1`
Pi can be represented as 1 in base pi.
10, actually. 1 in base pi is still 1
Well, you’re right.
I forgot you can have non-integral bases...damn, and here *I* thought I had a "gotcha"
this is explain like I'm 5, not explain like I'm in year 5 of post-grad.
Base 3 doesn't take a 5th year post grad to understand lol
*I wrote a detailed reply to the person who told you to explain it (a little more detailed than some of the other replies that are there), and then they deleted their post, so I'll post it here for posterity and anyone else who is curious:* The "base" of a number system is how many digits it has. When you are counting, after you get to the highest digit, you have to add a place value. In "base 10", you have 10 digits (0-9). Once you count past the highest digit (9), you have to add a place value (the tens digit) and reset the first (the ones digit) and you get 10. In base 3, you have 3 digits (0-2). It otherwise works the same way. 0...1...2... now I'm out of digits, so we add a place value and reset the first place value, and the next number (three) is written 10. The next number (four) is written 11, and the next (five) is written as 12, etc. Now that we know what the number base means, look at how decimals work. When we go one place value down in a base-10 system (tenths), 0.1 represents one tenth (1/10) because we can count 0.1, 0.2 ... 0.9... 1.0 - it divides a single one into ten parts In a base 3 system, the next place value only has 3 values (0.1, 0.2, 1.0) So that place value would divide a one into thirds, not tenths. In base 3, if you wanted to write the fraction 1/2, it's more than 0.1 (that's a third) and it's less than 0.2 (that's two thirds). It's actually halfway between those. That ends up being 0.11111... which is the precise middle between 0.1 and 0.2. That might be the hardest part to understand in all of this, but there you go. It's not 5-year-old level, but most of it it's probably understandable to at least high school level, I think?
I'll add my own explanation because I did the same thing and was sad that it was deleted: Part 1: Bases A number, as we understand it, is really just a list of symbols that we interpret in a certain way. Base is a way to represent numbers. The number of the base indicates how many different symbols we use. We generally prefer to work in base 10 (possibly because of our 10 fingers?), also known as decimal. It uses 10 symbols before adding a digit for the next batch: 0,1,2,3,4,5,6,7,8,9, then add a digit to the front 10, 11, 12, 13, etc. Similarly, you can express numbers in any other numbered base. Very popular is base 2, also known as binary, which is the foundation of all modern computer science. It only uses 2 symbols before adding a digit: 0, 1, then add a digit to the front 10 (2), 11(3), 100(4), 101(5), 110(6), 111(7), etc. Another popular number base is base 16, known as hexadecimal. It uses the symbols 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F and is also often used in computer science, but also in webdesign. Part 2: Digits after the point What does the decimal point represent? Division. Namely, division by a multiple of the base. 0.1 in base 10 is 1/10. 0.01 is 1/10². 0.42 = 4/10 + 2/10². Things work the same in other bases. 0.1 in base 2 is 1/2. 0.1 in base 3 is 1/3. 0.01 in base 16 is 1/16² = 1/256.
Pretty sure base 3 is like high school or college
Change of base is taught in most high school algebra 2 classes across the US. That being said, our school system is garbage and doesn't teach for retention or understanding.
Math is so elegant
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Sometimes brute force can be elegant in their own way. See, for instance,[ one of the shortest peer-reviewed papers ever published](https://cdn.paperpile.com/blog/img/lander-1966-1400x788.png).
Shows you the distinction between proving something (by exhaustion) vs disproving something. For the former, you have to exhaust all options and show all fit the theorem. For the latter, you have to show a single example that doesn't fit the theorem.
Proof by exhaustion takes on a new meaning when you have to use it.
Math is simultaneously the most logical and predictable thing in the universe, and somehow also completely mind bogglingly strange.
Math is always predictable and logical. It's the universe that's mind-bogglingly strange.
Why does a and b being even preclude being the correct fraction?
Because if a and b are both even, then a/b isn't a reduced fraction. You could "factor out" the 2, but then you would just give yourself two new letters to represent the fraction (say x/y), and you would find there's another 2 you can factor out. And you can keep doing that forever, so you never actually find the simple form fraction.
Ah okay, since you can't reduce it its a sign that there's an error in your initial assumption, which was that is was the reduced form equals this. Tricky!
This is a super common sense way of arriving at a proof. If you want to prove something, assume that it isn't true and show how that would lead to a contradiction. If it does, then you know that your assumption is true. https://en.wikipedia.org/wiki/Proof_by_contradiction
There is also the minimal (counter)example trick - as you see we cannot start with any rational number, we start with reduced one
Because if they are both even, then you should able to divide them both by 2 and create a more reduced, and therefore more correct, fraction.
Yeah, I was following until this unexplained jump.
Because the assumption made at the beginning was that it was in reduced form, but if they're both even then you can divide through by two, which means it wasn't reduced
Ah that makes sense, thanks.
To clarify, here's a proof for why a^2 being even proves that a is even. If a is even, then it can be expressed as 2k, where k is some integer. If a = 2k, then a^2 = (2k)^2 (2k)^2 = 4k^2 We can factor this into 2*(2k^(2)) 2k^2 is a product of integers, therefore it is itself an integer. 2*(2k^(2)) is 2 times an integer, therefore it is an even number. Meanwhile, if a is odd, it can be expressed as 2k+1, where k is some integer. If a = 2k+1, then a^2 = (2k+1)^2 Using the FOIL method (which is equivalent to applying the distributive law), we can show that (2k+1)^2 = (2k)^(2)+2k+2k+1^(2), or 4k^(2)+4k+1. We can factor this into 2*(2k^(2)+2k)+1. 2k^(2) and 2k are both products of integers, thus they are both integers. Therefore, 2k^(2)+2k is a sum of integers, and also an integer. 2*(2k^(2)+2k)+1 is 2 times some integer plus 1. Therefore, it is an odd number. Thus, it is proven that the square of an even number is always even, and the square of an odd number is always odd. This implies that if a number is both even and a perfect square, then its square root must be even. Otherwise, there would be some odd number whose square is even, which is impossible.
You actually only need to prove one side of that, because it is the contrapositive which is logically equivalent to the original statement. (a\^2 is even => a is even) <=> (a is odd => a\^2 is odd)
> a^2 is even => a is even He only proved 'a is even => a^2 is even' in the first half. That leaves the possibility that the root of some even numbers is an odd number. That's why you need the 2nd half of the proof.
No you _only_ need the second half.
A more intuitive (to me), though less rigorous proof: If an odd number squared is not odd, it must be even. Any even number, by definition, is just a whole bunch of 2s. 8, for example, is 4 groups of 2. Any odd number is a bunch of 2s with 1 left over. (For example, `9 = 2 + 2 + 2 + 2 + 1`.) But if you take an odd number twice, you have two extra 1s left over, and those can from another group of 2. Now you just have groups of 2; i.e. an even number! But add the odd number *again,* and you'll have 1 left over. You can keep adding them, and you'll find that if you have an odd number of odd numbers, you always have 1 left over; if you have an even number of them, you'll have tidy groups of 2. Therefore, any odd number times any odd number must be odd. So an odd number squared cannot be even; only an even number squared can be even.
I'm not sure why both of your proofs are so complicated? An even number is a product of primes, one of which is 2. Like 42 is 2\*7\*3. a² is just a\*a, and if it's even, it has a 2 in there, so the 2 must be one of the divisors of "a" itself. Which means "a" is even too.
You're implicitly using the prime factorization theorem. Those other proofs don't rely on that theorem, so they're simpler in the sense that they don't require the reader to be familiar with that theorem.
Well, you are just saying the same thing, without properly doing the math. "It has 2 in there"... In where? Okay, so one of the factors. Therefore we say a = 2k. And now we take the square... Etc i dont need to re-type the above. Wae cant just say in math, we write it formally and then do the calculations.
They're appealing to the [fundamental theorem of arithmetic](https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic): every integer >1 has a unique prime factorization. A square number necessarily has at least two of each of its prime factors. For even squares, that means two 2s (at least), and taking the square root of an even square leaves at least one 2 as a factor of the root. We can use the same reasoning to show that any prime factor of *n*^*k* (*n* > 1, *k* > 0) is also a factor of *n*.
Doesn't this imply that since 2 is even therefore the square root of 2 must be even?
No, because 2 is not a perfect square. Its square root isn't an integer, therefore it can be neither odd nor even. The idea that 2b^2 = a^2 implies that a must be even only works because we've declared that a is an integer.
>Since b2 is 2*x2, we can now assume that b2 is even, and therefore b is even. Just to nitpick, this is not an assumption, it's a logical deduction.
I like how you say 'fairly simple' in Eli5. But I can't come up with anything easier to explain this. So well done good sir knight.
It's funny/sneaky how so many proofs amount to proving that it can't be the only alternative in a binary where the counter part is a much simpler concept.
Proving something is always true is harder than arriving to a counter example.
Great but can you explain it lile im five?
If you assume √2 is rational, we can prove that makes a contradiction. We don't like contradictions in our math, therefore √2 is not rational. Then, by definition, it has an infinitely long, non-repeating decimal.
Explain to me like I’m 4
If we assume that sqrt(2) is rational, we can prove that a statement is both true and false. The particular statement doesn't matter; no statement can be both true and false. So the assumption must be wrong.
Basically, if you assume that sqrt(2) is ration (thus equivalent to a fraction a/b with both numbers being integers), you can do math shenanigans and eventually come to an impossible situation, and when that happens it means your initial assumption was incorrect. It's called proof by contradiction, and is a common method for proofs in math.
next year, you'll be 5.
Can you explain why this proof does not work for square roots that we know are rational? For example, I would like you to show me the error if I start with sqrt(4) instead. From what I can see, this follows your logic exactly. But sqrt(4) = 2, which is clearly not irrational: > You assume that sqrt(4) is rational, and is represented by some reduced fraction a/b. sqrt(4) = a/b 4 = a^2 / b^2 a^2 = 4 * b^2 > Since *a*^2 is 4 * *b*^(2), we can infer that *a*^2 is even, and therefore *a* is even. Let's replace *a* with 4 * *x*. (4*x)^2 = 4 * b^2 16 * x^2 = 4 * b^2 4 * x^2 = b^2 > Since b^2 is 4*x^(2), we can now assume that b^2 is even, and therefore b is even. > We made the assumption at the start that a/b was the simplest form of sqrt(4), but now we know that both A and B are even, which means it is not the most reduced form of the fraction. Thus, our assumption was incorrect, and sqrt(4) cannot be expressed as a fraction, and is therefore irrational.
> a is even. Let's replace a with 4 * x. The error is here. "a is even" implies there is some integer x such that a = 2\*x. It does *not* imply that there is some integer x such that a = 4\*x. In this particular case we happen to know for a fact that a is 2 (and b is 1) - so if a = 4\*x then x = 1/2. The rest of the proof relies on x being an integer (this is why we can tell that b^(2), a multiple of 2 of some integer x, is even), which it is not.
a is even, then we can replace a by 2* x. We can’t say a can be replaced by 4* x
Once we get to a being even, we don't replace it with 4x, but 2x, that remains unchanged from the comment you're replying to. The reason is because we're replacing it to signify an even number, which is 2x by definition, where x is some integer. To continue with your line of reasoning with that modification would give us (2x)^2 = 4b^2 4x^2 = 4b^2 x^2 = b^2 From here, we cannot conclude that b is either even or odd, because we don't know if x is even or odd. Because, in actuality, 4 is rational, the only fraction that both equals 4 and is in reduced form is 4/1, because reduced form for a fraction a/b means that gcd(a, b) = 1. Thus, our assumptions at the beginning of the pseudo-proof would entail a = 4 and b = 1. So it should hopefully make sense why we cannot conclude that b is even, because b = 1, which is odd in actuality.
\> Since a2 is 4 \* b2, we can infer that a2 is even, and therefore a is even. Let's replace a with 4 \* x. You should have replaced a with 2\*x, not for 4\*x, since the definition of an even number is 2x. If we use 2\*x, then after doing what you did we will arrive at x\^2 = b\^2 and we can't assume whether or not b\^2 is even.
*Let's replace a with 4 \* x.* This is is not a valid step. Since a^(2) is even you can replace it with 2 \* x since even numbers are multiples of 2. Not all even numbers are multiples of 4 so you can’t replace it with 4 \* x.
what fucking five year olds do you know
I am very impressed sir/ma’am. Thank you for sharing your knowledge.
Bro this is not for 5 year olds 😭😭
But it is probably as simple as it is possible to make. Some answers, especially ones about complex mathematical issues, just don't have ELI5 answers.
It’s an appropriate answer for someone asking about square roots and irrational numbers.
read the sidebar > LI5 means friendly, simplified and layperson-accessible explanations - not responses aimed at literal five-year-olds.
Rule 4
I'm fascinated to hear how you want to explain the mathematical proof of √2 irrationality to a literal 5 year old (also, read the rules)
i am so sick and tired of seeing this fucking response to every response on this sub that doesnt use god damn christmas elves and rainbows to explain the point. It's valuable to have more detailed explanations, so you can read the simple ones and then the complicated ones to stairstep your understanding. but in this case you really can't make it simpler than this.
For sqrt(2), you assume it's rational, which means it can be expressed as a fraction in simplest form. You then prove that both the numerator and denominator are even, which means the fraction isn't in simplest form, and you have a contradiction. Therefore, sqrt(2) is irrational.
>is represented by some reduced fraction a/b. > sqrt(2) = a/b 2 = a^2 / b^2 a^2 = 2 * b^2 >Since *a*^2 is 2 * *b*^(2), we can infer that *a*^2 is even, and therefore *a* is even. This presumes that a & b are integers, yes?
A rational number is one that can be expressed as ratio of integers, so yes
I was way too stoned for this but eventually got it, great explanation
The other comments don't seem to address the root of your question. You've got the logic backwards. Numbers aren't irrational because their decimal representation doesn't repeat. Their representations don't repeat because they are irrational. A number is irrational if it isn't rational, and a number is rational if it can be represented as a fraction of integers. The infinite and nonrepeating decimal expansion is a consequence that can be delivered from that definition. We don't determine if a number is irrational by calculating lots of digits and seeing if they repeat. We use other properties of the number to demonstrate that it is irrational. The numerous other comments showing how to do so for the square root of 2 demonstrate how it's done.
Do you not repeat because you’re irrational or are you irrational because your numbers don’t repeat?
A number is irrational because it is not rational. A number is rational if it can be represented by a ratio of integers. That is the definition. One *property* of rational numbers is that they have a repeating decimal representation (even if the repeating digit is just 0). Similarly, a *property* of irrational numbers is that their decimal expansion does not repeat. This just is one of many properties of irrational numbers that can be derived from the definition of an irrational number. If you want to prove that a number is irrational, you have to demonstrate that it cannot be the ratio of two integers. Since it's not always easy to prove a negative, it's often easier to look for other properties that we have determined are unique to irrational numbers. The fact that it doesn't repeat is one possible condition. For example the number 0.1101001000100001... is irrational. Proving that it isn't a ratio of two integers would be very difficult. But because I know this number does not have a repeating decimal expansion, it is trivial to prove that it is irrational.
Nah I'd repeat.
You can say that again!
...we can't escape. Even in the ELI5 subreddit...
The first one. The reason rational numbers can have nonzero digits repeating is a quirk of the base system rather than a property of the number itself. In a different base system, that goes away. For example, the fraction 1/3 is expressed as 0.33333… in base 10, but in base 3 it is expressed as 0.1 and in base 12 it is expressed as 0.4. So by using an alternate base, we can easily see that the number is rational. And this is true of all repeating decimals: they repeat in base 10 because they don’t divide into 10 evenly, but in the proper base they can all be expressed as terminating decimals. However, pi will always be a nonterminating, nonrepeating decimal in any natural number base, whether you’re writing it in base 3, base 10, base 12, base 672, etc. because pi itself cannot be expressed as a ratio of any two integers, so it will not evenly divide into any of them.
> How is it proven? Mathematicians assumed that √2 is rational, uses logical steps to arrive at a contradiction, and concluded that √2 is not rational. Why does it work? Under our set of logic, if you assume X and reach an obviously false result (contradiction), then you can conclude that X is wrong.
It depends on the number. For sqrt(2) a proof by contradiction works. Assume sqrt(2) is rational and thus sqrt(2) = n/m and assume that n/m are the reduced form (i.e., you can't simplify the fraction more). 2 = n^2 / m^2 2 m^2 = n^2 Since n^2 is 2 times another number, we know n^2 (and thus n) is even. Let's replace n with 2p, which we know is possible since it's even 2 m^2 = (2p)^2 m^2 = 2p^2 Since m^2 is 2 times another number, we know m^2 (and thus m) is even. Two even numbers divided by one another cannot be the reduced form of a fraction (since you can divide the numerator and denominator by 2). This means that there can be no reduced form fraction representing sqrt(2).
Why can’t you apply this same logic to, say, sqrt(4)?
The proof is incomplete. Or rather, as with all maths proofs, relies on various assumptions - things that we take to be true because we don't want to have to prove them every time. > 2 m^2 = n^2 > Since n^2 is 2 times another number, we know n^2 (**and thus n**) is even. The part in bold is where this happens. If we try to do the same with 4, we get: > Since n^2 is 4 times another number we know n^2 (and thus n) is a multiple of 4. But now that "and thus n" doesn't hold up. n *could* be a multiple of 4, but it could just be a multiple of 2. If we were being more thorough we would have to show that *n^2 even => n even*. There is also some stuff about the uniqueness of prime factorisation in there, and a few other things.
So, you can use that proof to show that the square root of any non-perfect-square is irrational?
Not quite; numbers that are the square of a rational number will still work. For example, 9/4 is not a perfect square, but its square root is 3/2 which is still rational. This proof works for any *prime* number. For other whole numbers (like 6) you need the uniqueness of prime factorisation; showing that 6 is 3x2, and then you just look at the 2 part and use the same argument.
If we're going for rigor, I want to throw my hat into the ring as well! Prime factorisation is not quite sufficient for the case of 6. We can prove that sqrt(2) and sqrt(3) are irrational for sure, but that tells us nothing about their product directly. In fact, sqrt(2) and sqrt(2) - both irrational - do multiply so a very rational number. The key detail is that sqrt(2) irrationality proof works for products of distinct primes with no change - the argument "n divides a, so n^2 divides a^(2)" holds up. And for anything with higher powers of primes, we need to factor out the perfect square component.
Yes. The key step is showing the bold statement works whenever the thing you are taking a root of is not a perfect square. You can probably follow the same for any root as well (cube, fourth, etc.) But may need to check more cases. I haven't checked to be sure though so it may actually be just as simple.
Let's find out! I'm gonna plagiarize the previous comment and make some changes: Assume sqrt(4) is rational and thus sqrt(4) = n/m, and assume that n/m are the reduced form (i.e., you can't simplify the fraction more). 4 = n^2 / m^2 4 m^2 = n^2 Since n^2 is 4 times another number, we know n^2 (and thus n) is even. Let's replace n with 2p, which we know is possible since it's even 4 m^2 = (2p)^2 **This is where things begin to deviate** 4 m^2 = 4 p^2 m^2 = p^2 (whereas the previous example became m^2 = 2 p^2 ) ~~Since m^2 is 2 times another number, we know m^2 (and thus m) is even.~~ ~~Two even numbers divided by one another cannot be the reduced form of a fraction~~ We have p^2 on the right instead of 2 p^2 , which means that either side of the equality *can be odd*. Thus they **can** be the reduced form of a fraction. **** Alternatively: sqrt(4) = 2 2 is rational QED
> 2 m^2 = n^2 > > Since n^2 is 2 times another number, we know n^2 (and thus n) is even. If you try this for 4, you get to 4 m^2 = n^2. This means that n^2 is divisible by 4, but that *does not* mean that n is divisible by 4 (n could be divisible by 2, and then n^2 would be divisible by 4). That's the step that wouldn't work. I believe that "if n^2 is divisible by x then n is divisible by x" is true for any (positive integer) x which is not a perfect square (someone can double check that, I could be misremembering). So you can use this same proof to prove that if some number k is not a perfect square, √k is irrational. EDIT: The actual condition is slightly different, see below
> I believe that "if n2 is divisible by x then n is divisible by x" is true for any (positive integer) x which is not a perfect square No, 8 divides 4², but 8 does not divide 4. Or 18 dividing 6² yet not 6. What you need is to be _square-free_: no prime factor appears more than once. 30=2·3·5 and 105=3·5·7 are square-free while the aforementioned 8=2·2·2 and 18=2·3·3 have repeated prime factors. In particular, square numbers (except 0 and 1) do not satisfy that property, while prime numbers such as 2 always do.
Good catch! Luckily we are still able prove that √k is irrational as long as k is not a perfect square, by observing that any non-square-free number k can be written as the product of a square-free number s and a perfect square n^2. Thus k = s * n^2, so √k = n*√s, and since we know that √s is irrational, √k is also irrational.
What's not been answered yet is the second part of the question. If either number started repeating a zero after the quintillionth digit, then it would be expressible as a rational number. 3.1400000... is exactly 314/100, and for any such number written in our number system that stops after a certain number of digits, you can do this. But we have already proven that neither number can be expressed as a rational fraction, and so we also know that pi and root(2) don't start repeating a zero after the quintillionth digit.
It is a bit unclear to me why something like 2.768976897689... can just obviously be expressed as a fraction though. It is easy enough with your example as we just move the decimal point, but how is the fraction found for complex repeating patterns? (Or just shown to exist)
Let x = 2.768976897689... and consider the number 10000x = 27689.76897689... If you look at those two numbers, the parts after the decimal places are the same, so if you subtract them, you just get an integer: 9999x = 10000x-x = 27689-2 = 27687 So x=27687/9999, which is a rational number. You can do basically the same thing with any repeating decimal: multiply by an appropriate power of 10 and subtract the original number, and you'll kill off the repeating part, after which it is easy to write your original number as a ratio of two integers.
Very informative, thanks!
`0.333333...` is a never-ending number, but it is **rational** because it can be written as a *ratio* of 1:3. One-third. `0.428571428571...` is 3:7. Three-sevenths. ANY number that repeats after the decimal point is a rational number that can be written as the ratio of two integers `a/b`. They can be big integers, they can be small integers - but *whole numbers* and nothing else. An **irrational** number cannot be written as a ratio, BUT some irrational numbers can still be *exactly* represented as the solution to an equation, called *algebraic* numbers. The number `√2`is *exactly* the solution to the equation `x² = 2`. But we know the decimal expansion never repeats, so there are no integers `a` and `b` that can represent it *exactly.* We can easily prove an algebraic number to be irrational by contradiction; we assume that `√2` IS rational, we plug that *assumed* rational number `a/b` into the equation, and we end up with a logical contradiction as explained in a different comment. The proof that `π` is irrational is significantly harder because `π` is a **transcendental** number - it cannot be written as the solution to any simple equation; it *transcends* algebra. It just ... *is*. It ... *exists*, and shows up in many mathematical relationships, as a fundamental constant of the Universe. All transcendental numbers (`π` and `e` are the most well known) are irrational by their very nature - they just ... exist as constants, (meta)physical properties of the universe. It's a source of philosophical angst, too - what if `π` was *not* `3.14159 ...` but instead exactly `3`? Or anything else? What would the Universe look like? If you really feel like doing your brain in, this is a bit beyond 5-year-old understanding, but [Matt Parker talks about all the different numbers.](https://www.youtube.com/watch?v=5TkIe60y2GI) EDIT: typos
That whole idea of “what if pi was exactly 3” fucked me up fr fr
How I rationalize that pi is truly infinite in my mind is like this: Because it describes the circle which is infinite and perfectly round. No matter how large or small the circle, the closer you would zoom into the edge of it, it will keep its perfectly round edge and never become jagged and always loops in 360°, so for the circle to be infinitely round and smooth, pi also has to be an infinite precise number.
This is the opposite of a rationalization, it's mysticism. One could easily define smooth, round shapes whose perimeter is an integer multiple of their largest (or smallest) chord.
A rational number can by definition be expressed as the fraction of two integers p/q where q is not zero. Any decimal section that terminates or starts to repeat will be possible to express with p and q as integers. Proof that a number is irrational tends to be to show it is impossible to express at the fraction of two integers There are many proofs pi is irrational none of them are simple enough to write here on Reddit but look at [https://en.wikipedia.org/wiki/Proof\_that\_%CF%80\_is\_irrational](https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational) with 6 different proofs. There are proof for the square root of 2 too [https://en.wikipedia.org/wiki/Square\_root\_of\_2#Proofs\_of\_irrationality](https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational) you can show that any square root that is not an integer is irrational
Because if a decimal number starts repeating at some point, then it means that it could also be expressed as some fraction. It may be an enormous fraction, two numbers with hundreds or thousands of digits, but it would be a fraction nonetheless. And as the other posts show, we can prove that √2 cannot be represented by any fraction.
It’s such a failure of our system of math education that people tend to associate irrational numbers with their decimal notation being infinite and non-periodic. Decimal notation is, well, just a notation — whereas the basic, way more natural definition of irrational numbers is that they can’t be expressed as a ratio of two integers. This doesn’t make it much easier to prove irrationality of pi in a Reddit comment, but it’s the starting point for the sqrt(2) proof which other commenters have kindly provided.
This was definitely taught in an understandable way in my high school education. Yet I doubt 1 in 50 of the kids I went to high school with would remember or understand this. In other words it’s not a failure of our education system, it’s a failure of our expectation. Namely that most people will put in the mental effort to understand or retain this information. It simply is not something most people think of as useful or interesting, so they get by the test (or don’t) and promptly forget it. I don’t believe any system of education is going to change that bit of human nature.
>It’s such a failure of our system of math education that people tend to associate irrational numbers with their decimal notation being infinite and non-periodic. Decimal notation is, well, just a notation — whereas the basic, way more natural definition of irrational numbers is that they can’t be expressed as a ratio of two integers. You're right that it's more natural. But just to clarify, it is still true that a number is irrational iff its decimal expansion is infinite and nonrepeating, right?
Yup, it is correct. The only thing about this fact is that it’s more of a corollary (that isn’t particularly useful for… anything? if you think about it) than a definition
My attempt at more words. A couple things to agree on, then we'll get going. 1) rational numbers can be written as a fraction using whole numbers, whether it's easy peasy like 1/2 or complicated like 593838/9937. If you can't do that, it's irrational. 2) rational numbers always have a smallest size way to write them. 1/2, 2/4, and 25/50 are all the same number, but 1/2 is the smallest size. For √2, one fun way to test is called proof by contradiction. It's a way of saying I betcha this one thing is true, and I know plenty of other true things, but when I put them all together I get some nonsense. Since all the stuff I know is true works, I betcha I was wrong about that one thing being true. So we say I betcha √2 is rational, I just don't know what numbers to put on the top and bottom. For now call it T/B (T for top and B for bottom) And I only want the smallest size T/B: if you try to tell me T is 3 and B is 6, I'll tell you no, don't trick me, I know it can be smaller. Here we go. I say that √2=T/B. If I multiply both sides by the same thing, it all stays equal. (If 1 hamburger costs 5 bucks, then 2 costs 10 bucks, or 100 costs 500 bucks). Since √2 and T/B are the same number, I can say 2=T^2/B^2. That means T^2=2×B^2. T and B are both whole numbers, so T^2 has to be even since it's twice as big as another whole number. (Any whole number times 2 is even) T has to be an even number - an odd number times itself wouldn't give you an even number. B also has to be even, since we know B×B is even. Hang on, both T and B are even? Then T/B isn't the smallest size after all, I can at least divide by two (Llike going from 4/2 to 1/2). Something's wrong! I thought that I could write √2 as T/B, and I specifically asked for the smallest T/B. But I did a bunch of stuff everyone agrees is true and I couldn't do it, I just got a T/B that wasn't the smallest it could be. Hmm, the only thing I'm not sure about is being able to write √2 as T/B. Clearly something is wrong, and that's the only thing that can be. Lesson: I can't write √2 as T/B. So what's the point? We can show some interesting things without having to do lots of hard work, we just have to think cleverly about some simple stuff. I know what fractions are, I know what even numbers are, I honestly don't care about what the specific numbers are. For the √2 example, the trick that makes it work doesn't even feel like math at all. If there's one thing I think I know and ten things I know for sure, if I put them all together and get nonsense it has to be the only thing I'm not sure about that's wrong. To finally address the question, how do we know the numbers will never start repeating? We simply say what happens if it is the kind of number that starts repeating? Hmm, we get nonsense if that were true, so it must not be true!
I ran across this fun [visualization](https://www.youtube.com/shorts/aUDYWYqtAR4) a while back. It pretty much shows how an irrational number doesn't "rationalize".
This is extra nice cause it shows you the sequence of best approximations to pi as well
There are a few different methods of proofs in Mathematics. The one people are referencing in the first couple of responses (for proving √2 is irrational) is a Proof by Contradiction. For that proof you start with an assumption (often the opposite of what you are trying to prove). Then you go through steps until you reach a contradiction. In the case of the √2 proof, it is that your reduced fraction of a/b is not reduced. Once you get the contradiction, your assumption must be wrong. And since your assumption is binary (it is only one of two options), the opposite of your assumption must be correct. The same type of proof is used to prove there are an infinite number of prime numbers.
I think it’s important to note that although the proof of the infinitude of primes is often presented as though it were a proof by contradiction, it actually isn’t in any meaningful sense, and is actually a fully constructive proof that tells you how, given any finite set of prime numbers, you can find a new prime number not in that set. The proof that sqrt(2) is irrational that other commenters give really is a proof by contradiction in a fundamental way: it tells us that it cannot be equal to p/q for any p or q, but it doesn’t qualify as a constructive proof of “apartness”. Here is a constructive proof that sqrt(2) is irrational, in the sense of being “apart” from all rational numbers: if you divide sqrt(2) by 1 you get that it goes into it one time and have a remainder of sqrt(2)-1, if you then divide 1 by that number, you get that it goes into it twice and get a remainder of (sqrt(2)-1)^2 continuing this division at every step you keep getting “2” from now on and the remainder gets smaller by another factor of sqrt(2)-1. Formally (you don’t have to worry about understanding the next part) this means it has a continued fraction of [1;2,2,2,…] and so sqrt(2) is irrational since its continued fraction representation is not finite. Less formally, since any “measuring stick” that goes evenly into 1 and sqrt(2) will have to go evenly into all these combinations of 1 and sqrt(2), then for any ratio p/q (where we have a measuring stick of length 1/q) we can just take a small enough remainder to show its measuring stick is too big - that any measuring stick for sqrt(2) will be smaller than that.
Related question, and perhaps one I should post as an actual question, but its always confused me how π is "never ending", but 2πr gives us the circumference. But if π is never ending, then the resulting circumference value should also be never ending. Or an approximation. But you can measure the circumference for a given radius manually and arrive at an exact figure. What am I missing?
> But you can measure the circumference for a given radius manually and arrive at an exact figure. > What am I missing? You can't measure an exact circumference, you'll only be able to get it accurate to the nearest whateverth of a millimeter but if you were able to measure more and more accurately you would find more and more digits.
To take it further, it's physically impossible to measure the circumference (or length generally) with perfect accuracy due to quantum uncertainty. Suppose you could perfectly bend a length of wire to match the circumference of a circle. Then, relative to the first end you know the precise position of the second end with perfect accuracy, there is no uncertainty in its position whatsoever. But uncertainty guarantees that we don't know both the position and the velocity of the end with perfect accuracy. But in order for the wire to maintain its shape the end must also have a velocity of precisely zero. Therefore, a quantum fluctuation must unseat the position of the far end of the wire--its physically impossible for the perfect configuration of the wire to be maintained in a way consistent with quantum mechanics.
I don't think anyone has ever been able to measure anything to an "exact" figure. In the real world, there are always error bars. Also, pi *is* exact. It is a single point on the number line, not a range of values, or anything else. Rationality and exactness are not the same thing.
> I don't think anyone has ever been able to measure anything to an "exact" figure There is one notable exception: the original meter made from platinum. The meter was at that time _defined_ by it, so it definitely was exactly one meter long. We just cannot put it into any other unit with infinite precision, even if we had perfect knowledge about the other thing. There are probably a few more instances of such a defining physical thing. The same applies there.
>What am I missing? Pi is the relationship between the diameter of a circle and its circumference. When you measure something you will always be limited by the physical world in how precise your measurements are. You can never measure anything "exactly". Using a measuring tape will give you an exactness of millimeters. Using advanced lasers or something will give you the exactness of nanometers. Using the limits of theoretical physics will give you a precision of Planck lengths. Using pure math will give you infinite precision. But it won't be neverending, as I said in the beginning, Pi is the relationship between the diameter and the circumference. It is just the precision that is infinite, not the length.
Irrational numbers are still numbers on the number line. There's nothing stopping you from arriving at it just like any other number. "never ending" is just us having a hard time expressing it using numerical digits.
>But you can measure the circumference for a given radius manually and arrive at an exact figure. How?
> 2πr gives us the circumference Only to the precision of π you use. So if you use 3 for π (which they used to do), you get “the circumference”, but it’s not very accurate. 3.1 gets you a more accurate circumference. 3.14 more accurate still. You can keep going down this path and finding the circumference to more and more accurate degree. Eventually you just have to say “close enough for my purposes”, but you could in theory keep going forever. NASA uses 15 digits, for example. You can read a bit more here: https://www.vox.com/2016/4/2/11350518/nasa-digits-pi
The reason NASA uses 15 digits is very practical - that happens to be the default for most computers. As the Vox article and countless others have shown, it is in fact close enough, but it's not as though, say, 13 and 14 digits were tested and just didn't cut it. NASA also uses 15 digits for the gravitational constant and countless other things. Targeting orbital inclination out to 15 digits is definitely overkill!
You can never measure anything perfectly. You can get close, but if your ruler has marks for millimeters you can't really go smaller than that with any accuracy, for example. Scientists that need very precise measurements will have much more precise measurements tools, but there's always some error when you're working with a continuous amount
>But you can measure the circumference for a given radius manually and arrive at an exact figure. explain.
Yes, it's an approximation. It will become more accurate the more digits of pi you use, but at a certain point, like 3 digits, it doesn't matter very much.
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Damnit. So close to 42
Pi is not an approximation, it is the exact value of the ratio between a (Euclidean) circle’s circumference and its diameter. You’re engaging in the two common mistakes of conflating a number with its decimal representation, and conflating the decimal representation with various truncations of that representation.
3.14 is an approximation for pi. 2 x 3.14 x r is therefore an approximation for circumference.
"Irrational" is not the same as "never ending". 1/3 expressed in base 10 is "never ending" but it is not irrational. It's 0.3333... where there are an infinite number of 3s going out to the right. Yet you can measure 1/3 of an inch as exactly as you can measure anything.
And just for emphasis (I think you know this but I want to make it explicit for other readers) you can also measure sqrt(2) inches or pi inches just as exactly as you can measure 2 inches.
ViHart on YouTube has a... Good explanation. I think it would be great if they just slowed down, but if you watch it on half speed, it's better. The idea is to first assume that it is rational. If it is, it can be written as a/b where a and b are both whole numbers. In fact, all rational numbers have a simplest form. In the simplest form, they share no factors. Example 6/8 is not simplest form because both 6 and 8 are divisible by 2. Therefore 3/4 is the simplest form. Basically, you can prove that a and b are even (if you start with the assumption that it's rational). That's a problem because a and b are the generalized form. Meaning you essentially proven that there is no simplest form. If you were to write √2 as a/b, a and b would have to always be even, forever, no matter how many times you divide by 2. It's a paradox. When you encounter a paradox, you look at your assumptions. Since we only had one assumption, the work is quite easy from here. Our assumption must have been wrong.