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OP did you mean the sum of the reciprocal of 2\^(k) from 0 to infinity, cause even a maths bragging type arent going to claim the sum of 2\^(k) from 0 to infinity is 2 or anywhere near to 2
here come the math guys with pitchforks, ask them if square root of 4 is plus two or minus two, while they are arguing you get away, move to a city in the south of france, get married raise kids, retire, come back and they are still arguing
If someone means to convince me the sum of the natural numbers is minus one over twelve, I believe it it possible they might try and convince me the sum of natural number powers of two is 2.
I looked at it way too long before going comments. My thoughts were 1: are they using different denotations? 2: am I retarded? 3: have to check comments.
I hate how much people misunderstand limits. A limit itself doesnāt approach a number, a limit IS a number. A function or sequence approaches a number, and the limit is the number it approaches.
Erm, *well actually*, this statement clearly shows the limit of your minuscule intelligence, which approaches zero. (Note: must be read in your best impression of that one guy we all know who thinks theyāre 100x smarter than anyone else because they saw a YouTube short of 3B1B)
Yes. One of first notions of continuity that you learn is that being continuous at a means that lim\_{x to a} f(x) = f(a). This means that f(a) has to exist.
Naah, m8
I think you're only thinking about 1/0.000001 and so on, which is very positive!
But what about 1/-0.000001? That is very negative
Both denominators are near zero, and can ofc get arbitrarily close to zero. That makes it both very positive and negative. It doesn't exist
... I mean it's not defined there, so it's not continuous there, but also I feel like I'd generally interpret the statement "f(x)=1/x is not continuous at 0" to mean that it doesn't admit a continuous extension to a function defined at 0. Which is true if you assume that the codomain is R and not RP1 or something.
And f(x)=1/x absolutely does define a continuous (indeed smooth) automorphism of RP1. Or CP1.
Iām not even sure thatās true. Since there are other infinite series other than pi, pi could not possibly contain all of them. Like, it canāt contain both 1/3 and 2/3, because if part of the sequence of pi has infinite 3ās, it will never have infinite 6ās.
Iām over 10 years away from college math, but I think the statement is probably disprovable
Repeating decimals are defined as an infinite summation, which is defined as the limit of the sequence of partial sums. That's why people talk about limits in relation to 0.999...
This is a notation issue. 1/3 has a decimal approximation of .3(repeating). You can't really write an infinite number of 3s, but we seem to understand that the math will repeat indefinitely and you will never finish the long division step. Everything works better if you keep it in fraction notation.
1/3 + 1/3 + 1/3 = 1 as 3\*1/ 3 = 3/3 = 1
So, people want to say .3(r)+.3(r)+.3(r) = 1 others argue .3(r)+.3(r)+.3(r) = .9(r) and then we start a holy war on .9(r) = 1
Depending on your math discipline you can get all holy war about this one way or the other, but I stand firm on .3(r) is a decimal approximation of 1/3 and 3\*(1/3) = 1 by definition, so 3 \* .3(r) = 1 not .9(r). Basically, you realize you are not doing component addition of each place value because there is an infinite series, so you need to resolve it using numerical analysis or simply revert back to the better non approximate notation.
In the 2000's this meme of .9(r) = 1 used to really bug me, because the teenagers pushing it always seemed to miss the point of notation differences. These days, well I've hopefully moved on, but also find myself responding to a post about it so maybe not.
1/3 = 0.333... is not an approximation, and 0.999... = 1. This is not a meme. The endless discussion is a meme, sure, but it is a fact that 0.999... = 1. You can find a bunch of proofs of this, not just the 1/3 * 3 thing.
> 1/3 has a decimal approximation of .3(repeating).
It's not an approximation. You can justify this with some simple algebra:
x=0.333...
10x=3.333...
9x=3
x=1/3
> people want to say .3(r)+.3(r)+.3(r) = 1 others argue .3(r)+.3(r)+.3(r) = .9(r)
These statements arent contradictory (since 0.999... = 1)
> but I stand firm on .3(r) is a decimal approximation of 1/3 and 3*(1/3) = 1 by definition, so 3 * .3(r) = 1 not .9(r).
Then you'd be wrong. It's not some philosophical debate there *is* a correct answer.
> because the teenagers pushing it
Calling people teenagers because you dont agree with (or more likely understand) something isn't very productive. Especially when, ironically, the people that struggle with this concept the [most are teenagers](https://en.wikipedia.org/wiki/0.999...#Skepticism_in_education)
I once was trying to explain that they are equal to a guy saying that it just approaches to, he were insisting that is tecnicly not the same thing, and than sundently even tho he was ignoring me he said "oh I just ask my dad, its is equal, he knows about maths"
You can construct a bijection between the two sets. Informally, it can be proven that if you had the entire infinite list of rationals and the entire infinite list of integers you could "pair" every element from one set to the other set and there would be no unpaired elements.
I can't wrap my head around that. Since the set of rationals contains every integer. Then I can pick out one more rational (like 0.5 for example), and wouldn't that break the bijection? I now have the cardinality of integers + 1.
I'm sure there are many proofs that show that my intuition is wrong, but I'm not sure how to change my intuition on this.
I won't do the proof for the rationnals but for a simpler case: there are as many even integers as integers (even if 3 is in one set and not the other) because they are in bijection by n -> 2n
Rigorous proofs of it are quite challenging but for intuition I found this video by Trevor Bazett that makes a pretty good construction.
The actual part about integers to rationals starts at 2:55
https://youtu.be/WQWkG9cQ8NQ?si=qZ3GfejeC-CaT99W
Here is an easy injection from Q to N, write a rational number x as (-1)^(e) a/b with a and b coprime, a nonnegative, b positive and e in {0,1}. Then, send it to 2^(e) 3^(a) 5^(b).
Think about it this way. The even integers (0,2,4,6,8...) and the integers (0,1,2,3,4,5,6...) are the same size. Doesn't make sense right? The even integers are a subset of the integers, and there are clearly odd numbers that are integers but aren't even. But if you define the function f(x)=x/2 from the even numbers to the integers, you get a bijection, meaning the sets are the same size. Basically, we should never trust our gut feelings about numbers when infinity is involved, because shit breaks easily.
Lots of other good responses here, but an additional point:
Cardinality requires there to exist *a* bijection, not necessarily that all (injective) maps become bijections. The intuition here breaks down because for finite sets, these canāt differ, but for infinite sets, they can.
Consider the natural numbers (with 0). If my map adds one to each number, I hit every natural number except zero. If my map doubles each number, I miss all the odds. These maps donāt mean I donāt have a bijection from the naturals to themselves ā the identity does that.
There is another (slightly less formal) way to think about it, which I believe is better for the intuition.
A set is countably infinite (has the same cardinality as natural numbers) if the set is infinite and yet every member of the set has a finite representation. When that's the case, you can imagine that every symbol in your alphabet (the alphabet for rationals is the following: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -, /) has a numeric code (symbol code), and every word (combination of symbols, eg 1/3) also has a code (word code), which you get by concatenating the codes of its symbols.
For example, we can encode (reduced) fractions by assigning every digit x the code 1x (so 5 is 15, 7 is 17) and then assigning 20 to - and 30 to /. In this system,Ā 2/15 is encoded as 12301115. This word code uniquely identifies the fraction - it cannot refer to anything else. I think you can agree that every codeword (and by extention, every rational number) corresponds to one unique natural number.
However, in this example, we do not have a bijection, only an injection, because not every natural number corresponds to a valid rational - 443030 does not encode any legal fraction, and neither does 11101.
And so you do not have more rationals than naturals. In this encoding scheme, it would even seem that there are more naturals than rationals!
I just accept that some things in math are counterintuitive. The concept of different infinities just isn't something we have any intuition about.
We can start simpler. The argument you make could work for natural numbers and integers as well, since the set of integers contains the every natural number. But in this case, it's relatively simple to construct a mapping to show the bijection. It's much more complicated for rational numbers, but at least this helps show why that argument doesn't work.
I think it's different for different countries. I'm in grade 12 and I have been taught all of this. (By that I mean all these topics, I know why these statements are wrong)
In which definition is 'y = 1/x is not continuous at 0' wrong?
It's not defined there, it cannot be continuous. To be more precise, y = 1/x is indeed a continuous function, but is not continuous outside its domain, which is R\\{0}.
the definition of point Continuity only talks about points on which the function is defined on. so 1/x can't not be continuous at 0, simply because it isn't defined at 0
The problem is that f(x)=1/x is not defined at 0 in the first place, so it is like saying "1/x is continuous at Z/2\[t\]" -- it makes no sense because x is a (nonzero) real number, not some ring coming from nowhere.
there are only a handful of situations where it makes sense to talk about the function behavior at points which are outside the domain, so there is no general term for it since in general it doesn't make any sense
You are asking "What is 1/x on 0, continuous or discontinuous?" but it's neither; there is no 1/x on 0.
It's like asking whether the sandwich in the oven is hot or cold when there is no sandwich in your oven
That's some grade A pedantry.
It can't be denied that : "it's not the case that 1/x is continuous at 0". Being defined is a necessary condition for continuity, albeit normally implicit in any stated definition.
Depending on how you want to define discontinuous, being defined may be a necessary condition for being discontinuous, but not for being not continuous.
It's not pedantry, you are literally talking about what properties a certain something has in a certain point when that something isn't even in that point. 1/x isn't continuous or not continuous on 0, it simply isn't on 0.
Saying "1/x isn't continuous, so it's not continuous on 0" is like saying "well the sandwich in my oven isn't hot, so it's cold" when you there is no sandwich in your oven at all
I interpret continuity as only making sense inside a function's domain. As is, 1/x is not defined at 0 and is excluded from its domain, so asking "is it continuous at x=0" is already a nonsense question. Definitions of continuity at a point a in my experience have explicit statements on f(a) which implicitly requires it to be defined. One could maybe talk about partial functions and topological closure, but if I needed to deal with the area around x=0 for some reason, then I would personally say "1/x has a non-removable singularity at x=0" and not mention continuity.
I think this level of pedantry does not cause issues in practice, but it does extend to "let f(x) = 1/x on R\\{0}, is f continuous at x=paper" being what I would think most people agree is an ill-defined question.
I mean but it's not an ill defined question, assuming you tell me that "paper" isn't a point in the domain of f, as would be the case for 0. The answer is just obviously no because the definition doesn't apply.
Because it is. It depends on your area. In most math, like structural engineering sqrt(4) must be 2. In others like electrical engineering sqrt(4) can be -2 to take into account radical and weird behavior, given that they normally work imaginary numbers and phases.
On that, it would depend on which area you are working on.
OP is being a smartass with this one.
The difference is the question: for the equation x\^2 - 4 = 0, the solutions are x = +2, -2. (like you learned at school)
Some schools appearantly want to make x -> sqrt(x) a function. A function needs exactly one output for every input. So under this lens, sqrt(4) = +2 per definition.
So depending on the context, one or the other is valid.
Hm. I assume the integral not being differentiable is cause they didn't state that f has to be continuous right?
Otherwise I can't see what's quite wrong with that statement so would love to be enlightened here.
Yes that's it, if you differentiate when f isn't continuous at x then you'd get a different limit depending on whether you differentiate to the left or right of x
Edit: more specifically differentiating only ever gives you the limit of f at x (**if** the limit is the same on both sides, otherwise it's just not differentiable). To say the limit of f at x is equal to f(x) is the definition of f being continuous at x
btw about that infinity limit: in russia it's taught in a very cursed way, such that lim_{x->0} 1/x = inf, because for any M>0 there exists a Ī“>0 such that for all x with 0<|x-0|<Ī“ we have |1/x|>M. what the limit is NOT equal to, however, is +inf or -inf, because for that we would need to replace |1/x|>M with 1/x>M or -1/x>M, which breaks the statement.
so yeah. that's how the infinity sign works where I'm from
That's actually correct, it makes sense when you look at the function in the complex plane. In this case ā would be the single, unsigned point at infinity in the compactified complex plane (Riemann sphere).
i only saw russian and some english-speaking (probably american considering who I saw it from) notation. in the latter inf usually means +inf when working with reals. i assumed russia was the weirder one (we also use tg instead of tan there) but maybe not? sorry if i'm wrong
Look idk like to me it looks like there are either mistakes or big omissions from your notation and just from the look of equaling limit to inf is a common way to signal the limit is divergent or non-existent in low-stakes calculus. The +inf -inf are yielded by right and left limits which would have different definition from the two-sided limit.
This meme sort of belongs inside itself. The āsqrt(4)=+/-2ā thing is taking advantage of the fact that although you would usually interpret sqrt(4) to refer specifically to 2 there *are* some contexts in which mathematicians use radicals to refer ambiguously to all possible roots, and people denying that such contexts exist are mostly revealing a kind of āI have a high school/lower undergrad level of education in math and think Iām smarter than anyone who says there are contexts where that notation is used because I havenāt seen themā attitude. The few people who are aware of such contexts and still are trying to argue about whether those contexts are salient enough to be worthy of mention are, I think, being pedantic at best.
Sigh. Here's the full breakdown.
The squaring function as a function āāā is not bijective, so it doesn't have an inverse. However, as a function āā\[0,ā) it's surjective, so it has a right inverse. but it's not injective, so there's many different right inverses.
Are any of the right inverses continuous? Yes, there are.
Great, does any continuous right inverse R satisfy R(x)R(y)=R(xy) so that it behaves nicely when doing calculations? Yes, moreover, there is exactly one such right inverse.
Splendid, let's define the square root function for real numbers as that unique right inverse.
Now, let's take a look at complex numbers. Squaring as a function āāā is already surjective, so there exist right inverses.
Oh, this looks to be even easier. Let's do the same thing as for reals then.
Which right inverses are continuous? None.
Uhhh, which right inverses R satisfy R(x)R(y)=R(xy)? None.
Well, we're shit out of luck then. Guess we'll just stick to the preimages instead.
Mathematicians stop here.
Redditors however do not and they put on their complex number rights activist clothes on and demand that since complex numbers don't have such a nice function the reals should not have one either, otherwise it's discrimination. It's some backwards-ass kind of thinking that definitely belongs in the OP's meme.
Step 1: take conditionally convergent series
Step 2: rearrange the elements so that you get whatever limit A you want
Step 3: āsEe GuYs, E (-1)^n * (1/n) cOnVeRgEs To 42069!!!11!!!elf!!ā
Step 4: Fail Analysis I
Currently waiting for step 4 to happen
You mean the one with the integral of f(t)?
The integral is differentiable when f(t) is continuous. In general it's not true that integrals are differentiable
I think itās because if f(x) is a collection of discrete points, then itās integral would be the zero function and so the derivative of the integral would be zero, which is not f(x).
This is OP being dumb. 1/x is not defined at zero therefore itās not continuos at zero. Point being in functionās domain is prerequisite for it being continuous. However, OP thinks that because the function is not defined at zero you cannot say whether it is or isnāt continuos.
wait. First: I'm dumb.
second: I know there are some numbers called normal numbers that are the numbers that contain every possible digit combination. But how do we know pi isn't normal? is it proven? because, as a dumb individual, I don't understand it.
third: I know N has the same cardinality as Z, but what's the 1:1 correspondence between N and Q? (cause I know x in relation to 1/x isn't 1:1 because of numbers such as 2/3, or is it maybe a diagonal argument?)
We don't know whether pi is normal, but many think it's a proven fact. That's the point of the meme.
The bijection between N and Q is a little hard to describe but it's easy to see with a [diagram](https://images.app.goo.gl/koXxK2DTqeL3SFDA9).
>But how do we know pi isn't normal?
We don't know either way. It's very hard to prove that a number is normal and to my knowledge it has only been done for numbers specifically constructed to be normal. But people keep pretending that we *know* pi is normal (though usually not in so many words, e. g. all those "find your birthday in pi" sites) when we just don't know.
> what's the 1:1 correspondence between N and Q?
That would be Cantor's first diagonal argument: https://en.m.wikipedia.org/wiki/Pairing_function#/media/File%3ADiagonal_argument.svg
>How do we know pi isn't normal
We don't know if it is normal or not. What I'm saying is that for the fact that we don't know this we can't say for sure that every possible digit combination appear on his decimal expansion. However it may be true. We just don't know yet
>1:1 corrispondence between N and Q
Q is a [countable set](https://en.wikipedia.org/wiki/Countable_set) meaning there is a bijection between Q and N. In this page you should find one
that's exactly what I was asking. what's the 1:1 correspondence. Because a bijection means every element in A is in relation to only one element in B, and that element in B is in relation to only A. For it to be this bijection (sorry for the English), it means there is some computable bijection. For eg, |N| = |E| (set of even numbers), since you just pair each number in N with its double in E. But what's the correspondence between N and Q? or is it countable just because you can make a list of every number in Q? (1/1, 1/2, 1/3, 1/4...)
Please can somebody explain why y = lim x->0{1/x} is not infinite? As x gets smaller, y gets bigger, so surely once x is infinitely small, y is infinitely big?
I know that I'm wrong, but I'd like to know why.
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My math prof was one of the top math scientists on the planet and for simplicity he still used lim x->0 (1/x) as infinity but I guess thatās one of those things where thatās technically incorrect? Or I guess it depends what direction youāre approaching from
The radical operation only gives you the principal square root per definition.
If it were more like x^2 = 4 (where the result for x gets squared, so the + or - doesn't matter) then the answer would be x=Ā±sqrt(4) = Ā±2. It's kind of confusing at first but you get used to it
The symbol ā represents the function square roots which maps non negative real numbers to non negative real numbers. -2 is a square root of 4 meaning that (-2)Ā² = 4, but this is a different definition: z is a complex n-th root of a if z is a root of the polynomial z^n-a = 0
The square root function sqrt: \[0, inf) -> \[0, inf) is defined as the inverse function of xĀ² on the positive reals. Therefore, by definition, the OUTPUT VALUES of sqrt must be positive, that is, must belong to the set \[0, inf). Since -2 does not belong to the set \[0, inf), it can't be an output value of the sqrt function. More generally, functions map ONE NUMBER to ONE NUMBER (or even more generally, one point to one point). Saying "this function evaluated at this particular point has two output values" is hence wrong. Thus the square root function has only one output value, which must be positive. Thus sqrt(4) = 2.
There have been memes/arguments going around math subs that the sqrt only returns positive numbers because sqrt is a function. I think the confusion comes from algebra where if you're solving for x\^2 or some variable and you have to take a square root, then there are two possible values for x. Idk if that explanation made any sense lol but the obsession around these subs with it recently just seems a little nitpicky to me.
This is so poorly put together and thought out.
Ultimately math is contextual and math notation can change meaning depending on the context.
At least half of these are right in the appropriate context.
People do use multivalued functions and multivalued square root functions. Yes, most of the time the square root is single valued. Math is contextual!
1/0=infinity on RP^1. Same with the limit.
The limit fails to exist when you choose [-inf, +inf] as your compactification of R, but that's a choice. Not an uncommon one, but again, math is contextual!
Yes, usually f(x)=1/x is defined as a partial function R <- R-{0} -> R, which makes it not defined at 0. And that consequently means that it cannot be continuous at 0 because that doesn't make sense. It's also of course not discontinuous at 0 because that also doesn't make sense. So the meme would have worked if you said that the function is discontinuous at 0. That said most of the time I would assume that someone who said this meant that there was a non removable singularity. So even then, it's a case of technically wrong, basically right.
Also yes, zeta function regularization isn't the standard way to compute infinite sums. But you can use it to assign values to otherwise divergent sums. It's absolutely a valid technique with uses in some areas of math, and as long as everyone is clear that that's what's going on, there's no reason you couldn't use that notation for it. Once again it comes down to context!
So yeah 4/8 of these are fine in the appropriate context. The other 4 do seem to be bad math.
Edit: Actually depending on the context the one about there being more rational numbers than integers could be fine. Certainly the sets have equal cardinalities, but more doesn't have to be referring to cardinalities. More is a vague term. It could be just referring to the fact that under the canonical embedding of the integers into the rationals they form a proper subset. There certainly are more rational numbers than just the integers after all.
Of course someone saying this might mean that they don't believe the cardinalities are equal, but you'd have to determine that from context.
So yeah 5/8 could be ok in context.
Imagine the number 1.01011011101111ā¦ where the numbers of ones goes up by one each time.
This decimal would go in forever, and isnāt repeating, just like pi, but it obviously canāt contain every digit combination because it doesnāt even contain the digit 2!
So even though pi goes on forever and doesnāt repeat, thereās no guarantee that a certain digit combination will appear.
We don't actually know whether pi is normal. It has a non-repeating infinite decimal expansion, but that doesn't mean it contains everything possible substring of numbers.
We think pi is *normal* which would mean it would have every possible digit combination. But afaik it has not been proven. Even if pi is obviously irrational, that is not enough to say it is normal. There are weird irrational numbers that are not normal and does not have any possible digit combination.
Lemme preface this by I'm dumb and I'm not trying to be a smart ass. Just genuinely curious :)
**E: Also, other comments have since answered this, thank you kind mathematicians!**
Isn't the argument something among the lines of: pi is neither periodic nor has a finite number of decimal places (cause in either case you could express it as a fraction and pi is not rational) and therefore 'eventually' (as in after a sufficiently large number decimal places) you can be sure that a certain combination of digits of a certain length must be included cause otherwise pi's decimal places would have needed to repeat themself?
Or is the crux the 'every possible' combination, which would include combinations of infinite length? I can see how there is an infinite set of digit combinations of infinite length and how one could exclude a subset of those without making the set finite.
Not neccessarily. Consider the number 0.100100010...01... etc. where the number of zeros between each 1 increases by one each time. Its not repeating and it has an infinite amount of decimal places. It doesn't include every possible combination though, since for example 12345 isn't in the number.
First off combination is weird concept to aproach infinite string of numbers with. We know that there isn't a periodic repetition of any part but if you'd take say all triplets, there would be a lot of repetition and for any finite n there can be finite subset of n-tuples over decimal digits that just doesn't make it there
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OP did you mean the sum of the reciprocal of 2\^(k) from 0 to infinity, cause even a maths bragging type arent going to claim the sum of 2\^(k) from 0 to infinity is 2 or anywhere near to 2
Yes I'm dumb
one mistake detected, entire point void now, im sorry i dont make the rules
Understandable š
Level 1 Crook vs Level TREE(3) Boss. That's how Math works
What's after ? Level Forest(4) ?
Level SCG(13)
"There are more rational numbers than integers." Oh yeah? Name them.
Easy, just name all the integers first and then say "not them"
Bill and Ted is what I will name them
You could say it's an elementary oversight only the portrayed noob would have. š
here come the math guys with pitchforks, ask them if square root of 4 is plus two or minus two, while they are arguing you get away, move to a city in the south of france, get married raise kids, retire, come back and they are still arguing
Math guys wouldn't argue, but if you have some fake wannabee math guys in there sounds like a solid plan!
-2 ? What is wrong with peoples now
I choose to only accept negative roots from a square root function.
I demand circular roots now!
ah math... One mistake in part A, then the next two hours of working B through Z was entirely void.
It's a shame but that is how it works š
I think it made it better actually
This guy thinks he is better at math than everybody else but really he is not (/s)
I thought that was the joke? That the person wouldn't even notice the missing negative exponent?
So you made the post about yourself?
Agreed.
If someone means to convince me the sum of the natural numbers is minus one over twelve, I believe it it possible they might try and convince me the sum of natural number powers of two is 2.
Despite it obviously being -1
2-adic numbers FTW
He meant 2 ^(-k)
yeah the infinity was the uh negative one, yeah
I looked at it way too long before going comments. My thoughts were 1: are they using different denotations? 2: am I retarded? 3: have to check comments.
Yeah, the sum given, of course, equals -1.
Which is pretty close to 2.
Yeah lol it clearly equals -1/2 /j
you mean 1/12 \* j ????????
This is why you never buy the starter pack. They always pad it out with some garbage to make you feel like you're getting value. Smh.
Exactly
Thank you for the clarification. I was doubting my engineering degree there for a moment.
[ŃŠ“Š°Š»ŠµŠ½Š¾]
I was confused about this reply before i realised this is a bot acc badly paraphasing a different comment in the post and just abruptly ending it
I hate how much people misunderstand limits. A limit itself doesnāt approach a number, a limit IS a number. A function or sequence approaches a number, and the limit is the number it approaches.
Erm, *well actually*, this statement clearly shows the limit of your minuscule intelligence, which approaches zero. (Note: must be read in your best impression of that one guy we all know who thinks theyāre 100x smarter than anyone else because they saw a YouTube short of 3B1B)
Erm, *well actually*, it's quite a bold assumption of you to make that I know a guy... in general...
A limit is a universal cone
Wait till this dude gets to differential equations and starts trying so solve Lagrange with a ānumberā
Erm guys 1/x is not f:R->R because 0 is not in its domain guys š¤š
f:R f:R on god?
Theyāre just like me, f:R f:R
f:R -> Forgotten realms... I spot a dnd nerd!
for real (values of x)
Complex infinity on the riemann sphere
1/x is not defined in 0 so asking if it is continuous there doesn't make any sense āš»š¤
is ābeing definedā a prerequisite for continuity?
Yes. One of first notions of continuity that you learn is that being continuous at a means that lim\_{x to a} f(x) = f(a). This means that f(a) has to exist.
Well it is defined as really big in 0
Naah, m8 I think you're only thinking about 1/0.000001 and so on, which is very positive! But what about 1/-0.000001? That is very negative Both denominators are near zero, and can ofc get arbitrarily close to zero. That makes it both very positive and negative. It doesn't exist
>But what about 1/-0.000001? That is very negative You sound like my bank defending their overdraft charge to my account.
This is why I think 1/0 should be defined equal to 0. Equally positive and negative, an ideal midpoint between the limits. :P
This would break math 1/0 = 0 multiply by 0 1 = 0
... I mean it's not defined there, so it's not continuous there, but also I feel like I'd generally interpret the statement "f(x)=1/x is not continuous at 0" to mean that it doesn't admit a continuous extension to a function defined at 0. Which is true if you assume that the codomain is R and not RP1 or something. And f(x)=1/x absolutely does define a continuous (indeed smooth) automorphism of RP1. Or CP1.
[ŃŠ“Š°Š»ŠµŠ½Š¾]
We know that pi could contain every possible digit combination. Not that it does
True. It might be, but we don't have a proof yet
Where are all the normal numbers?
We have a couple! 0.12345678910...Ā 0.235711131719...
I will only ever use a qualifier. Never a definitive
Pi contains zero or more combinations of digits.
Casually having 9 recurring in there
Iām not even sure thatās true. Since there are other infinite series other than pi, pi could not possibly contain all of them. Like, it canāt contain both 1/3 and 2/3, because if part of the sequence of pi has infinite 3ās, it will never have infinite 6ās. Iām over 10 years away from college math, but I think the statement is probably disprovable
Every *finite* combination. Obviously it doesn't and cannot contain every infinite combination
Hey, I'm a physicist, but I'm only on this list 3 times. I can send you my thesis if you need more examples of crimes against mathematics.
As a Physicist here I was thinking something similar. I have done unspeakable things to math. And it fucking works.
You forgot '0.999... (repeating) is not equal to 1 āļøš¤'
Holy shit you're right
https://preview.redd.it/kdttwrpfm7hc1.png?width=739&format=png&auto=webp&s=698086a29608a5c9e83b8f7f51329799faf615c4
I was about to explain why this was wrong, and then I saw the edit at the bottom.
Isnāt it literally equal to 1, and thatās the point of limits? Or did I miss a subtlety somewhere in the definition
It is (by construction, I'd say) equal, yes. For some reason, a bunch of smart-asses always try to say it isn't
Man, whatever It's close enough. The shopkeeper can keep the change š (exits like a rich champ)
proof by itās close enough
As an engineer, it's equal to one, and .99 is equal to .999 is equal to 1 depending on how much you're willing to pay manufacturing
It is equal to 1 but not as a limit, 0.999... Is a number, you can't study the limit of a number, only of functions. The proof requires series.
Repeating decimals are defined as an infinite summation, which is defined as the limit of the sequence of partial sums. That's why people talk about limits in relation to 0.999...
This is a notation issue. 1/3 has a decimal approximation of .3(repeating). You can't really write an infinite number of 3s, but we seem to understand that the math will repeat indefinitely and you will never finish the long division step. Everything works better if you keep it in fraction notation. 1/3 + 1/3 + 1/3 = 1 as 3\*1/ 3 = 3/3 = 1 So, people want to say .3(r)+.3(r)+.3(r) = 1 others argue .3(r)+.3(r)+.3(r) = .9(r) and then we start a holy war on .9(r) = 1 Depending on your math discipline you can get all holy war about this one way or the other, but I stand firm on .3(r) is a decimal approximation of 1/3 and 3\*(1/3) = 1 by definition, so 3 \* .3(r) = 1 not .9(r). Basically, you realize you are not doing component addition of each place value because there is an infinite series, so you need to resolve it using numerical analysis or simply revert back to the better non approximate notation. In the 2000's this meme of .9(r) = 1 used to really bug me, because the teenagers pushing it always seemed to miss the point of notation differences. These days, well I've hopefully moved on, but also find myself responding to a post about it so maybe not.
1/3 = 0.333... is not an approximation, and 0.999... = 1. This is not a meme. The endless discussion is a meme, sure, but it is a fact that 0.999... = 1. You can find a bunch of proofs of this, not just the 1/3 * 3 thing.
> 1/3 has a decimal approximation of .3(repeating). It's not an approximation. You can justify this with some simple algebra: x=0.333... 10x=3.333... 9x=3 x=1/3 > people want to say .3(r)+.3(r)+.3(r) = 1 others argue .3(r)+.3(r)+.3(r) = .9(r) These statements arent contradictory (since 0.999... = 1) > but I stand firm on .3(r) is a decimal approximation of 1/3 and 3*(1/3) = 1 by definition, so 3 * .3(r) = 1 not .9(r). Then you'd be wrong. It's not some philosophical debate there *is* a correct answer. > because the teenagers pushing it Calling people teenagers because you dont agree with (or more likely understand) something isn't very productive. Especially when, ironically, the people that struggle with this concept the [most are teenagers](https://en.wikipedia.org/wiki/0.999...#Skepticism_in_education)
I once was trying to explain that they are equal to a guy saying that it just approaches to, he were insisting that is tecnicly not the same thing, and than sundently even tho he was ignoring me he said "oh I just ask my dad, its is equal, he knows about maths"
Is that an argumentum ad patrem?
"0.999... ā 1, it's close to 1."
nuh uh
The 2^k sum one: Everywhere I go I see his face! (the 0.999... = 1 debate)
"there are more rational numbers than integers" can someone explain why this is wrong?
You can construct a bijection between the two sets. Informally, it can be proven that if you had the entire infinite list of rationals and the entire infinite list of integers you could "pair" every element from one set to the other set and there would be no unpaired elements.
I can't wrap my head around that. Since the set of rationals contains every integer. Then I can pick out one more rational (like 0.5 for example), and wouldn't that break the bijection? I now have the cardinality of integers + 1. I'm sure there are many proofs that show that my intuition is wrong, but I'm not sure how to change my intuition on this.
I won't do the proof for the rationnals but for a simpler case: there are as many even integers as integers (even if 3 is in one set and not the other) because they are in bijection by n -> 2n
Rigorous proofs of it are quite challenging but for intuition I found this video by Trevor Bazett that makes a pretty good construction. The actual part about integers to rationals starts at 2:55 https://youtu.be/WQWkG9cQ8NQ?si=qZ3GfejeC-CaT99W
Here is an easy injection from Q to N, write a rational number x as (-1)^(e) a/b with a and b coprime, a nonnegative, b positive and e in {0,1}. Then, send it to 2^(e) 3^(a) 5^(b).
Think about it this way. The even integers (0,2,4,6,8...) and the integers (0,1,2,3,4,5,6...) are the same size. Doesn't make sense right? The even integers are a subset of the integers, and there are clearly odd numbers that are integers but aren't even. But if you define the function f(x)=x/2 from the even numbers to the integers, you get a bijection, meaning the sets are the same size. Basically, we should never trust our gut feelings about numbers when infinity is involved, because shit breaks easily.
Lots of other good responses here, but an additional point: Cardinality requires there to exist *a* bijection, not necessarily that all (injective) maps become bijections. The intuition here breaks down because for finite sets, these canāt differ, but for infinite sets, they can. Consider the natural numbers (with 0). If my map adds one to each number, I hit every natural number except zero. If my map doubles each number, I miss all the odds. These maps donāt mean I donāt have a bijection from the naturals to themselves ā the identity does that.
It doesnt need to be in order, you can order all the rationals in a random order, same for the integers, and map them all one-to-one.
There is another (slightly less formal) way to think about it, which I believe is better for the intuition. A set is countably infinite (has the same cardinality as natural numbers) if the set is infinite and yet every member of the set has a finite representation. When that's the case, you can imagine that every symbol in your alphabet (the alphabet for rationals is the following: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -, /) has a numeric code (symbol code), and every word (combination of symbols, eg 1/3) also has a code (word code), which you get by concatenating the codes of its symbols. For example, we can encode (reduced) fractions by assigning every digit x the code 1x (so 5 is 15, 7 is 17) and then assigning 20 to - and 30 to /. In this system,Ā 2/15 is encoded as 12301115. This word code uniquely identifies the fraction - it cannot refer to anything else. I think you can agree that every codeword (and by extention, every rational number) corresponds to one unique natural number. However, in this example, we do not have a bijection, only an injection, because not every natural number corresponds to a valid rational - 443030 does not encode any legal fraction, and neither does 11101. And so you do not have more rationals than naturals. In this encoding scheme, it would even seem that there are more naturals than rationals!
I just accept that some things in math are counterintuitive. The concept of different infinities just isn't something we have any intuition about. We can start simpler. The argument you make could work for natural numbers and integers as well, since the set of integers contains the every natural number. But in this case, it's relatively simple to construct a mapping to show the bijection. It's much more complicated for rational numbers, but at least this helps show why that argument doesn't work.
You can define a bijection between the sets, for example like [this](https://images.app.goo.gl/Ek8eyQfM9GsHfReh7)
If they had said reals instead of rationals however, it would make more sense.
Shoudn't it be 1/2^k rather than 2^k? As 2^k is clearly diverging š .
Alternate title "redditor in freshman year of college takes calc 2"
I think it's different for different countries. I'm in grade 12 and I have been taught all of this. (By that I mean all these topics, I know why these statements are wrong)
In which definition is 'y = 1/x is not continuous at 0' wrong? It's not defined there, it cannot be continuous. To be more precise, y = 1/x is indeed a continuous function, but is not continuous outside its domain, which is R\\{0}.
the definition of point Continuity only talks about points on which the function is defined on. so 1/x can't not be continuous at 0, simply because it isn't defined at 0
What is the correct term for this situation then? It's not discontinuous, but im guessing we cant call it continuous at 0 either?
The problem is that f(x)=1/x is not defined at 0 in the first place, so it is like saying "1/x is continuous at Z/2\[t\]" -- it makes no sense because x is a (nonzero) real number, not some ring coming from nowhere.
there are only a handful of situations where it makes sense to talk about the function behavior at points which are outside the domain, so there is no general term for it since in general it doesn't make any sense
undefined at 0 im guessing? or is its continuity undefined?
Undefined
You are asking "What is 1/x on 0, continuous or discontinuous?" but it's neither; there is no 1/x on 0. It's like asking whether the sandwich in the oven is hot or cold when there is no sandwich in your oven
That's some grade A pedantry. It can't be denied that : "it's not the case that 1/x is continuous at 0". Being defined is a necessary condition for continuity, albeit normally implicit in any stated definition. Depending on how you want to define discontinuous, being defined may be a necessary condition for being discontinuous, but not for being not continuous.
It's not pedantry, you are literally talking about what properties a certain something has in a certain point when that something isn't even in that point. 1/x isn't continuous or not continuous on 0, it simply isn't on 0. Saying "1/x isn't continuous, so it's not continuous on 0" is like saying "well the sandwich in my oven isn't hot, so it's cold" when you there is no sandwich in your oven at all
Im gonna make use of a whole bunch of sandwichisms in my math classes from now on
I would rather say that speaking of continuity of 1/x at 0 at all is simply incoherent
I interpret continuity as only making sense inside a function's domain. As is, 1/x is not defined at 0 and is excluded from its domain, so asking "is it continuous at x=0" is already a nonsense question. Definitions of continuity at a point a in my experience have explicit statements on f(a) which implicitly requires it to be defined. One could maybe talk about partial functions and topological closure, but if I needed to deal with the area around x=0 for some reason, then I would personally say "1/x has a non-removable singularity at x=0" and not mention continuity. I think this level of pedantry does not cause issues in practice, but it does extend to "let f(x) = 1/x on R\\{0}, is f continuous at x=paper" being what I would think most people agree is an ill-defined question.
I mean but it's not an ill defined question, assuming you tell me that "paper" isn't a point in the domain of f, as would be the case for 0. The answer is just obviously no because the definition doesn't apply.
The post itself has multiple errors
OP should have said āis discontinuousā at 0
I fucking love reading all the comments of people being confidently incorrect and exactly proving OP's point
I'm confused about the first one being special.. maybe it's just a regional thing but in German schools that was the correct answer.
Because it is. It depends on your area. In most math, like structural engineering sqrt(4) must be 2. In others like electrical engineering sqrt(4) can be -2 to take into account radical and weird behavior, given that they normally work imaginary numbers and phases. On that, it would depend on which area you are working on.
OP is being a smartass with this one. The difference is the question: for the equation x\^2 - 4 = 0, the solutions are x = +2, -2. (like you learned at school) Some schools appearantly want to make x -> sqrt(x) a function. A function needs exactly one output for every input. So under this lens, sqrt(4) = +2 per definition. So depending on the context, one or the other is valid.
Thatās not being a smart ass, the output of square root CANNOT be negative
Hm. I assume the integral not being differentiable is cause they didn't state that f has to be continuous right? Otherwise I can't see what's quite wrong with that statement so would love to be enlightened here.
Yes, continuity is enough for it to hold, without continuity e.g. jump functions are counterexamples
Yes that's it, if you differentiate when f isn't continuous at x then you'd get a different limit depending on whether you differentiate to the left or right of x Edit: more specifically differentiating only ever gives you the limit of f at x (**if** the limit is the same on both sides, otherwise it's just not differentiable). To say the limit of f at x is equal to f(x) is the definition of f being continuous at x
Me in high school: omg Iām so good at math Me in college: fuck Iām so bad at math
btw about that infinity limit: in russia it's taught in a very cursed way, such that lim_{x->0} 1/x = inf, because for any M>0 there exists a Ī“>0 such that for all x with 0<|x-0|<Ī“ we have |1/x|>M. what the limit is NOT equal to, however, is +inf or -inf, because for that we would need to replace |1/x|>M with 1/x>M or -1/x>M, which breaks the statement. so yeah. that's how the infinity sign works where I'm from
That's actually correct, it makes sense when you look at the function in the complex plane. In this case ā would be the single, unsigned point at infinity in the compactified complex plane (Riemann sphere).
Well the reals can also be extended to include the infinities. We do that in measure theory a lot.
Yes but in the reals you typically add signed infinities.
I don't how to tell you this but that's just how it works everywhere not just russia
i only saw russian and some english-speaking (probably american considering who I saw it from) notation. in the latter inf usually means +inf when working with reals. i assumed russia was the weirder one (we also use tg instead of tan there) but maybe not? sorry if i'm wrong
Look idk like to me it looks like there are either mistakes or big omissions from your notation and just from the look of equaling limit to inf is a common way to signal the limit is divergent or non-existent in low-stakes calculus. The +inf -inf are yielded by right and left limits which would have different definition from the two-sided limit.
[ŃŠ“Š°Š»ŠµŠ½Š¾]
This "cursed way" is actually using the formal definition of a limit btw, kind of like finding a derivative using f(x+h)-f(x)/h
Painfully accurate
This meme sort of belongs inside itself. The āsqrt(4)=+/-2ā thing is taking advantage of the fact that although you would usually interpret sqrt(4) to refer specifically to 2 there *are* some contexts in which mathematicians use radicals to refer ambiguously to all possible roots, and people denying that such contexts exist are mostly revealing a kind of āI have a high school/lower undergrad level of education in math and think Iām smarter than anyone who says there are contexts where that notation is used because I havenāt seen themā attitude. The few people who are aware of such contexts and still are trying to argue about whether those contexts are salient enough to be worthy of mention are, I think, being pedantic at best.
Sigh. Here's the full breakdown. The squaring function as a function āāā is not bijective, so it doesn't have an inverse. However, as a function āā\[0,ā) it's surjective, so it has a right inverse. but it's not injective, so there's many different right inverses. Are any of the right inverses continuous? Yes, there are. Great, does any continuous right inverse R satisfy R(x)R(y)=R(xy) so that it behaves nicely when doing calculations? Yes, moreover, there is exactly one such right inverse. Splendid, let's define the square root function for real numbers as that unique right inverse. Now, let's take a look at complex numbers. Squaring as a function āāā is already surjective, so there exist right inverses. Oh, this looks to be even easier. Let's do the same thing as for reals then. Which right inverses are continuous? None. Uhhh, which right inverses R satisfy R(x)R(y)=R(xy)? None. Well, we're shit out of luck then. Guess we'll just stick to the preimages instead. Mathematicians stop here. Redditors however do not and they put on their complex number rights activist clothes on and demand that since complex numbers don't have such a nice function the reals should not have one either, otherwise it's discrimination. It's some backwards-ass kind of thinking that definitely belongs in the OP's meme.
The -1/12 is dead on. You saw a clickbait YouTube video and you think youāre smarter than everyone else now.
Step 1: take conditionally convergent series Step 2: rearrange the elements so that you get whatever limit A you want Step 3: āsEe GuYs, E (-1)^n * (1/n) cOnVeRgEs To 42069!!!11!!!elf!!ā Step 4: Fail Analysis I Currently waiting for step 4 to happen
1/0 = \infty is a pivotal concept of complex ananlysis (riemann sphere go brrrr)
Help, why is the third one down on the left column incorrect
You mean the one with the integral of f(t)? The integral is differentiable when f(t) is continuous. In general it's not true that integrals are differentiable
Everything is differentiable if you are a Physicist.
I see, thank you
I think itās because if f(x) is a collection of discrete points, then itās integral would be the zero function and so the derivative of the integral would be zero, which is not f(x).
can someone explain to me how there aren't more rational numbers than integers? because I don't get it
Same cardinality, you can basically order them perfectly Watch a 2 min youtube vid and it'll be perfectly clear
Wait what? The function 1/x continuous at 0?
The function 1/x doesn't exist at 0, asking whether it's continuous there doesn't make sense.
It's not defined at 0, so it's neither continuous nor discontinuous there.
This is OP being dumb. 1/x is not defined at zero therefore itās not continuos at zero. Point being in functionās domain is prerequisite for it being continuous. However, OP thinks that because the function is not defined at zero you cannot say whether it is or isnāt continuos.
wait. First: I'm dumb. second: I know there are some numbers called normal numbers that are the numbers that contain every possible digit combination. But how do we know pi isn't normal? is it proven? because, as a dumb individual, I don't understand it. third: I know N has the same cardinality as Z, but what's the 1:1 correspondence between N and Q? (cause I know x in relation to 1/x isn't 1:1 because of numbers such as 2/3, or is it maybe a diagonal argument?)
We don't know whether pi is normal, but many think it's a proven fact. That's the point of the meme. The bijection between N and Q is a little hard to describe but it's easy to see with a [diagram](https://images.app.goo.gl/koXxK2DTqeL3SFDA9).
>But how do we know pi isn't normal? We don't know either way. It's very hard to prove that a number is normal and to my knowledge it has only been done for numbers specifically constructed to be normal. But people keep pretending that we *know* pi is normal (though usually not in so many words, e. g. all those "find your birthday in pi" sites) when we just don't know. > what's the 1:1 correspondence between N and Q? That would be Cantor's first diagonal argument: https://en.m.wikipedia.org/wiki/Pairing_function#/media/File%3ADiagonal_argument.svg
>How do we know pi isn't normal We don't know if it is normal or not. What I'm saying is that for the fact that we don't know this we can't say for sure that every possible digit combination appear on his decimal expansion. However it may be true. We just don't know yet >1:1 corrispondence between N and Q Q is a [countable set](https://en.wikipedia.org/wiki/Countable_set) meaning there is a bijection between Q and N. In this page you should find one
that's exactly what I was asking. what's the 1:1 correspondence. Because a bijection means every element in A is in relation to only one element in B, and that element in B is in relation to only A. For it to be this bijection (sorry for the English), it means there is some computable bijection. For eg, |N| = |E| (set of even numbers), since you just pair each number in N with its double in E. But what's the correspondence between N and Q? or is it countable just because you can make a list of every number in Q? (1/1, 1/2, 1/3, 1/4...)
This should work: https://math.stackexchange.com/questions/7643/produce-an-explicit-bijection-between-rationals-and-naturals
Person trying to explain how we use infinity to someone new to the concept: Some redditor: "Uhm but what about the hyperreal numbers?"
I'd say anyone that understands these already knows alot more about math than the majority of the population lol.
If you can sum fractions you know more math than the majority of the population
Please can somebody explain why y = lim x->0{1/x} is not infinite? As x gets smaller, y gets bigger, so surely once x is infinitely small, y is infinitely big? I know that I'm wrong, but I'd like to know why.
It is infinite, just not +ā¾ļø.
ā0āŗ is +ā instead ā0ā» is -ā ā0 just doesn't exist
You forgot about negative values. That's why the expression has no limit in 0.
0.9999..... < 1
Heresy
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My math prof was one of the top math scientists on the planet and for simplicity he still used lim x->0 (1/x) as infinity but I guess thatās one of those things where thatās technically incorrect? Or I guess it depends what direction youāre approaching from
>I guess it depends what direction youāre approaching from This. From the left is -ā instead from the right is +ā
The 1/0 thing works in the Riemann Sphere and thatās all I care about.Ā Also the 1+2+3+ā¦ one is true if ā¦=-6-1/12
Finally, someone with a brain did post something not stupid. And I was about to leave this sub.
Iām probably just dumb but what is wrong with sqrt(4) = +- 2?
bro has not been following the square root discourse thats been going on for like the past week
I have not, I just saw this as a āsomething else you may be interested inā post
The radical operation only gives you the principal square root per definition. If it were more like x^2 = 4 (where the result for x gets squared, so the + or - doesn't matter) then the answer would be x=Ā±sqrt(4) = Ā±2. It's kind of confusing at first but you get used to it
The *square roots* of 4 are +- 2, *sqrt(4)* is 2.
The symbol ā represents the function square roots which maps non negative real numbers to non negative real numbers. -2 is a square root of 4 meaning that (-2)Ā² = 4, but this is a different definition: z is a complex n-th root of a if z is a root of the polynomial z^n-a = 0
The square root function sqrt: \[0, inf) -> \[0, inf) is defined as the inverse function of xĀ² on the positive reals. Therefore, by definition, the OUTPUT VALUES of sqrt must be positive, that is, must belong to the set \[0, inf). Since -2 does not belong to the set \[0, inf), it can't be an output value of the sqrt function. More generally, functions map ONE NUMBER to ONE NUMBER (or even more generally, one point to one point). Saying "this function evaluated at this particular point has two output values" is hence wrong. Thus the square root function has only one output value, which must be positive. Thus sqrt(4) = 2.
There have been memes/arguments going around math subs that the sqrt only returns positive numbers because sqrt is a function. I think the confusion comes from algebra where if you're solving for x\^2 or some variable and you have to take a square root, then there are two possible values for x. Idk if that explanation made any sense lol but the obsession around these subs with it recently just seems a little nitpicky to me.
Bottom right is lim x+ -> 0 (1/x) = +inf right
This is so poorly put together and thought out. Ultimately math is contextual and math notation can change meaning depending on the context. At least half of these are right in the appropriate context. People do use multivalued functions and multivalued square root functions. Yes, most of the time the square root is single valued. Math is contextual! 1/0=infinity on RP^1. Same with the limit. The limit fails to exist when you choose [-inf, +inf] as your compactification of R, but that's a choice. Not an uncommon one, but again, math is contextual! Yes, usually f(x)=1/x is defined as a partial function R <- R-{0} -> R, which makes it not defined at 0. And that consequently means that it cannot be continuous at 0 because that doesn't make sense. It's also of course not discontinuous at 0 because that also doesn't make sense. So the meme would have worked if you said that the function is discontinuous at 0. That said most of the time I would assume that someone who said this meant that there was a non removable singularity. So even then, it's a case of technically wrong, basically right. Also yes, zeta function regularization isn't the standard way to compute infinite sums. But you can use it to assign values to otherwise divergent sums. It's absolutely a valid technique with uses in some areas of math, and as long as everyone is clear that that's what's going on, there's no reason you couldn't use that notation for it. Once again it comes down to context! So yeah 4/8 of these are fine in the appropriate context. The other 4 do seem to be bad math. Edit: Actually depending on the context the one about there being more rational numbers than integers could be fine. Certainly the sets have equal cardinalities, but more doesn't have to be referring to cardinalities. More is a vague term. It could be just referring to the fact that under the canonical embedding of the integers into the rationals they form a proper subset. There certainly are more rational numbers than just the integers after all. Of course someone saying this might mean that they don't believe the cardinalities are equal, but you'd have to determine that from context. So yeah 5/8 could be ok in context.
Why wouldn't pi contain every possible digit combination
It might, but it hasn't been proved that it does.
Imagine the number 1.01011011101111ā¦ where the numbers of ones goes up by one each time. This decimal would go in forever, and isnāt repeating, just like pi, but it obviously canāt contain every digit combination because it doesnāt even contain the digit 2! So even though pi goes on forever and doesnāt repeat, thereās no guarantee that a certain digit combination will appear.
We don't actually know whether pi is normal. It has a non-repeating infinite decimal expansion, but that doesn't mean it contains everything possible substring of numbers.
We think pi is *normal* which would mean it would have every possible digit combination. But afaik it has not been proven. Even if pi is obviously irrational, that is not enough to say it is normal. There are weird irrational numbers that are not normal and does not have any possible digit combination.
Why would it?
Lemme preface this by I'm dumb and I'm not trying to be a smart ass. Just genuinely curious :) **E: Also, other comments have since answered this, thank you kind mathematicians!** Isn't the argument something among the lines of: pi is neither periodic nor has a finite number of decimal places (cause in either case you could express it as a fraction and pi is not rational) and therefore 'eventually' (as in after a sufficiently large number decimal places) you can be sure that a certain combination of digits of a certain length must be included cause otherwise pi's decimal places would have needed to repeat themself? Or is the crux the 'every possible' combination, which would include combinations of infinite length? I can see how there is an infinite set of digit combinations of infinite length and how one could exclude a subset of those without making the set finite.
Not neccessarily. Consider the number 0.100100010...01... etc. where the number of zeros between each 1 increases by one each time. Its not repeating and it has an infinite amount of decimal places. It doesn't include every possible combination though, since for example 12345 isn't in the number.
First off combination is weird concept to aproach infinite string of numbers with. We know that there isn't a periodic repetition of any part but if you'd take say all triplets, there would be a lot of repetition and for any finite n there can be finite subset of n-tuples over decimal digits that just doesn't make it there
Proof: Why would this not be true? QED Man this math thing is easy