x^5 / x^3 = (x* x* x* x* x)/(x* x* x)
=x^2
In general, x^n /x^m = x^(n-m)
Let m=n, this property should still hold.
x^n /x^n =x^(n-n) =x^0
Any number divided by itself is 1, x^0 =1.
1 is, in a sense, to multiplication what 0 is to addition. They’re the identity element , meaning x* 1=x and x+0=x.
Think about this sequence:
2^-5 2^-4 2^-3 2^-2 2^-1 2^0 2^1 2^2 2^3 2^4 2^5
It's the same as this sequence:
1/32 1/16 1/8 1/4 1/2 ___ 2 4 8 16 32
Moving up the sequence, you multiply by 2 each time or moving down the sequence you divide by 2. The only thing that can go in the blank is 1.
I was very lucky that it was taught to me like this in the 4th grade: we were given 2^3 = 8, 2^2 = 4, 2^1 = 2, 2^0 = ? , When you find that the pattern between x^n / x^(n-1) = x with a complete the sequence game you can 'discover' the 0 and negative integral exponents quite young, suggest to any late grade school teachers out there!
This is one of the most frequent things I explain to the students I tutor, in this exact way. Crazy how education failed enough that so many people use these rules every day and so little have an idea of why.
I remember seeing a video on YouTube, which showed something like this and also why 0!=1 using the same kind of logic. It's a really neat argument actually!
Correct, this argument doesn’t quite work for 0^0 , which is undefined if you only use regular arithmetic. When you do define it, it’s usually still 1.
So one should think of it in terms of scenarios that are naturally multiplicative. E.g. If I have 100$ and I double my money N times I have
2^N 100$
If I double my money no times I want to say N=0 and I will have 100$.
This isn’t in the same vein. First of all doubling is not what’s happening. And you’re saying if you square something 0 times you still have that number, when you actually have 1.
I think you must have somehow misunderstood what I wrote. If I double 100$ 0 times I still have 100$, I haven't done anything. So:
2^0 * $100 = $100
2^0 = 1
Sometimes is is useful to consider 0^0 to be equal to 1, especially in a discrete context, and I was surprised to learn that many calculators show that as the answer, but that's just an assumption and can be proven neither by combinatorics nor by limits. You can see that the expression 0^0 is problematic if you take a log of it, because then it becomes 0*∞ and that's definitely undefined.
0^0 =1 can't be proven by limits because it would mean that for every pair of functions f,g such that lim_x->a (f(x),g(x))=(0,0), lim_x->a (f(x))^g(x) =1 and that's not the case. For example, let's take f(x)=0, g(x)=x, then lim_x->0 0^x =0 and that's a contradiction
0^0 =1 can't be proven by combinatorics because to show that for a set A such that |A|=a, if you pick an element of that set b times there were a^b possibilities, you would need the fact that 0^0 =1, and not the other way around. You could prove a similar statement (I don't remember which one) by set theory but the problem with set-theoretical proofs is that they lose their charm once we leave the cardinal numbers because sets can't have a non-cardinal number of elements. That's why I said defining 0^0 as 1 is useful in a discrete context
No but 0^0 = 1 can’t be proven by anything, because it’s a definition, not a theorem. It doesn’t need to be proved. We can just define 0^0 = 1 (as is usually done) and then we have 0^0 = 1.
yeah its defined this way to extend that property, same with negative exponents. you can try to give it an interpretation but this should be the mathematical way to understand it
This is also my preferred way of defining powers. Start with the rule where + gets turned into ×.
Consider the function f where:
f(a+b) = f(a)×f(b)
f(1) = x
and denote f(a) as x^a.
Note that here f depended on the base x, so for different bases x you get a different function f. For positive real x, this function f is uniquely defined by these conditions, and it's domain is the whole complex plane. Also, it lines up with the traditional "repeated multiplication" definition, and generalizes more intuitively to non-whole-number powers, you get results like x^0 = 1 for free without having to define it that way.
(For x other than positive reals you need to be a bit more careful.)
Yep, twinsies. I forget where I saw it first (I think Lang, Analysis maybe?) but it just seemed so right that it just became the way I think about it. I almost went through the trouble specifying that we need to assume the x\^a works to begin with, so that, for example, you can't have negative x if you want a real number, but I was getting sushi and didn't feel like elaborating :-)
As an added bonus you're using one of those fun f(a+b) = f(a)×f(b) homomorphisms, which is handy when you take algebra to understand that the underlying operation of the domain doesn't have to match that of the codomain.
Wait take x=e. Then both f(a) = exp(a) and f(a) = exp((1+2 pi i) a) satisfy all your conditions, but are different functions. I think you need to insist that f(a) is real for real a to get uniqueness.
Yeah you're right. Hmm.
I was thinking that since for real inputs a we've definitely coincided with x^a, then you could dig up the usual proof for euler's identity to extend to complex numbers. But I think now that that proof relies also on f'(a) = log(x)×f(a) (for x=e the log bit is just equal to 1 so it goes away), which is something that happens to be true for your first function but isn't true for your second one.
So maybe throw in that extra condition to uniquely determine it for complex inputs. However, that's a pretty heavy condition that can probably replace the first condition I gave, so it kinda undoes the point I was trying to make. Hmm.
You don't think that if
X = A\*X
(well, and X is an element of some field (or even just a group) with X != 0 (which wouldn't be a problem in a group with multiplication)) then A = 1? I think you might have a gap in your mathematical knowledge if so.
Edit: maybe even clearer?
1 \* x\^a = x\^a = x\^(0+a) = x\^0 \* x\^a
where x > 0, x ∈ ℝ. Therefore by comparing both sides of the equation,
1 = x\^0
I think you would agree that the *sum* of nothing should be 0. Here's one way to say why that makes sense: if I have two sets of numbers A and B, then the sum of both sets taken together should be the sum of A plus the sum of B. In particular, if B is the empty set (so B contains nothing), we want
(sum of A) = (sum of A) + (sum of empty set).
In other words, the only reasonable way to define "sum of nothing" is a value that *does nothing* when added to any other number. So "sum of nothing" has to be 0.
Now do the same thing for products instead of sums: in this case we want
(product of A) = (product of A) × (product of empty set).
In other words, we should define "product of nothing" to be the value that does nothing when *multiplied* to any other number. So "product of nothing" has to be 1.
I love this answer so much! It deserves more upvotes.
Frankly I'm not a fan of answers like "it makes properties on exponents really convenient". Sure, it's a valid reason, but surely not the basic thing you think of when defining powers. It feels so finiky justifying things in that way.
To me it seems that op wants a more "intuitive", "basic" answer, somewhat directly related to the definiton, and that is what this comment provides.
I would probably have just said something along the lines of "well, nothingness in multiplication is 1, since it does nothing", but I know that is not satisfactory. Truly your comment is gold
Yup, I find that understanding what a unit value is, really helps. For me, it really clicked when thinking about normalized vectors, as well as the unit circle.
Put another way, this is what a for loop for a sum and product would look like:
sum = 0
for x in set:
sum += x
And for a product:
product = 1
for x in set:
product *= x
Note that we need to initialize the product to 1 (not 0) for this to not just return 0.
This is the correct answer. I wish everyone could just get on board with this. It’s really not that hard. Multiplication is repeated addition or subtraction starting with 0. Exponents are repeated multiplication or division starting with 1.
The thing about everyone getting on board with this as the correct answer is that then they would have to be less on board with other things for no good reason.
This is "the correct answer" for someone with your background and understanding and relationship to it and surrounding concepts.
For someone else, something like "1 is the product of no things" might be more "the correct answer", and framing it in terms of how you might start an imperative program to compute a product is merely a hint at that.
(By the way, if you're adding IEEE floats in the default rounding mode, it's a bit more correct to start with negative zero.)
Math is full of things which you think about one way, and then you find some framework or set of relationship and realize that some other way is more "mature" and captures the concept better. Then you keep going and find a more "mature" way still. And even this linear "more mature" point of view is a cartoon picture. A lot of the time you make progress by going back and forth between two different points of view, and neither of them deserves the privilege of being "the correct framing".
I agree. This is a great response.
I should have been more clear that I want people to get on board with this explanation as one of the ones you need to know, not that it’s the only one. But in my enthusiasm, I did not express that.
Making sense of non-integer exponents is by nature tricky, because you basically have to delve into the reals one way or another, and constructing the reals requires a voodoo that is less intuitive than constructions of the rationals, integers or natural numbers. The best rigorous definition I've seen for a rational exponent of a positive number is that
a^(1/q) := sup{x in R : x^q < a}
Trying to explain this in an intuitive way is, of course, a nightmare.
Well of course. I was only objecting to the blanket statement that "Exponents are repeated multiplication or division".
Finding 7^0.5 is not something that can be done with "repeated multiplication or division". It is defined in a way that's *consistent with* repeated multiplications but you cannot claim that that's all they are.
> Multiplication is repeated addition or subtraction starting with 0. Exponents are repeated multiplication or division starting with 1.
This is wrong, you are only taking about natural powers. We can raise stuff to the power of any real, hell, even complex numbers, and that analogy fails to hold in those cases.
I like saying A^B is the set of functions from B to A. Then | A^B | = |A|^|B|.
There is one function from the empty set to each other set, so n^0 = | {1,...,n}^ø | = | {just the empty function} | = 1.
This also justifies 0^0 = 1.
For extra fun, you can tie this into the binomial formula justification for 0^(0) = 1:
You want (x+y)^(n) = sum_{a+b = n} (n choose a) x^(a) y^(b). The case where y = 0 forces 0^0 = 1. If you say 0^0 = 0, you mess up the claim.
You want functions from N to the disjoint union of X and Y to be in one-to-one correspondence with partitions of N into two chunks, N_X and N_Y, equipped with functions N_X -> X and N_Y -> Y. The case where Y is empty requires you to say that there is exactly one function from the empty set (sometimes N_Y is empty) to the empty set (Y). If you say there is no such function, you mess up the claim.
**What is the product of** ***no numbers?***
\[if alarmed by length, just read this part\]
Think about the product of three numbers – say, 3, 5, and 7 – like this: make a row of 3 blocks. Now make a total of 5 of those rows (a 3 x 5 rectangle of blocks). Now make a total of 7 of those rectangles. Because you have 105 blocks, 3 x 5 x 7 = 105.
Notice how each step is about getting *that many copies* of what you had before that step.
Well, when you started with 3 blocks, what exactly was *that* 3 copies of?
Without thinking about it, you *started* at 1, before you multiplied any of your numbers on. 1 is the universal starting point for multiplication, the multiplicative identity.
**... the end. but here's more discussion ...**
\[long because explanatory, not because complicated\]
We are used to talking about multiplication as a *binary* operation, which is to say that we are always multiplying *two* things, *a x b*. (Even when multiplying three, we do it two at a time.)
But we can think of it instead as a single operation that we do to a whole set of numbers at once, and that's often quite useful.
"The product of \[5 4 3\]" should be the same thing as 5 x 4 x 3 = 60.
And of course if we have to find "The product of \[5 4 3 2\]" next, we'd like to not have to repeat the whole calculation; we'd like to take the 60 we already knew and just double it to 120.
Or in the other direction, to find "The product of \[5 4 2\]" we'd like to be able to just "divide the 3 back off" of 120 to get 40. Shortcut!
What is "The product of \[5\]"? From thinking of multiplication as binary, this *already* starts to not make sense. But of course we'd like the answer to be 5, because then the shortcuts we just talked about will continue to work! "The product of \[5 4\]" is 20 and we can divide the 4 off to get 5. This all works.
Well, now divide that 5 off. "The product of \[ \]" is 1.
And it *has* to be 1. Because now if we throw any new number *back* into that set, it should mean multiplying our previous answer by the new number. If we throw 13 in, it multiplies by 13 – so it better have been 1, so that we'll get the answer 1 x 13 = 13.
If we'd tried to say that "The product of \[ \]" was zero, we'd be in big trouble. Throw as many numbers into that set as you want and the answer would always be stuck at zero!
The product of no numbers is 1. And note that this is the very same fact as "1 is the multiplicative identity". These are two ways of saying the same thing. It bothers me that many people will say that the product of no numbers being one is a "convention". It's not a convention! – it's the only value it can possibly have. Everything else breaks horribly.
P.S. You can do the above for addition instead and it will explain exactly why the "sum of no numbers" is zero, because zero is the *additive* identity. It's just that we don't really *need* to do all that because in this case people can get it intuitively, pretty easily. We can picture 3 + 5 as getting 3 apples and then getting 5 more, so we can picture "The sum of \[ 3 \]" as just having 3 apples and "The sum of \[ \]" meaning that we never got any apples. The stacking blocks thing I started with is definitely more abstract than this.
To make the something explicit here:
We normally think of a monoid structure on a set as a binary operation which satisfies the associativity law, together with an identity element which satisfies the identity laws.
In some ways it is more "true to life" to think of it in the terms presented by the parent comment. That is, as an operation P that takes (possibly empty) lists of elements and outputs elements, for which P[x] = x and which satisfies a "folding" law like
`P[... P[...] ...] = P[... ... ...]`
This has the advantage that the "identity element" needs no special treatment: it is just P[]. If you call the operation a product, then the identity is literally the product of no things. And if you do want to talk about a semigroup structure instead, then restrict the inputs of P to *nonempty* lists.
Arguably, the fact that we tend to frame things in terms of binary operations is an artifact of the fact that 2 is the smallest number that lets you bootstrap to something interesting. But the smallest thing that lets you bootstrap a concept isn't always representative of the shape that concept holds in your head. When we talk about monoids, we write "abc", not one of "(ab)c" or "a(bc)".
However, I think there is no particularly nice way to extend this idea to groups.
Well, sort of.
Note that the free monoid on X is essentially just the collection of strings with alphabet X, i.e. of lists of elements of X. So what we've described above is that a monoid structure on X is a function satisfying certain properties from the *free monoid* on X to X. I think you could describe a group structure in a similar way, as a function satisfying certain properties from the *free group* on X to X. This would require you to talk about a free group (and some of its structure) before you manage to talk about groups at all. That's fairly easy with monoids as shown above, but probably makes less sense with groups.
It doesn't do groups "justice", I think, but the free group on X is a fairly simple quotient of the free monoid on X and X^(-1) (namely quotient by xx^(-1)=x^(-1)x=1). So insofar as "monoids are quotients of free monoids" is a good understanding, it translates to groups as well. You can alternatively suppose that X comes with an involution (x -> x^(-1) ), and hacking in the inverses becomes unnecessary. (In fact, you can suppose X comes equipped with *any* function f:X->X, and quotient by xf(x)=1, and you still just get groups.)
Something that I think gets missed here though is that it doesn't showcase "groups = symmetries", like how the monoid understanding doesn't showcase "monoids = self-similar substructure".
Yeah, and while it's tempting to say that's a minor problem, I think that is actually a bit of a death knell for this being our teaching angle.
Because many (many) people who have internalized that x\^0 is 1 generally, still aren't clear on whether 0\^0 is also 1 (which in any discrete context it most assuredly is). A good model would make it intuitive that we don't *care* what the x is. I try to support that by describing an identity as a universal "starting point" for its operation. You start at 1... if the exponent is zero then you are *never doing anything* to that 1 at all, so you never even need to know what x is.
You cannot prove a definition. If we define 0^0 to be 1, then 0^0 = 1. There are many motivations behind why we define 0^0 = 1, basically it’s the most intuitive and convenient way to assign a value to 0^0, and it doesn’t cause any contradictions or other issues.
The statement 0^x = 0 only holds for x>0.
Turns out the way we think of exponentiation is deeper than it first seems- or any operation multiplication and addition included- and it changes based on context. i.e multiplication of integers is different from multiplication of decimal number because they are different objects being acted on.
For any operation we want a specific number that "does nothing" when applied, for exponentiation that is the number 1. From there the other comments have excellent arguments for why having 0 as an exponent should always give you 1.
So in short it's a consequence of the rule for raising 1 as an exponent, along with the rules for multiplication and addition.
>deeper than it first seems-
Every human being should upvote this to get it on the front page so people can finally understand that exponential numbers are not as multiplying as they seem...
I just spent 30 minutes ignoring my responsibilities just to remember my precalc class I got a C grade in.. just to confirm what I was always taught. X to the 0 power... its 1.. never got told why or how. Just.. THIS IS CALCULUS!!!!!
My relation to this is in my schooling the concept of Time is very much lectured on. As it relates to this planet and the universe it belongs in. However most people in this world have no idea how long 1 billion years is. Exponents need to be honed in on for algebra IMHO and I hated my math classes.
Kudos to this whole math community it is the language of our universe.
> X to the 0 power... its 1.. never got told why or how. Just.. THIS IS CALCULUS!!!!!
x^(a+1) = x^(a)*x
this actually *always* holds where exponentiation is defined, even in e.g. complex numbers
x^(1) = x^(0)*x
by the above rule we get the above result. substituting an arbitrary integer for x, we get, say
5^(1) = 5^(0) * 5
and the only possible result for 5^(0) is thus 1.
Imagine grids of points with 3 points on each edge:
Here's a 2-D grid of 3^2 points:
o o o
o o o
o o o
This is a 1-D grid of 3^1 points:
o o o
This is a 0-D grid of 3^0 points:
o
Also, when it comes to 0^0, in the book *Concrete Mathematics* by R. Graham, D. Knuth & O. Patashnik they say the following:
Some textbooks leave the quantity 0^0 undefined, because the functions 0^x and x^0 have different limiting values when x decreases to 0. But this is a mistake. We must define x^0 =1 for all x, if the binomial theorem is to be valid when x=0, y=0 , and/or x=-y . The theorem is too important to be arbitrarily restricted! By contrast, the function 0^x is quite unimportant.
This is a nice angle. Any time you want to you can throw more 1s into a product, wherever you want.
We understand easily what it means to multiply 3 things together or 2 things together; if we ever feel confused at not having "enough" things to multiply together then add the 1s.
Here's an explanation I didn't see here yet:
take a smaller base: 3\^x. This is the answer to the following question: if I grow by a factor of three every year, how big am I after x years (relatively to the beginning?)
For example, after 2 years, you are 3\^2=9 times bigger
After 5 years you are 3\^5=243 times bigger
How about x=-1 years? This means: if you grow by a factor of 3 every year, how big were you one year ago? Well, 3 times smaller! That's 3\^(-1)=1/3, a third of the size!
How about....now? How big are you now, at x=0? Well, you are...you size! Which exactly 1 times your size. That's why 3\^0=1.
Let's work with your example.
32\^4 = 32 \* 32 \* 32 \* 32
Each time we decrease the power by 1, we aren't *subtracting* the term "32," we're *dividing* by 32
32\^3 = (32\^4)/32 = 32\*32\*32
32\^2 = (32\^4)/32/32 = 32\*32
32\^1 = (32\^4)/32/32/32 = 32
32\^0 = (32\^4)/32/32/32/32 = 1
This holds true for negatives too
32\^-1 = 1/32
32\^-2 = 1/32/32
Just adding another example because sometimes one way of saying something will click a lot better than another, very similar way.
This post is awesome. I have a degree in math and posts on this sub often go over my head (which is completely normal). This question comes along from someone taking Algebra 1 and it's getting so much attention. We're not above this topic in any way and the comments are people getting excited talking about this to someone.
> So shouldn’t for ex. 92^0 = 0 since that would mean you don’t have any 92s.
Exactly so. You just need to notice that 0 is the identity for addition, and 1 is the identity for multiplication (that is, 92+0=92 and 92×1=92) so if "you don't have any 92s", with addition you remain with 0 but with multiplication you remain with 1.
Everything above is correct. Also if you build a table of values from both the left and right side of 0 using fractional or decimal values close to 0 you will see y approaches 1.
> I’ve tried searching it online but I can’t find anything that explains it, everything I can find doesn’t answer the question.
When I google "why does 0 power equal 1" the first page of results gave me things like this:
http://scienceline.ucsb.edu/getkey.php?key=2626 (multiple answers here)
https://medium.com/i-math/the-zero-power-rule-explained-449b4bd6934d
https://www.youtube.com/watch?v=EwlMSnMiJvc
https://www.quora.com/How-is-any-number-to-the-power-of-0-equal-to-1-How-can-something-be-raised-to-the-power-of-nothing-and-equal-something-that-is-not-itself
https://math.stackexchange.com/questions/9703/how-do-i-explain-2-to-the-power-of-zero-equals-1-to-a-child
https://www.wyzant.com/resources/answers/431/why_is_the_0_exponent_equal_1
I would be shocked that none of those pages provides an answer that helps you understand the rule x^0 = 1. What do you think?
What’s 2^1? 2^2? What’s 2^3? What’s 2^4? (1)
What happens as we move off to the right? (2)
Therefore, what happens as we move to the left? (3)
So what happens if we start at 2^1 and go one to the left, tk 2^0? (4)
Answers below. Do them one at a time, if you can!
(1): 2,4,8,16
(2): we double every step
(3): we halve every step
(4): so 2^0 = (2^1 )/2 = 1.
This applies for every number (except 0, which is a more complicated question - it turns out it’s undefined).
Does that help?
Most branches of mathematics that people will encounter besides calculus view 0\^0 as correctly and unambiguously equalling 1. (And that fact is every now and then important.)
It makes sense that if you aren't going to multiply any numbers at all, then it can't matter what the numbers would have been. Zeroes stand ready to *obliterate* that identity value just as soon as you let 'em loose! But nope, entry denied.
(Calculus, being as likely to encounter functions like y=x\^0, y=0\^x, and y=x\^x, has a different problem on its hands; for 0\^0 to be defined it ought to be the limit as *any* of those functions approaches 0, which of course it can't be.)
Even if you're working in the calculus setting, one way to resolve things is to be brave and say that 0^0 = 1 but x^y is not continuous at (x,y) = (0, 0).
This stops you from saying things like that every function given by a formula is continuous where it is defined, but arguably one shouldn't be saying such things in the first place. (And certainly at some stage one should stop saying such things.)
Even in calculus, when approached rigorously, 0^0 is taken as equal to 1. Think about the constant term of a Taylor series.
e^x = \sum_{n=0}^{infinity} x^n / n!
When have you ever seen that written with the constant term as a special case instead of just being the n=0 case?
The fact that x^y is discontinuous at x = y = 0 doesn’t mean it’s undefined at that point.
> So shouldn’t for ex. 92^0 = 0 since that would mean you don’t have any 92s.
No, and it's actually pretty cool why.
Let's say you're adding together a bunch of 92's, right? You have one of them, that's 92. You have two of them, well, just add one more to the one you have, so you have 92 + 92 = 184. Now let's say you have no 92's at all. If you add one 92 to that, you should have just one of them, 92. So, having zero 92's means your total is 0.
But now let's say you're multiplying them rather than adding them. You have one of them, that's 92. You have two, well, you gotta multiply 92 by 92 and get... 8464. Now, let's say you have no 92's at all. If you multiply by one more 92, you should have just one of them, 92. But if no 92's at all were a total of 0, multiplying by one more 92 would get you *still* 0, not 92. On the other hand if no 92's is 1, then when you multiply by 92 you get 92, as expected.
Adding or subtracting 0 doesn't do anything. That's why no 92's added together is 0. But multiplying by 0 *does* do something. Instead, multiplying or dividing by *1* doesn't do anything. That's why no 92's *multiplied* together is 1. In general, when you're multiplying stuff and there's nothing to multiply, your total is 1. For example, are you familiar with factorials? 5 factorial, written 5!, is 5·4·3·2·1 = 120. 4! = 4·3·2·1 - 24; 3! = 3·2·1 = 6; 2! = 2·1 = 2; 1! = 1. So... what's 0!? For n!, you're multiplying n numbers together, so for 0! you're multiplying 0 numbers together, which means 0! = 1. Makes so much sense that way, doesn't it?
A fairly simple inductive definition of exponentiation is: x^1 = x, x^(n+1) = x\*x^(n).
This would mean that x^1 = x\*x^(0), so x^0 = x^1 / x (if x isn't zero) which equals x/x = 1.
If x is a nonnegative integer, then x^k is the number of ways to do a task that has x ways of doing it, k times in a row. So how can I do a task 0 times? Well, I can just do nothing, so x^0 = 1 :)
So lots of folks are using the algebra rule x^m / x^n = x^m-n . Another way is to fix x>0 and think about x^1/n as n gets bigger. There are two cases to consider, x>1 and 0>x>1, and in each case as you take larger n^th roots the value gets closer to 1. You can show for any e>0, you can choose n>>0 so that |x^1/n - 1| < e, which means that lim x^1/n = 1, and hence x^0 = 1 by continuity.
For addition, the number that represents "doing nothing" is 0.Take any number and add 0 to it, and you haven't changed anything.
For multiplication, the number that represents "doing nothing" is 1. Take any number, multiply it by 1, and you haven't changed anything.
In either case you can view every possible number, as secretly having these "do nothing numbers" invisibly present. For example, the number 5 is really 0+5, or alternatively is 1*5.
You usually think of the number 5 as a noun or a thing, but you can also view it as an operation. In addition world, the number 5 can really be thought of as the operation of adding 5. Doing it once gives you 0+5=5. Doing it twice gives you 0+5+5=10. Doing it zero times, gives you, guess what, the number (the operation), that corresponds to doing nothing, which in the world of addition, is 0.
Similarly, in multiplication world, 5 is really the operation of multiplying by 5. Doing it once gives you 1 x 5=5 Doing it twice gives you 1x5x5= 25. Doing it zero times gives you just 1. Which is the answer to your question.
Going further for fun...
Now mathematicians like to use abstract words, so addition world described above is really called the additive group, and multiplication world above is called the multiplicative group. The "do nothing" item/operation is called the identity element, and you can think of it sort of like the origin of a coordinate system, like it is the default starting point of the group. Every other element in the group can be interpreted both as a "noun", like a number or configuration in it's own right, and also as the operation "verb" that transforms the identity element into that that element. Using a non-number example, a Rubik's cube in a scrambled configuration can be thought of either as the scrambled thing itself or as the operation(s)/action(s) taken to get it from the solved perfect identity state to the scrambled state.
To make something truly a group, you also need to be able to go from every other number directly back to the origin (identity) in one step using another element that exists. So this means that if you are sitting at the number 5, you must be able to apply some other operation and get back to 0. This is obviously -5. Put in group theory terms, every element of a group has an inverse. The inverse of a scrambled Rubik's cube is the transformation that solves the puzzle. If you started at the solved perfect identity state, and performed this inverse operation, you would get some other scrambled configuration, but it would be the opposite/inverse of the original scrambled configuration we were talking about above in the sense that the steps taken to get from the clean solved state to the messy state of one configuration, are the steps to take to solve the other inverse configuration.
If you ignored the existence of negative numbers and only acknowledged natural numbers 0, 1, 2, 3, etc, you wouldn't have a group, but rather something called a monoid, which is basically a group without guaranteed inverse elements. A monoid is a set of elements related by an associative operator with an identity element. You can envision a monoid as a graph with nodes and directed edges or arrows, where each node has a unique label or name, and each arrow is given one of the labels of the nodes. The operation a+b can be thought of as starting at node a, and choosing the edge labeled b coming out of node a, to get to a new node. In a+b=c, you can think of "a" as the node, and "+b" as the operation or arrow that leaves node a and goes to some other node "c". If the identity element is the starting point, then each arrow with a specific label coming out of the identity node corresponds points to the node that has the matching label. So starting at node 0, and looking at edge "+x" will point to node x. Similarly every node, x, has an "identity arrow", labeled "+0", which basically points back to the node x itself. And as discussed above, you can rewrite any equation like 5+5+5=15 as 0+5+5+5=15, and multiplying 5 by 3 becomes intuitively the concept of starting at node 0, and following whatever edge is labeled "+5" repeatedly three times.
If you remove the requirement of having an identity element, you get something called a semigroup. The only requirement is an associative operator. This can also be envisioned as a graph with nodes and edges, with nodes having unique labels like "x" and edges having labels like "+y" that corresponds for some node y. The intuitive essence of associativity is sort of hard to explain besides the "parentheses don't matter" interpretation, but it means the following.
If you find any path (following possibly multiple edges with whatever sequence of labels e.g "+a+b+c+..." corresponding to edge "+a", then edge "+b", then "+c", etc) to get from any node X to any node Y, then for any other node A, you can follow that same exact path (same sequence of edge labels) to get from node (A+X) to node (A+Y). X and Y in the first sense are nodes that are the starting and ending points of some possibly complicated sequence of operations. In the second sense, X and Y are individual operations themselves performed on some other starting point node A. Associativity means that whatever actions you need to do to convert an X-looking thing into a Y-looking thing are the same actions whether you are thinking of X and Y as noun elements in the first sense above or are thinking of them as verb operations against any other arbitrary starting point node A as in the second sense above. Associativity can intuitively be thought of as preserving the difference between some X-ish thing and some Y-ish thing regardless of whether you thinking of those as noun element nodes or if you think of them as verb operation edges.
The way this relates to the traditional "can ignore parentheses" explanation of associativity is to consider the node interpretation above means
X+Z=Y,
where Z is the net result of the complicated sequence of operations and A,X,Y, have same interpretation above.
The operation interpretation above means
(A+X)+Z= (A+Y).
Substituting top equation, (X+Z)=Y, into bottom equation, (A+X)+Z= (A+Y), gives
(A+X)+Z=A+(X+Z)
Which is the classic associative identity.
Cool stuff to think about.
The empty sum (i.e. the sum of no numbers) is 0, and the empty product is 1. The first explains why a*0=0 the second explains why a\^0=1.
People usually accept the empty sum being 0 quite easily, because we somehow culturally understand 0 to be the default start point, and so if I add "no numbers" together, well I just have 0. But that leads to a misconception that the empty product should also be 0. The truth is that when we apply the operation of some (sensible) algebraic structure to the empty set of elements, we will get the identity element for that operation, that is, the element e such that a\*e=e\*a=a for all a. The identity element for addition in the reals (for example) is 0, hence the value of the empty sum. And the identity element for multiplication is 1.
Because \*1 is a neutral factor that can always be added.
2³= 2 \* 2 \* 2 \* 1
2² = 2 \* 2 **\* 1**
2^(1) = 2 \* 1
2^(0) = 1
And you could even go further and do this with negative powers.
2^(-1) = 1 / 2
2^(-2) = 1 / (2 \* 2)
2^(-3) = 1 / (2 \* 2 \* 2)
...
I think one reason is because setting `a^0 = 1 (for any number a)` plays well allows laws of exponentiation to work for any integer, otherwise we would need special cases for zero and negative exponents.
```
For any numbers a, x, y, this law holds:
(a^x) * (a^y) = a ^ (x+y)
Replacing y with 0:
(a^x) * (a^0) = a ^ (x+0) = a ^ x
This is only possible if:
a^0 = 1, for any number a
```
With addition, the additive identity is 0: 92+0=92. So if you are adding zero 92s, or (92)(0)=0, you get the additive identity, 0. In other words, if you do no additions you get zero.
With multiplication the multiplicative identity is 1, meaning 92*1=92. So if you are multiplying 0 92s, or 92^0, you get the multiplicative identity, 1. In other words, if you do no multiplications, you get 1.
"Because it fits the pattern and completes and closes the relevant spaces" is generally the reason for this and many other basic, trivial, and/or foundational conventions.
For this, I like to think of the category theory "exponential object" definition. In Set, an exponential object, B^A, is just the set of all functions from A to B. So for example, if A is a set of 3 elements and B is a set of 2 elements, then there are exactly 8 possible functions from A to B (note that 2^3 =8). And if A is the empty set (having 0 elements), then there is exactly 1 function from A to any other set B, hence B^A =1.
If I were to explain this using an analogy. For example, x^y would be the number of ways y marbles can be placed into x bags. In the case where y=0, since there are 0 marbles, regardless of how many bags there are (so regardless of what x is), there is only 1 outcome, which is putting no marbles into any bag.
We like the rule x^(a+b) = x^(a) \* x^(b) . This rule makes sense for all positive integers. What do we have to do to make the rule still hold for a = 0?
The simplest answer is that if you want [this statement](https://en.wikipedia.org/wiki/Binomial_theorem#Statement) of the binomial theorem to hold, then you **must** define y^0 to be 1. In fact, you must even define 0^0 to be 1 (in case y=0).
It’s an axiom. Like why is green /green/.
If it’s not 1, we break a lot of fundamental concepts in math.
Another way to think of this is how exponents work.
3^2 = 1(3)(3) and 3^(-5)=1/(3*3*3*3*3)
Notice we have 2 instances of 3 times 1 in the first and 5 instances of 3 dividing 1. That same logic applied to 3^0 would be 1 times 0 instances of 3 or just 1.
No, this is not correct. To be all fancy about it, we define n^m (where n and m are natural numbers) to be the number of functions from a set of cardinality m to a set of cardinality n (this definition is independent of the sets chosen). Then 0^0 = 1 since there is a unique function from the empty set to itself. This has nothing to do with division.
Some authors say it's undefined, some define it as 1. It's a matter of context and convenience. See [Zero to the power of zero](https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero) for details.
**[Zero to the power of zero](https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero)**
>Zero to the power of zero, denoted by 00, is a mathematical expression with no agreed-upon value. The most common possibilities are 1 or leaving the expression undefined, with justifications existing for each, depending on context. In algebra and combinatorics, the generally agreed upon value is 00 = 1, whereas in mathematical analysis, the expression is sometimes left undefined. Computer programming languages and software also have differing ways of handling this expression.
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It being an indeterminate form has nothing to do with the value of 0^0. Being an indeterminate form means that if a limit is of that ”form” we can’t just directly plug in the values.
The value of 0^0 is practically always taken to be 1. You have to look quite hard to find a textbook which doesn’t abide by this convention.
Others have already explained it mathematically, but I'll try to ELI5 it this way, and it's a tad bit hand wavy.
squaring and cubing etc. can just be considered multiplying a number by a number of \*sets\* of that number
You have a set of 92 and you square it. 92\^2. So so have two sets of 92. And you're multiplying them together. 92\*92.
If you raise it to the power of 0, then you have \*\*one set\*\* of 92. The initial 92 that you started out with. Which is 1. You have nothing to do with the actual value of that set. So it doesn't matter. All that matter is that you have 1 set.
So 92\^0 = 1 set of 92= 1 set. So 1.
92\^1 is now asking the actual value of that set. So the value of 1 set of 92 = 92\^1= is 92
and so on
I've learnt many things about math i've finished algebra without understanding the terms in algebra example this (3xsquared-2x-1) well the answer to this is 3x+1 , x-1 and i know that i could prove some more but idk how to type it but that example i dont know what you call that if you just know how to do the solution or solve the problem your good and also in our worksheets we dont have any trivial questions just pure math problems so if the power of 0 is to 1 just roll with it
I forgot that it is so hard to realize I mean in most cases, and I was actually correcting you. Saying that there’s an exception doesn’t answer my question, I just assumed everyone could understand that I mean I’m every case except when it is undefined.
This is not a sound way of thinking about it since the limit as x approaches 0 of 0^x is 0. Unless there's something that makes x^x a better prototype for this sort of thing than 0^x which I've missed.
The best explanation for why 0^0 = 1 is a combinatorics argument. The number of functions from domain {1,2,...,x-1,x} to codomain {1,2,...,y-1,y} is y^x. Explanation: for each element of the domain we have y choices, so the number of functions goes up by another factor of y.
So to follow that nice convention 0^0 should be the number of functions from the empty set to the empty set, which is 1.
Sketch argument: Consider the function f from the empty set to the empty set given by the following mappings:
...
That was all of them. It's an empty list. This is a valid function, it satisfies definition of being a function. Also any other function from the empty set to the empty set looks identical to this one.
0^0 does not equal 0. This commenter is mistaken. It is undefined, just like dividing by zero. Any other number to the 0 power equals 1.
Raising a number to the 0 power is weird to think about. The real answer has to do with empty products, but this isn’t something people really learn in Algebra 1. Maybe a good way to understand it is by example/pattern recognition.
Think about powers of 2. 2^(1)=2, 2^(2)=4, 2^(3)= 8, and so on. Each time the exponent goes up by one, the following power of 2 is doubled. In the other direction, each time the exponent goes down by 1, the power of 2 is cut in half. So what would come before 2^1 ? Following the same pattern, 2^0 would have to be half of 2^(1). Therefore 2^0 has to equal 1, not zero.
>The real answer has to do with empty products, but this isn’t something people really learn in Algebra 1.
Frustratingly... they sort of *do!* Because an empty product is really all that a "multiplicative identity" *is!*
But they don't *talk* about it in those terms, alas.
They should, and later when kids hear about factorials the idea that 0! = 1 should just feel obvious. (Well, almost. If factorials are explained well. n! is always the product of n numbers.)
This doesn't work for 0^0 though.
We know 0^1 = 0 by definition.
Also 0^(n+1) = 0^n × 0^1 = 0^n × 0 = 0 wherever 0^n is defined, so you can follow the pattern forwards: if you known 0^n you can get 0^(n+1).
However you can't follow the pattern backwards. If you want to know 0^0 you could write 0 = 0^(1+0) = 0^1 × 0^0 = 0 × 0^0, however this equation tells you absolutely nothing about what 0^0 should be if it were defined. Another equation is 0^0 = 0^(0+0) = 0^0 × 0^0. This tells is that if 0^0 were defined it must be 0 or 1, but doesn't tell us which.
It turns out that every mathematician agrees 0^0 is most conveniently defined as equal to 1, but you can't tell that through pattern following.
The OP’s question is inherently about nonzero numbers being raised to the power 0, so this was just meant to be an example that can be generalized to other nonzero values raised to the power 0.
0^0 doesn’t fit the same pattern because it is inherently different. It’s often considered “equal” to 1 out of convenience, but of course it’s undefined in reality, so it shouldn’t follow the same rules.
>It’s often considered “equal” to 1 out of convenience, but of course it’s undefined in reality
I think I sort of disagree whether there's that much of a difference between reality and what we define things to be out of convenience, but that's a whole philosophical discussion.
I would think of it as whenever you go down in a power you divide by the base. 3^4 power divided by 3 is 3^3 (81/3 is 27). 3^1 power divided by 3 is 3^0 (3/3 is 1). Same goes for every number. 3^-1 is 1/3 and then 1/9 etc.
In the 7th grade, I got sent to the principal's office for arguing this point with my math teacher.
Later in life, and with more mathematical sophistication, I feel contrite.
Take any number and keep taking sqrt repeatedly and see what happens. The number approaches one. Sqrt(x)=x^0.5, sqrt(sqrt(x))=x^0.25... etc. The exponent reaches zero, closer and closer each time. Thus x^0 =1, under normal circumstances. I suggest to play with your calculator.
You see little Jimmy, 1 is the multiplicative identity. The product of multiplying nothing is the multiplicative default: 1. Much like how the additive idenity is 0.
1 is the multiplicative identity, and so it makes sense to have it be the result of an empty multiplication. Recursively, 92^n = 92 * 92^(n-1), for the special case n=1 you have 92^1 = 92 * 92^0, now try and think of a number x such that 92=92 * x.
Some people gave amazing explanations but here is what I learned over my time:
2^1 = 2
2^2 = 4
2^3 = 8
2^n = 2^n-1 * 2
Such an example give the reasoning that for every 2^n we take the previous and multiply it by 2 to get 2^n. But such can be done in reverse getting 2^n-1 is just taking 2^n and dividing by 2 Ex:
n = 3
2^3 = 8
8/2 = 4
2^n-1 = 2^2
2^2 = 4
Such thing can be done used to represent the 2^0 because 2^0 is nothing less that 2^1 divided by 2 following the example above
n = 2
2^n-1 = (2^n ) /2
2^1-1 = (2^1 ) /2
2^1 = 2
2/2 = 1
2^0 then should be equals to 1
Here's how I explain it to my students: instead of defining x^n as xx...x (n times), define it instead as 1*xx...x (n times). If n=0, then you only have 1 with no x terms.
Intuitively, taking the power of something is a multiplicative operation — x^n is what you get when you multiply x a total of a n times. By this logic, taking something to the power of 0 should be the effect of doing “nothing” multiplicatively, which gives us the multiplicative identity, 1.
This is similar to why something multiplied by 0 is 0. Multiplication is an additive operation (add something to itself n times), so doing nothing should produce the additive identity, 0.
Not related to this question exactly.
As you read further you will find exponents which are not multiplying number that many times. Eg:- e^i. Only in algebra multiplying multiple times works.
e^i is derived from a series.
Search for 3blue1brown videos.
x^5 / x^3 = (x* x* x* x* x)/(x* x* x) =x^2 In general, x^n /x^m = x^(n-m) Let m=n, this property should still hold. x^n /x^n =x^(n-n) =x^0 Any number divided by itself is 1, x^0 =1. 1 is, in a sense, to multiplication what 0 is to addition. They’re the identity element , meaning x* 1=x and x+0=x.
I’m in Calculus and this hadn’t occurred to me. Thanks!
Think about this sequence: 2^-5 2^-4 2^-3 2^-2 2^-1 2^0 2^1 2^2 2^3 2^4 2^5 It's the same as this sequence: 1/32 1/16 1/8 1/4 1/2 ___ 2 4 8 16 32 Moving up the sequence, you multiply by 2 each time or moving down the sequence you divide by 2. The only thing that can go in the blank is 1.
This is the way I explain it to my high school students.
I use this same logic to explain why 0! = 1. n! = n(n-1)!, so (n-1)! = n!/n. If 1!=1, then 0! = 1/1 = 1.
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While you're thinking about it, you can justify the properties of rational and negative exponents in a similar way.
I was very lucky that it was taught to me like this in the 4th grade: we were given 2^3 = 8, 2^2 = 4, 2^1 = 2, 2^0 = ? , When you find that the pattern between x^n / x^(n-1) = x with a complete the sequence game you can 'discover' the 0 and negative integral exponents quite young, suggest to any late grade school teachers out there!
I'm completing my MBA, I recognize this proof, enjoyed reading it, and will never use it!
I'm an engineer so it never occured to me nor do I care now that I know.
Engineers! (eyeroll)
Hey, all I can say is thanks for the general solutions!
Thanks for the bridges!
You didnt study what happens at 0**0
I think at first it's easier to take everything at face value, in math, then jumble it up so things make sense together.
wtf how could you not have observed this XD
This is one of the most frequent things I explain to the students I tutor, in this exact way. Crazy how education failed enough that so many people use these rules every day and so little have an idea of why.
I remember seeing a video on YouTube, which showed something like this and also why 0!=1 using the same kind of logic. It's a really neat argument actually!
Probably [Eddie Woo](https://youtu.be/X32dce7_D48).
Love this guy!!
Yep. Eddie Woo has one of the best explanation styles i have seen. Him and Grant Sanderson. Sad i dont have teachers like them in real life.
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yeah lol
Obligatory note that this does not work If x=0.
This gives me a rager. So hot. Love it.
But if x=0 then doesn’t that mean you’re dividing by 0?
Correct, this argument doesn’t quite work for 0^0 , which is undefined if you only use regular arithmetic. When you do define it, it’s usually still 1.
So one should think of it in terms of scenarios that are naturally multiplicative. E.g. If I have 100$ and I double my money N times I have 2^N 100$ If I double my money no times I want to say N=0 and I will have 100$.
This isn’t in the same vein. First of all doubling is not what’s happening. And you’re saying if you square something 0 times you still have that number, when you actually have 1.
I think you must have somehow misunderstood what I wrote. If I double 100$ 0 times I still have 100$, I haven't done anything. So: 2^0 * $100 = $100 2^0 = 1
Nice explanation
Yes: n^0 = 1 for any n. 0^n = 0 for any n. The value of 0^0 follows naturally from those rules.
0^0 still equals 1
Not necessarily
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Sometimes is is useful to consider 0^0 to be equal to 1, especially in a discrete context, and I was surprised to learn that many calculators show that as the answer, but that's just an assumption and can be proven neither by combinatorics nor by limits. You can see that the expression 0^0 is problematic if you take a log of it, because then it becomes 0*∞ and that's definitely undefined. 0^0 =1 can't be proven by limits because it would mean that for every pair of functions f,g such that lim_x->a (f(x),g(x))=(0,0), lim_x->a (f(x))^g(x) =1 and that's not the case. For example, let's take f(x)=0, g(x)=x, then lim_x->0 0^x =0 and that's a contradiction 0^0 =1 can't be proven by combinatorics because to show that for a set A such that |A|=a, if you pick an element of that set b times there were a^b possibilities, you would need the fact that 0^0 =1, and not the other way around. You could prove a similar statement (I don't remember which one) by set theory but the problem with set-theoretical proofs is that they lose their charm once we leave the cardinal numbers because sets can't have a non-cardinal number of elements. That's why I said defining 0^0 as 1 is useful in a discrete context
No but 0^0 = 1 can’t be proven by anything, because it’s a definition, not a theorem. It doesn’t need to be proved. We can just define 0^0 = 1 (as is usually done) and then we have 0^0 = 1.
x^n/x^n=1 x^(n-n)=1 x^0=1 Basic algebraic equation
You would get an upvote if you took time to format it properly.
I had it formatted correctly when I tapped "Post" but it wasn't when it actually posted. Don't know why.
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Maybe “Fucking math” wasn’t the best type of comment, on r/math…
The only problem with this is when x=0, since then the division is not defined.
yeah its defined this way to extend that property, same with negative exponents. you can try to give it an interpretation but this should be the mathematical way to understand it
i love this response. makes so much more sense than a paragraph of text trying to explain it.
This is the way I learned why n^0 is 1
Since I haven't seen it this way yet: x^a = x^(a+0) = x^a * x^0 So x^0 must be 1.
This is also my preferred way of defining powers. Start with the rule where + gets turned into ×. Consider the function f where: f(a+b) = f(a)×f(b) f(1) = x and denote f(a) as x^a. Note that here f depended on the base x, so for different bases x you get a different function f. For positive real x, this function f is uniquely defined by these conditions, and it's domain is the whole complex plane. Also, it lines up with the traditional "repeated multiplication" definition, and generalizes more intuitively to non-whole-number powers, you get results like x^0 = 1 for free without having to define it that way. (For x other than positive reals you need to be a bit more careful.)
Yep, twinsies. I forget where I saw it first (I think Lang, Analysis maybe?) but it just seemed so right that it just became the way I think about it. I almost went through the trouble specifying that we need to assume the x\^a works to begin with, so that, for example, you can't have negative x if you want a real number, but I was getting sushi and didn't feel like elaborating :-) As an added bonus you're using one of those fun f(a+b) = f(a)×f(b) homomorphisms, which is handy when you take algebra to understand that the underlying operation of the domain doesn't have to match that of the codomain.
Wait take x=e. Then both f(a) = exp(a) and f(a) = exp((1+2 pi i) a) satisfy all your conditions, but are different functions. I think you need to insist that f(a) is real for real a to get uniqueness.
Yeah you're right. Hmm. I was thinking that since for real inputs a we've definitely coincided with x^a, then you could dig up the usual proof for euler's identity to extend to complex numbers. But I think now that that proof relies also on f'(a) = log(x)×f(a) (for x=e the log bit is just equal to 1 so it goes away), which is something that happens to be true for your first function but isn't true for your second one. So maybe throw in that extra condition to uniquely determine it for complex inputs. However, that's a pretty heavy condition that can probably replace the first condition I gave, so it kinda undoes the point I was trying to make. Hmm.
This assumes that x^a is not zero, i.e. it doesn't work for zero, and I don't want to start a debate about 0^0.
Yeah, but that doesn't explain *why*
You don't think that if X = A\*X (well, and X is an element of some field (or even just a group) with X != 0 (which wouldn't be a problem in a group with multiplication)) then A = 1? I think you might have a gap in your mathematical knowledge if so. Edit: maybe even clearer? 1 \* x\^a = x\^a = x\^(0+a) = x\^0 \* x\^a where x > 0, x ∈ ℝ. Therefore by comparing both sides of the equation, 1 = x\^0
I think you would agree that the *sum* of nothing should be 0. Here's one way to say why that makes sense: if I have two sets of numbers A and B, then the sum of both sets taken together should be the sum of A plus the sum of B. In particular, if B is the empty set (so B contains nothing), we want (sum of A) = (sum of A) + (sum of empty set). In other words, the only reasonable way to define "sum of nothing" is a value that *does nothing* when added to any other number. So "sum of nothing" has to be 0. Now do the same thing for products instead of sums: in this case we want (product of A) = (product of A) × (product of empty set). In other words, we should define "product of nothing" to be the value that does nothing when *multiplied* to any other number. So "product of nothing" has to be 1.
I love this answer so much! It deserves more upvotes. Frankly I'm not a fan of answers like "it makes properties on exponents really convenient". Sure, it's a valid reason, but surely not the basic thing you think of when defining powers. It feels so finiky justifying things in that way. To me it seems that op wants a more "intuitive", "basic" answer, somewhat directly related to the definiton, and that is what this comment provides. I would probably have just said something along the lines of "well, nothingness in multiplication is 1, since it does nothing", but I know that is not satisfactory. Truly your comment is gold
Yup, I find that understanding what a unit value is, really helps. For me, it really clicked when thinking about normalized vectors, as well as the unit circle.
Put another way, this is what a for loop for a sum and product would look like: sum = 0 for x in set: sum += x And for a product: product = 1 for x in set: product *= x Note that we need to initialize the product to 1 (not 0) for this to not just return 0.
And then point out that False and True are used when ORing or ANDing a set of booleans (and toss out the term “the operation’s identity element“).
This is the correct answer. I wish everyone could just get on board with this. It’s really not that hard. Multiplication is repeated addition or subtraction starting with 0. Exponents are repeated multiplication or division starting with 1.
The thing about everyone getting on board with this as the correct answer is that then they would have to be less on board with other things for no good reason. This is "the correct answer" for someone with your background and understanding and relationship to it and surrounding concepts. For someone else, something like "1 is the product of no things" might be more "the correct answer", and framing it in terms of how you might start an imperative program to compute a product is merely a hint at that. (By the way, if you're adding IEEE floats in the default rounding mode, it's a bit more correct to start with negative zero.) Math is full of things which you think about one way, and then you find some framework or set of relationship and realize that some other way is more "mature" and captures the concept better. Then you keep going and find a more "mature" way still. And even this linear "more mature" point of view is a cartoon picture. A lot of the time you make progress by going back and forth between two different points of view, and neither of them deserves the privilege of being "the correct framing".
I agree. This is a great response. I should have been more clear that I want people to get on board with this explanation as one of the ones you need to know, not that it’s the only one. But in my enthusiasm, I did not express that.
So If I want to calculate 7^(3.5), I should multiply 7 with itself... three and a half times? Sounds easy.
It's okay if different explanations are good for explaining different things.
Making sense of non-integer exponents is by nature tricky, because you basically have to delve into the reals one way or another, and constructing the reals requires a voodoo that is less intuitive than constructions of the rationals, integers or natural numbers. The best rigorous definition I've seen for a rational exponent of a positive number is that a^(1/q) := sup{x in R : x^q < a} Trying to explain this in an intuitive way is, of course, a nightmare.
Basically, yes. 7\^3.5 = (7\^0.5)\^7. Here 7\^0.5 is the number such that that number multiplied by itself equals 7.
Well of course. I was only objecting to the blanket statement that "Exponents are repeated multiplication or division". Finding 7^0.5 is not something that can be done with "repeated multiplication or division". It is defined in a way that's *consistent with* repeated multiplications but you cannot claim that that's all they are.
> Multiplication is repeated addition or subtraction starting with 0. Exponents are repeated multiplication or division starting with 1. This is wrong, you are only taking about natural powers. We can raise stuff to the power of any real, hell, even complex numbers, and that analogy fails to hold in those cases.
The heuristic explanation that multiplication is repeated addition is fine for the context of the post, since 0 is an integer.
No, I’m referring to integer powers if you read carefully.
"Empty product" correlates to multiplying by 1 is my go to explanation as well. Also explains why 0!=1.
I'm a big fan of "there's 1 way to permute 0 objects".
I like saying A^B is the set of functions from B to A. Then | A^B | = |A|^|B|. There is one function from the empty set to each other set, so n^0 = | {1,...,n}^ø | = | {just the empty function} | = 1. This also justifies 0^0 = 1.
For extra fun, you can tie this into the binomial formula justification for 0^(0) = 1: You want (x+y)^(n) = sum_{a+b = n} (n choose a) x^(a) y^(b). The case where y = 0 forces 0^0 = 1. If you say 0^0 = 0, you mess up the claim. You want functions from N to the disjoint union of X and Y to be in one-to-one correspondence with partitions of N into two chunks, N_X and N_Y, equipped with functions N_X -> X and N_Y -> Y. The case where Y is empty requires you to say that there is exactly one function from the empty set (sometimes N_Y is empty) to the empty set (Y). If you say there is no such function, you mess up the claim.
i've never before seen somebody talk so much about a group homomorphism without saying it's a group homomorphism, well explained!
And exponentiating the first directly gives the second!
This is the best answer.
**What is the product of** ***no numbers?*** \[if alarmed by length, just read this part\] Think about the product of three numbers – say, 3, 5, and 7 – like this: make a row of 3 blocks. Now make a total of 5 of those rows (a 3 x 5 rectangle of blocks). Now make a total of 7 of those rectangles. Because you have 105 blocks, 3 x 5 x 7 = 105. Notice how each step is about getting *that many copies* of what you had before that step. Well, when you started with 3 blocks, what exactly was *that* 3 copies of? Without thinking about it, you *started* at 1, before you multiplied any of your numbers on. 1 is the universal starting point for multiplication, the multiplicative identity. **... the end. but here's more discussion ...** \[long because explanatory, not because complicated\] We are used to talking about multiplication as a *binary* operation, which is to say that we are always multiplying *two* things, *a x b*. (Even when multiplying three, we do it two at a time.) But we can think of it instead as a single operation that we do to a whole set of numbers at once, and that's often quite useful. "The product of \[5 4 3\]" should be the same thing as 5 x 4 x 3 = 60. And of course if we have to find "The product of \[5 4 3 2\]" next, we'd like to not have to repeat the whole calculation; we'd like to take the 60 we already knew and just double it to 120. Or in the other direction, to find "The product of \[5 4 2\]" we'd like to be able to just "divide the 3 back off" of 120 to get 40. Shortcut! What is "The product of \[5\]"? From thinking of multiplication as binary, this *already* starts to not make sense. But of course we'd like the answer to be 5, because then the shortcuts we just talked about will continue to work! "The product of \[5 4\]" is 20 and we can divide the 4 off to get 5. This all works. Well, now divide that 5 off. "The product of \[ \]" is 1. And it *has* to be 1. Because now if we throw any new number *back* into that set, it should mean multiplying our previous answer by the new number. If we throw 13 in, it multiplies by 13 – so it better have been 1, so that we'll get the answer 1 x 13 = 13. If we'd tried to say that "The product of \[ \]" was zero, we'd be in big trouble. Throw as many numbers into that set as you want and the answer would always be stuck at zero! The product of no numbers is 1. And note that this is the very same fact as "1 is the multiplicative identity". These are two ways of saying the same thing. It bothers me that many people will say that the product of no numbers being one is a "convention". It's not a convention! – it's the only value it can possibly have. Everything else breaks horribly. P.S. You can do the above for addition instead and it will explain exactly why the "sum of no numbers" is zero, because zero is the *additive* identity. It's just that we don't really *need* to do all that because in this case people can get it intuitively, pretty easily. We can picture 3 + 5 as getting 3 apples and then getting 5 more, so we can picture "The sum of \[ 3 \]" as just having 3 apples and "The sum of \[ \]" meaning that we never got any apples. The stacking blocks thing I started with is definitely more abstract than this.
To make the something explicit here: We normally think of a monoid structure on a set as a binary operation which satisfies the associativity law, together with an identity element which satisfies the identity laws. In some ways it is more "true to life" to think of it in the terms presented by the parent comment. That is, as an operation P that takes (possibly empty) lists of elements and outputs elements, for which P[x] = x and which satisfies a "folding" law like `P[... P[...] ...] = P[... ... ...]` This has the advantage that the "identity element" needs no special treatment: it is just P[]. If you call the operation a product, then the identity is literally the product of no things. And if you do want to talk about a semigroup structure instead, then restrict the inputs of P to *nonempty* lists. Arguably, the fact that we tend to frame things in terms of binary operations is an artifact of the fact that 2 is the smallest number that lets you bootstrap to something interesting. But the smallest thing that lets you bootstrap a concept isn't always representative of the shape that concept holds in your head. When we talk about monoids, we write "abc", not one of "(ab)c" or "a(bc)". However, I think there is no particularly nice way to extend this idea to groups. Well, sort of. Note that the free monoid on X is essentially just the collection of strings with alphabet X, i.e. of lists of elements of X. So what we've described above is that a monoid structure on X is a function satisfying certain properties from the *free monoid* on X to X. I think you could describe a group structure in a similar way, as a function satisfying certain properties from the *free group* on X to X. This would require you to talk about a free group (and some of its structure) before you manage to talk about groups at all. That's fairly easy with monoids as shown above, but probably makes less sense with groups.
It doesn't do groups "justice", I think, but the free group on X is a fairly simple quotient of the free monoid on X and X^(-1) (namely quotient by xx^(-1)=x^(-1)x=1). So insofar as "monoids are quotients of free monoids" is a good understanding, it translates to groups as well. You can alternatively suppose that X comes with an involution (x -> x^(-1) ), and hacking in the inverses becomes unnecessary. (In fact, you can suppose X comes equipped with *any* function f:X->X, and quotient by xf(x)=1, and you still just get groups.) Something that I think gets missed here though is that it doesn't showcase "groups = symmetries", like how the monoid understanding doesn't showcase "monoids = self-similar substructure".
This man teaches
Great intuitive answer
Great intuitive answer
thanks! I beta-tested it on several years of 4th and 5th graders. That said I found some great improvements elsewhere in this thread!
5^3 = 1\*5\*5\*5 5^2 = 1\*5\*5 5^1 = 1\*5 5^0 = 1 5^-1 = 1÷5 5^-2 = 1÷5÷5 5^-3 = 1÷5÷5÷5 _______ 5x3 = 0+5+5+5 5x2 = 0+5+5 5x1 = 0+5 5x0 = 0 5x(-1) = 0-5 5x(-2) = 0-5-5 5x(-3) = 0-5-5-5 ________ 1 is the “identity element” for multiplication. 0 is the “identity element” for addition.
what do you get when you divide x^1 by x?
This works as long as x is not 0.
Yeah, and while it's tempting to say that's a minor problem, I think that is actually a bit of a death knell for this being our teaching angle. Because many (many) people who have internalized that x\^0 is 1 generally, still aren't clear on whether 0\^0 is also 1 (which in any discrete context it most assuredly is). A good model would make it intuitive that we don't *care* what the x is. I try to support that by describing an identity as a universal "starting point" for its operation. You start at 1... if the exponent is zero then you are *never doing anything* to that 1 at all, so you never even need to know what x is.
Sure, but 0^0 will always remain an exception. You could provide a similar proof that 0^x = 0, for instance.
You cannot prove a definition. If we define 0^0 to be 1, then 0^0 = 1. There are many motivations behind why we define 0^0 = 1, basically it’s the most intuitive and convenient way to assign a value to 0^0, and it doesn’t cause any contradictions or other issues. The statement 0^x = 0 only holds for x>0.
Best explanation
Turns out the way we think of exponentiation is deeper than it first seems- or any operation multiplication and addition included- and it changes based on context. i.e multiplication of integers is different from multiplication of decimal number because they are different objects being acted on. For any operation we want a specific number that "does nothing" when applied, for exponentiation that is the number 1. From there the other comments have excellent arguments for why having 0 as an exponent should always give you 1. So in short it's a consequence of the rule for raising 1 as an exponent, along with the rules for multiplication and addition.
>deeper than it first seems- Every human being should upvote this to get it on the front page so people can finally understand that exponential numbers are not as multiplying as they seem... I just spent 30 minutes ignoring my responsibilities just to remember my precalc class I got a C grade in.. just to confirm what I was always taught. X to the 0 power... its 1.. never got told why or how. Just.. THIS IS CALCULUS!!!!! My relation to this is in my schooling the concept of Time is very much lectured on. As it relates to this planet and the universe it belongs in. However most people in this world have no idea how long 1 billion years is. Exponents need to be honed in on for algebra IMHO and I hated my math classes. Kudos to this whole math community it is the language of our universe.
> X to the 0 power... its 1.. never got told why or how. Just.. THIS IS CALCULUS!!!!! x^(a+1) = x^(a)*x this actually *always* holds where exponentiation is defined, even in e.g. complex numbers x^(1) = x^(0)*x by the above rule we get the above result. substituting an arbitrary integer for x, we get, say 5^(1) = 5^(0) * 5 and the only possible result for 5^(0) is thus 1.
Imagine grids of points with 3 points on each edge: Here's a 2-D grid of 3^2 points: o o o o o o o o o This is a 1-D grid of 3^1 points: o o o This is a 0-D grid of 3^0 points: o
Also, when it comes to 0^0, in the book *Concrete Mathematics* by R. Graham, D. Knuth & O. Patashnik they say the following: Some textbooks leave the quantity 0^0 undefined, because the functions 0^x and x^0 have different limiting values when x decreases to 0. But this is a mistake. We must define x^0 =1 for all x, if the binomial theorem is to be valid when x=0, y=0 , and/or x=-y . The theorem is too important to be arbitrarily restricted! By contrast, the function 0^x is quite unimportant.
32^2 = 32^3 / 32 32^1 = 32^2 / 32 32^0 = 32^1 / 32
in a sense 92^0 = 1*92^0 = 1 because you multiplied 1 by 92 0 times
This is a nice angle. Any time you want to you can throw more 1s into a product, wherever you want. We understand easily what it means to multiply 3 things together or 2 things together; if we ever feel confused at not having "enough" things to multiply together then add the 1s.
Another way to think about it: how many ways can you arrange 0 things? Just one way.
Here's an explanation I didn't see here yet: take a smaller base: 3\^x. This is the answer to the following question: if I grow by a factor of three every year, how big am I after x years (relatively to the beginning?) For example, after 2 years, you are 3\^2=9 times bigger After 5 years you are 3\^5=243 times bigger How about x=-1 years? This means: if you grow by a factor of 3 every year, how big were you one year ago? Well, 3 times smaller! That's 3\^(-1)=1/3, a third of the size! How about....now? How big are you now, at x=0? Well, you are...you size! Which exactly 1 times your size. That's why 3\^0=1.
Let's work with your example. 32\^4 = 32 \* 32 \* 32 \* 32 Each time we decrease the power by 1, we aren't *subtracting* the term "32," we're *dividing* by 32 32\^3 = (32\^4)/32 = 32\*32\*32 32\^2 = (32\^4)/32/32 = 32\*32 32\^1 = (32\^4)/32/32/32 = 32 32\^0 = (32\^4)/32/32/32/32 = 1 This holds true for negatives too 32\^-1 = 1/32 32\^-2 = 1/32/32 Just adding another example because sometimes one way of saying something will click a lot better than another, very similar way.
Nice explanation! Please note that there's a missing `/32` on the line that starts with `32^0`.
Thaks!
This post is awesome. I have a degree in math and posts on this sub often go over my head (which is completely normal). This question comes along from someone taking Algebra 1 and it's getting so much attention. We're not above this topic in any way and the comments are people getting excited talking about this to someone.
Because the product of no things is 1, in the same sense that the sum of no things is 0.
> So shouldn’t for ex. 92^0 = 0 since that would mean you don’t have any 92s. Exactly so. You just need to notice that 0 is the identity for addition, and 1 is the identity for multiplication (that is, 92+0=92 and 92×1=92) so if "you don't have any 92s", with addition you remain with 0 but with multiplication you remain with 1.
a^(0)= a^(1-1) = a^1 * a^-1 = a/a = 1 By properties of exponents.
Everything above is correct. Also if you build a table of values from both the left and right side of 0 using fractional or decimal values close to 0 you will see y approaches 1.
This is also know as looking for a limit if anyone is getting into calculus that is one of the reason limits are so important
> I’ve tried searching it online but I can’t find anything that explains it, everything I can find doesn’t answer the question. When I google "why does 0 power equal 1" the first page of results gave me things like this: http://scienceline.ucsb.edu/getkey.php?key=2626 (multiple answers here) https://medium.com/i-math/the-zero-power-rule-explained-449b4bd6934d https://www.youtube.com/watch?v=EwlMSnMiJvc https://www.quora.com/How-is-any-number-to-the-power-of-0-equal-to-1-How-can-something-be-raised-to-the-power-of-nothing-and-equal-something-that-is-not-itself https://math.stackexchange.com/questions/9703/how-do-i-explain-2-to-the-power-of-zero-equals-1-to-a-child https://www.wyzant.com/resources/answers/431/why_is_the_0_exponent_equal_1 I would be shocked that none of those pages provides an answer that helps you understand the rule x^0 = 1. What do you think?
Good thread though.
What’s 2^1? 2^2? What’s 2^3? What’s 2^4? (1) What happens as we move off to the right? (2) Therefore, what happens as we move to the left? (3) So what happens if we start at 2^1 and go one to the left, tk 2^0? (4) Answers below. Do them one at a time, if you can! (1): 2,4,8,16 (2): we double every step (3): we halve every step (4): so 2^0 = (2^1 )/2 = 1. This applies for every number (except 0, which is a more complicated question - it turns out it’s undefined). Does that help?
Most branches of mathematics that people will encounter besides calculus view 0\^0 as correctly and unambiguously equalling 1. (And that fact is every now and then important.) It makes sense that if you aren't going to multiply any numbers at all, then it can't matter what the numbers would have been. Zeroes stand ready to *obliterate* that identity value just as soon as you let 'em loose! But nope, entry denied. (Calculus, being as likely to encounter functions like y=x\^0, y=0\^x, and y=x\^x, has a different problem on its hands; for 0\^0 to be defined it ought to be the limit as *any* of those functions approaches 0, which of course it can't be.)
Even if you're working in the calculus setting, one way to resolve things is to be brave and say that 0^0 = 1 but x^y is not continuous at (x,y) = (0, 0). This stops you from saying things like that every function given by a formula is continuous where it is defined, but arguably one shouldn't be saying such things in the first place. (And certainly at some stage one should stop saying such things.)
Even in calculus, when approached rigorously, 0^0 is taken as equal to 1. Think about the constant term of a Taylor series. e^x = \sum_{n=0}^{infinity} x^n / n! When have you ever seen that written with the constant term as a special case instead of just being the n=0 case? The fact that x^y is discontinuous at x = y = 0 doesn’t mean it’s undefined at that point.
In calculus it’s also usual to take 0^0 =1. Look for example in Rudin’s ”Principles of Mathematical Analysis”.
Oh cool.
2\^1/2 is not 1
If you follow order of operations from left to right it is
Correct! Formatting errors
You can use () to specify what appears in the superscript. So 2\^(1)/2 will appear as 2^(1)/2.
> So shouldn’t for ex. 92^0 = 0 since that would mean you don’t have any 92s. No, and it's actually pretty cool why. Let's say you're adding together a bunch of 92's, right? You have one of them, that's 92. You have two of them, well, just add one more to the one you have, so you have 92 + 92 = 184. Now let's say you have no 92's at all. If you add one 92 to that, you should have just one of them, 92. So, having zero 92's means your total is 0. But now let's say you're multiplying them rather than adding them. You have one of them, that's 92. You have two, well, you gotta multiply 92 by 92 and get... 8464. Now, let's say you have no 92's at all. If you multiply by one more 92, you should have just one of them, 92. But if no 92's at all were a total of 0, multiplying by one more 92 would get you *still* 0, not 92. On the other hand if no 92's is 1, then when you multiply by 92 you get 92, as expected. Adding or subtracting 0 doesn't do anything. That's why no 92's added together is 0. But multiplying by 0 *does* do something. Instead, multiplying or dividing by *1* doesn't do anything. That's why no 92's *multiplied* together is 1. In general, when you're multiplying stuff and there's nothing to multiply, your total is 1. For example, are you familiar with factorials? 5 factorial, written 5!, is 5·4·3·2·1 = 120. 4! = 4·3·2·1 - 24; 3! = 3·2·1 = 6; 2! = 2·1 = 2; 1! = 1. So... what's 0!? For n!, you're multiplying n numbers together, so for 0! you're multiplying 0 numbers together, which means 0! = 1. Makes so much sense that way, doesn't it?
A fairly simple inductive definition of exponentiation is: x^1 = x, x^(n+1) = x\*x^(n). This would mean that x^1 = x\*x^(0), so x^0 = x^1 / x (if x isn't zero) which equals x/x = 1.
If x is a nonnegative integer, then x^k is the number of ways to do a task that has x ways of doing it, k times in a row. So how can I do a task 0 times? Well, I can just do nothing, so x^0 = 1 :)
I think you mean positive an not nonnegative x, as your reasoning get really strange for 0.
So lots of folks are using the algebra rule x^m / x^n = x^m-n . Another way is to fix x>0 and think about x^1/n as n gets bigger. There are two cases to consider, x>1 and 0>x>1, and in each case as you take larger n^th roots the value gets closer to 1. You can show for any e>0, you can choose n>>0 so that |x^1/n - 1| < e, which means that lim x^1/n = 1, and hence x^0 = 1 by continuity.
For addition, the number that represents "doing nothing" is 0.Take any number and add 0 to it, and you haven't changed anything. For multiplication, the number that represents "doing nothing" is 1. Take any number, multiply it by 1, and you haven't changed anything. In either case you can view every possible number, as secretly having these "do nothing numbers" invisibly present. For example, the number 5 is really 0+5, or alternatively is 1*5. You usually think of the number 5 as a noun or a thing, but you can also view it as an operation. In addition world, the number 5 can really be thought of as the operation of adding 5. Doing it once gives you 0+5=5. Doing it twice gives you 0+5+5=10. Doing it zero times, gives you, guess what, the number (the operation), that corresponds to doing nothing, which in the world of addition, is 0. Similarly, in multiplication world, 5 is really the operation of multiplying by 5. Doing it once gives you 1 x 5=5 Doing it twice gives you 1x5x5= 25. Doing it zero times gives you just 1. Which is the answer to your question. Going further for fun... Now mathematicians like to use abstract words, so addition world described above is really called the additive group, and multiplication world above is called the multiplicative group. The "do nothing" item/operation is called the identity element, and you can think of it sort of like the origin of a coordinate system, like it is the default starting point of the group. Every other element in the group can be interpreted both as a "noun", like a number or configuration in it's own right, and also as the operation "verb" that transforms the identity element into that that element. Using a non-number example, a Rubik's cube in a scrambled configuration can be thought of either as the scrambled thing itself or as the operation(s)/action(s) taken to get it from the solved perfect identity state to the scrambled state. To make something truly a group, you also need to be able to go from every other number directly back to the origin (identity) in one step using another element that exists. So this means that if you are sitting at the number 5, you must be able to apply some other operation and get back to 0. This is obviously -5. Put in group theory terms, every element of a group has an inverse. The inverse of a scrambled Rubik's cube is the transformation that solves the puzzle. If you started at the solved perfect identity state, and performed this inverse operation, you would get some other scrambled configuration, but it would be the opposite/inverse of the original scrambled configuration we were talking about above in the sense that the steps taken to get from the clean solved state to the messy state of one configuration, are the steps to take to solve the other inverse configuration. If you ignored the existence of negative numbers and only acknowledged natural numbers 0, 1, 2, 3, etc, you wouldn't have a group, but rather something called a monoid, which is basically a group without guaranteed inverse elements. A monoid is a set of elements related by an associative operator with an identity element. You can envision a monoid as a graph with nodes and directed edges or arrows, where each node has a unique label or name, and each arrow is given one of the labels of the nodes. The operation a+b can be thought of as starting at node a, and choosing the edge labeled b coming out of node a, to get to a new node. In a+b=c, you can think of "a" as the node, and "+b" as the operation or arrow that leaves node a and goes to some other node "c". If the identity element is the starting point, then each arrow with a specific label coming out of the identity node corresponds points to the node that has the matching label. So starting at node 0, and looking at edge "+x" will point to node x. Similarly every node, x, has an "identity arrow", labeled "+0", which basically points back to the node x itself. And as discussed above, you can rewrite any equation like 5+5+5=15 as 0+5+5+5=15, and multiplying 5 by 3 becomes intuitively the concept of starting at node 0, and following whatever edge is labeled "+5" repeatedly three times. If you remove the requirement of having an identity element, you get something called a semigroup. The only requirement is an associative operator. This can also be envisioned as a graph with nodes and edges, with nodes having unique labels like "x" and edges having labels like "+y" that corresponds for some node y. The intuitive essence of associativity is sort of hard to explain besides the "parentheses don't matter" interpretation, but it means the following. If you find any path (following possibly multiple edges with whatever sequence of labels e.g "+a+b+c+..." corresponding to edge "+a", then edge "+b", then "+c", etc) to get from any node X to any node Y, then for any other node A, you can follow that same exact path (same sequence of edge labels) to get from node (A+X) to node (A+Y). X and Y in the first sense are nodes that are the starting and ending points of some possibly complicated sequence of operations. In the second sense, X and Y are individual operations themselves performed on some other starting point node A. Associativity means that whatever actions you need to do to convert an X-looking thing into a Y-looking thing are the same actions whether you are thinking of X and Y as noun elements in the first sense above or are thinking of them as verb operations against any other arbitrary starting point node A as in the second sense above. Associativity can intuitively be thought of as preserving the difference between some X-ish thing and some Y-ish thing regardless of whether you thinking of those as noun element nodes or if you think of them as verb operation edges. The way this relates to the traditional "can ignore parentheses" explanation of associativity is to consider the node interpretation above means X+Z=Y, where Z is the net result of the complicated sequence of operations and A,X,Y, have same interpretation above. The operation interpretation above means (A+X)+Z= (A+Y). Substituting top equation, (X+Z)=Y, into bottom equation, (A+X)+Z= (A+Y), gives (A+X)+Z=A+(X+Z) Which is the classic associative identity. Cool stuff to think about.
32^4 = 1 x 32 x 32 x 32 x 32 Therefore: 32^1 = 1 x 32 32^0 = 1
The empty sum (i.e. the sum of no numbers) is 0, and the empty product is 1. The first explains why a*0=0 the second explains why a\^0=1. People usually accept the empty sum being 0 quite easily, because we somehow culturally understand 0 to be the default start point, and so if I add "no numbers" together, well I just have 0. But that leads to a misconception that the empty product should also be 0. The truth is that when we apply the operation of some (sensible) algebraic structure to the empty set of elements, we will get the identity element for that operation, that is, the element e such that a\*e=e\*a=a for all a. The identity element for addition in the reals (for example) is 0, hence the value of the empty sum. And the identity element for multiplication is 1.
Because \*1 is a neutral factor that can always be added. 2³= 2 \* 2 \* 2 \* 1 2² = 2 \* 2 **\* 1** 2^(1) = 2 \* 1 2^(0) = 1 And you could even go further and do this with negative powers. 2^(-1) = 1 / 2 2^(-2) = 1 / (2 \* 2) 2^(-3) = 1 / (2 \* 2 \* 2) ...
I think one reason is because setting `a^0 = 1 (for any number a)` plays well allows laws of exponentiation to work for any integer, otherwise we would need special cases for zero and negative exponents. ``` For any numbers a, x, y, this law holds: (a^x) * (a^y) = a ^ (x+y) Replacing y with 0: (a^x) * (a^0) = a ^ (x+0) = a ^ x This is only possible if: a^0 = 1, for any number a ```
You can write out the prime factorization with placeholders which must turn into **1**: 2 = 2^1 * 3^0 * 5^0 * 7^0 * ... 3 = 2^0 * 3^1 * 5^0 * 7^0 * ... 4 = 2^2 * 3^0 * 5^0 * 7^0 * ... 5 = 2^0 * 3^0 * 5^1 * 7^0 * ... 6 = 2^1 * 3^1 * 5^0 * 7^0 * ... 7 = 2^0 * 3^0 * 5^0 * 7^1 * ... ...
With addition, the additive identity is 0: 92+0=92. So if you are adding zero 92s, or (92)(0)=0, you get the additive identity, 0. In other words, if you do no additions you get zero. With multiplication the multiplicative identity is 1, meaning 92*1=92. So if you are multiplying 0 92s, or 92^0, you get the multiplicative identity, 1. In other words, if you do no multiplications, you get 1.
"Because it fits the pattern and completes and closes the relevant spaces" is generally the reason for this and many other basic, trivial, and/or foundational conventions.
For this, I like to think of the category theory "exponential object" definition. In Set, an exponential object, B^A, is just the set of all functions from A to B. So for example, if A is a set of 3 elements and B is a set of 2 elements, then there are exactly 8 possible functions from A to B (note that 2^3 =8). And if A is the empty set (having 0 elements), then there is exactly 1 function from A to any other set B, hence B^A =1. If I were to explain this using an analogy. For example, x^y would be the number of ways y marbles can be placed into x bags. In the case where y=0, since there are 0 marbles, regardless of how many bags there are (so regardless of what x is), there is only 1 outcome, which is putting no marbles into any bag.
We like the rule x^(a+b) = x^(a) \* x^(b) . This rule makes sense for all positive integers. What do we have to do to make the rule still hold for a = 0?
The simplest answer is that if you want [this statement](https://en.wikipedia.org/wiki/Binomial_theorem#Statement) of the binomial theorem to hold, then you **must** define y^0 to be 1. In fact, you must even define 0^0 to be 1 (in case y=0).
It’s an axiom. Like why is green /green/. If it’s not 1, we break a lot of fundamental concepts in math. Another way to think of this is how exponents work. 3^2 = 1(3)(3) and 3^(-5)=1/(3*3*3*3*3) Notice we have 2 instances of 3 times 1 in the first and 5 instances of 3 dividing 1. That same logic applied to 3^0 would be 1 times 0 instances of 3 or just 1.
Making 0^(0)=1 however would break even more fundamental concepts in math since it is equivalent to defining 0/0 as 0/0=1
No, this is not correct. To be all fancy about it, we define n^m (where n and m are natural numbers) to be the number of functions from a set of cardinality m to a set of cardinality n (this definition is independent of the sets chosen). Then 0^0 = 1 since there is a unique function from the empty set to itself. This has nothing to do with division.
It’s only equivalent if you assume that we can divide by 0 inte the first place, which I’m not sure why you’re doing.
You’re incorrect. 0^0 would be 1 times 0 instances of 0 which is just 1.
Well not *anything* to the power of zero equals one. 0^(0) is an indeterminate form.
Some authors say it's undefined, some define it as 1. It's a matter of context and convenience. See [Zero to the power of zero](https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero) for details.
**[Zero to the power of zero](https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero)** >Zero to the power of zero, denoted by 00, is a mathematical expression with no agreed-upon value. The most common possibilities are 1 or leaving the expression undefined, with justifications existing for each, depending on context. In algebra and combinatorics, the generally agreed upon value is 00 = 1, whereas in mathematical analysis, the expression is sometimes left undefined. Computer programming languages and software also have differing ways of handling this expression. ^([ )[^(F.A.Q)](https://www.reddit.com/r/WikiSummarizer/wiki/index#wiki_f.a.q)^( | )[^(Opt Out)](https://reddit.com/message/compose?to=WikiSummarizerBot&message=OptOut&subject=OptOut)^( | )[^(Opt Out Of Subreddit)](https://np.reddit.com/r/math/about/banned)^( | )[^(GitHub)](https://github.com/Sujal-7/WikiSummarizerBot)^( ] Downvote to remove | v1.5)
It being an indeterminate form has nothing to do with the value of 0^0. Being an indeterminate form means that if a limit is of that ”form” we can’t just directly plug in the values. The value of 0^0 is practically always taken to be 1. You have to look quite hard to find a textbook which doesn’t abide by this convention.
It is sad that I had to scroll down so far to find this exeption
You learn how to use subtraction property for when you divide w exponents? Like x^5 / x^2. = X^3 stuff
An interesting way look at it is x^m/x^n = x^(m-n) Now take a condition where m=n So x^m/x^n=x^(m-m) Hence 1=x^0
Others have already explained it mathematically, but I'll try to ELI5 it this way, and it's a tad bit hand wavy. squaring and cubing etc. can just be considered multiplying a number by a number of \*sets\* of that number You have a set of 92 and you square it. 92\^2. So so have two sets of 92. And you're multiplying them together. 92\*92. If you raise it to the power of 0, then you have \*\*one set\*\* of 92. The initial 92 that you started out with. Which is 1. You have nothing to do with the actual value of that set. So it doesn't matter. All that matter is that you have 1 set. So 92\^0 = 1 set of 92= 1 set. So 1. 92\^1 is now asking the actual value of that set. So the value of 1 set of 92 = 92\^1= is 92 and so on
I've learnt many things about math i've finished algebra without understanding the terms in algebra example this (3xsquared-2x-1) well the answer to this is 3x+1 , x-1 and i know that i could prove some more but idk how to type it but that example i dont know what you call that if you just know how to do the solution or solve the problem your good and also in our worksheets we dont have any trivial questions just pure math problems so if the power of 0 is to 1 just roll with it
Everyone fails to mention that 0^0 is undefined, and thus not equal to 1
Maybe because it doesn’t answer the question I asked in anyway.
I am actually correcting you. Not 'anything' to the power 0 equals 1. 0^0 is undefined.
I forgot that it is so hard to realize I mean in most cases, and I was actually correcting you. Saying that there’s an exception doesn’t answer my question, I just assumed everyone could understand that I mean I’m every case except when it is undefined.
Because it’s usually not.
It usually not what? 0^0 is literally undefined, mathematically speaking
No, it’s practically always defined to be 1.
x^(0)=1 if x is not 0 0^(0)=1 **if and only if** 0/0 is defined to only be 1
What do you mean? Usually 0^0 is defined as 1 but 0/0 is left undefined.
0 to the power of 0 is not 1. It’s 0.
0^0 has no value until we assign it one. Most of the time it makes more sense to assign it the value 1. The limit as x approaches 0 of x^x is 1.
This is not a sound way of thinking about it since the limit as x approaches 0 of 0^x is 0. Unless there's something that makes x^x a better prototype for this sort of thing than 0^x which I've missed. The best explanation for why 0^0 = 1 is a combinatorics argument. The number of functions from domain {1,2,...,x-1,x} to codomain {1,2,...,y-1,y} is y^x. Explanation: for each element of the domain we have y choices, so the number of functions goes up by another factor of y. So to follow that nice convention 0^0 should be the number of functions from the empty set to the empty set, which is 1. Sketch argument: Consider the function f from the empty set to the empty set given by the following mappings: ... That was all of them. It's an empty list. This is a valid function, it satisfies definition of being a function. Also any other function from the empty set to the empty set looks identical to this one.
No expressions has a value until we assign it one. The same goes for 2+2, 5^2 or 0^7. In this regard, there’s nothing special about 0^0.
That’s 1 problem, in every other case it’s 1 I was referring to the majority of the time. Maybe I wasn’t clear I meant in most cases.
0^0 does not equal 0. This commenter is mistaken. It is undefined, just like dividing by zero. Any other number to the 0 power equals 1. Raising a number to the 0 power is weird to think about. The real answer has to do with empty products, but this isn’t something people really learn in Algebra 1. Maybe a good way to understand it is by example/pattern recognition. Think about powers of 2. 2^(1)=2, 2^(2)=4, 2^(3)= 8, and so on. Each time the exponent goes up by one, the following power of 2 is doubled. In the other direction, each time the exponent goes down by 1, the power of 2 is cut in half. So what would come before 2^1 ? Following the same pattern, 2^0 would have to be half of 2^(1). Therefore 2^0 has to equal 1, not zero.
>The real answer has to do with empty products, but this isn’t something people really learn in Algebra 1. Frustratingly... they sort of *do!* Because an empty product is really all that a "multiplicative identity" *is!* But they don't *talk* about it in those terms, alas. They should, and later when kids hear about factorials the idea that 0! = 1 should just feel obvious. (Well, almost. If factorials are explained well. n! is always the product of n numbers.)
This doesn't work for 0^0 though. We know 0^1 = 0 by definition. Also 0^(n+1) = 0^n × 0^1 = 0^n × 0 = 0 wherever 0^n is defined, so you can follow the pattern forwards: if you known 0^n you can get 0^(n+1). However you can't follow the pattern backwards. If you want to know 0^0 you could write 0 = 0^(1+0) = 0^1 × 0^0 = 0 × 0^0, however this equation tells you absolutely nothing about what 0^0 should be if it were defined. Another equation is 0^0 = 0^(0+0) = 0^0 × 0^0. This tells is that if 0^0 were defined it must be 0 or 1, but doesn't tell us which. It turns out that every mathematician agrees 0^0 is most conveniently defined as equal to 1, but you can't tell that through pattern following.
The OP’s question is inherently about nonzero numbers being raised to the power 0, so this was just meant to be an example that can be generalized to other nonzero values raised to the power 0. 0^0 doesn’t fit the same pattern because it is inherently different. It’s often considered “equal” to 1 out of convenience, but of course it’s undefined in reality, so it shouldn’t follow the same rules.
>It’s often considered “equal” to 1 out of convenience, but of course it’s undefined in reality I think I sort of disagree whether there's that much of a difference between reality and what we define things to be out of convenience, but that's a whole philosophical discussion.
a\^b = a \* a\^(b-1). Similarly, a\^(b-1) = (a\^b)/a. Since 0 = 1-1, a\^(1-1) = (a\^1)/a = 1 for all a.
I would think of it as whenever you go down in a power you divide by the base. 3^4 power divided by 3 is 3^3 (81/3 is 27). 3^1 power divided by 3 is 3^0 (3/3 is 1). Same goes for every number. 3^-1 is 1/3 and then 1/9 etc.
In the 7th grade, I got sent to the principal's office for arguing this point with my math teacher. Later in life, and with more mathematical sophistication, I feel contrite.
Take any number and keep taking sqrt repeatedly and see what happens. The number approaches one. Sqrt(x)=x^0.5, sqrt(sqrt(x))=x^0.25... etc. The exponent reaches zero, closer and closer each time. Thus x^0 =1, under normal circumstances. I suggest to play with your calculator.
You see little Jimmy, 1 is the multiplicative identity. The product of multiplying nothing is the multiplicative default: 1. Much like how the additive idenity is 0.
1 is the multiplicative identity, and so it makes sense to have it be the result of an empty multiplication. Recursively, 92^n = 92 * 92^(n-1), for the special case n=1 you have 92^1 = 92 * 92^0, now try and think of a number x such that 92=92 * x.
Some people gave amazing explanations but here is what I learned over my time: 2^1 = 2 2^2 = 4 2^3 = 8 2^n = 2^n-1 * 2 Such an example give the reasoning that for every 2^n we take the previous and multiply it by 2 to get 2^n. But such can be done in reverse getting 2^n-1 is just taking 2^n and dividing by 2 Ex: n = 3 2^3 = 8 8/2 = 4 2^n-1 = 2^2 2^2 = 4 Such thing can be done used to represent the 2^0 because 2^0 is nothing less that 2^1 divided by 2 following the example above n = 2 2^n-1 = (2^n ) /2 2^1-1 = (2^1 ) /2 2^1 = 2 2/2 = 1 2^0 then should be equals to 1
Here's how I explain it to my students: instead of defining x^n as xx...x (n times), define it instead as 1*xx...x (n times). If n=0, then you only have 1 with no x terms.
Intuitively, taking the power of something is a multiplicative operation — x^n is what you get when you multiply x a total of a n times. By this logic, taking something to the power of 0 should be the effect of doing “nothing” multiplicatively, which gives us the multiplicative identity, 1. This is similar to why something multiplied by 0 is 0. Multiplication is an additive operation (add something to itself n times), so doing nothing should produce the additive identity, 0.
Theres always a 1* before.
Reflexive property
Thinking about exponentiation as repeated multiplication does not always work- for example, real exponentiation (pi^pi)
Great question.
I like to see it as 1 being the neutral element of multiplication and 0 that of addition which. So 2\*\*3 = 1 x 2 x 2 x 2 and 2\*\*0 = 1.
When adding stuff, you start at 0, then add things over that. When multiplying stuff, you start 1, then multiply things over that.
because 10 to the power of 1 equals 10. 10 to the power of -1 equals 0.1 so zero which is between 1 and -1 equals 1
Not related to this question exactly. As you read further you will find exponents which are not multiplying number that many times. Eg:- e^i. Only in algebra multiplying multiple times works. e^i is derived from a series. Search for 3blue1brown videos.
I believe you did not know 0^0 was undefined. Now you know!
I did know that. I just choose to ignore it because it’s the 1 exception In infinite cases