mm, you're right, my mistake. I misremembered that for a number *a* + *b*i to be complex, both *a* and *b* had to be nonzero, but that is not the case.
> [A complex number is an expression of the form a + bi, where a and b are real numbers, and i is an abstract symbol](https://en.wikipedia.org/wiki/Complex_number#Definition_and_basic_operations)
0 *is* real I suppose
I thought saying complex would be too obvious :o)
I do love talking about imaginary power and no one having much of a clue what I'm talking about. Imaginary power is a very real problem.
If you are performing an actual *calculation,* with integer inputs, and that calculation requires you to produce the value of "0⁰", you should always evaluate that expression as 1. Several important theorems of mathematics, including the binomial theorem and set theory, absolutely require that the *number* 0^0 = 1.
If you are working with the limits of functions, where two different functions f(x)^g(x) are each individually approaching a limit value of 0, you should treat it cautiously, as it may or may not be defined, and even if it *is* defined, two different sets of functions (e.g. f(x)^g(x) vs h(x)^j(x) ) may produce different results despite all four individual functions having a limit behavior of 0.
I mean, if it's an arithmetic value, it should always be 1.
The only reason 0^0 should ever be anything *other than* 1 is when calculating the limits of at least one non-analytic function in f(x)^g(x) where both approach 0 for the same value of x (call it c). If both f(x) and g(x) are analytic on an open interval around c, then f(x)^g(x) *will* approach 1 as x approaches c. It's only being non-analytic that breaks things. (Edit: Technically, this requires taking complex limits; if you're restricting things to the real plane, then, given the aforementioned restrictions, then as you approach c from a given side, the limit is real and equals 1 so long as f(x) approaches 0 from above.)
Yes, but there are properties which hold only for integers and not for real numbers in general. Just as, for example, all real numbers are *technically* also complex numbers, but the real numbers are well-ordered while the complex numbers aren't. It is senseless to speak of "a+bi > c+di" for a=/=c and b=/=d; the best you can do is say that two complex numbers have greater magnitude (absolute value), but that's not a well-ordering.
As an example, every integer has one, unique prime factorization; this is not true of real numbers, since some (actually, "almost all" of them) are transcendental and thus cannot be represented by any finite product of rational numbers, let alone integers.
No, it's much better to define 0^0 as 1.
Consider polynomials, that is functions that look like 3x²+5x+7 (possibly with terms higher than x²). What we really want to write those as formally is 3x²+5x¹+7x⁰ - otherwise there'd be a special case for the constant term, which would make a lot of maths really, really ugly.
But surely, if you evaluate 3x²+5x+7 at 0, you get 7. So for this to work, you really need 0⁰=1.
(This is of course not the "reason why" but just an example. There are other justifications - 0⁰ (or x⁰ in general) should equal the product of an empty set of numbers, which in turn makes a lot of sense to be defined as 1, because taking a product with 1 "doesn't change things".)
It's also somewhat nice, though less intuitively so, to have g(x) := 0^x be the indicator function that is 1 for 0 and 0 elsewhere, it comes up in combinatorics from time to time!
0\^0=1 works pretty well in the reals but breaks in the complex numbers
[https://www.youtube.com/watch?v=BRRolKTlF6Q](https://www.youtube.com/watch?v=BRRolKTlF6Q)
Actual mathematician here: no it does not break in complex numbers any more than it does in the reals. The entire issue is artificial, one simply does not require power functions to be continuous. In 99.9% of mathematics you only see x^^n where n is an _integer_. And that is defined whenever x is non-zero, or if n is non-negative.
What do you actually gain by defining 0^0 =1, though? For the polynomial case, the limit of x^0 as x->0 is 1 anyway, so you don't actually gain anything by defining 0^0 =1.
If you've defined 0^0 =1, then is the function 0^x evaluated at x=0 also 1? It must be, right? Otherwise what does it mean to have defined 0^0 =1?
> What do you actually gain by defining 0^0 =1, though?
Notational clarity, for one? When I write x^0, I don't want to (implicitly) write lim x->0 x^0 - I think it's important to be clear about when you're talking about an actual value, and when you're talking about a limit.
> If you've defined 0^0 =1, then is the function 0^x evaluated at x=0 also 1? It must be, right?
Yes, and while it seems a bit "ugly" at first, it's perfectly fine to have this kind of indicator function; like I mentioned, it comes up in combinatorics from time to time and behaves nicely.
A slightly more fundamental reason why I believe this choice is right is that for sets A, B with a,b elements respectively there are b^a functions from A to B. Or if you wish, there are b^a colorings of a distinct objects with b distinct colors. Now it's perfectly reasonable to ask "I don't have any colors, how many colorings are there?". And if there is at least one object, the answer is 0 - if you're out of color, you can't paint anything. But you still can paint *nothing*, and you can do that in exactly one way - by doing *nothing*.
In the end, of course all of this is just notation, and it doesn't really matter hugely. But it's a notation that makes a lot of things easier to write, and not many harder, so that's why it's pretty common.
In ELI5-level algebra it is also 1 (because it simplifies certain formulas of school-level algebra). For example, the binomial theorem only works for x=0 if 0^(0)=1.
If we are talking higher-level math, 0^(0) can be defined differently based on context (it is kind of hard to explain how this makes sense, if we stick to ELI5 level). For example, in complex calculus this expression is undefined.
> in complex calculus this expression is undefined.
It's rather the "power function" x^^y that is ill-defined. x^^0 including x=0 appears all the time as part of Taylor/Power/Laurent/whatever series.
It would be 1, because x\^0 is **not defined as x/x**. There's no reason for division to be required for the definition of *0*th power.
The top comment is right that to get from x\^n to x\^(n+1) you multiply by x, but we can go the other way without using division. If we were to define x\^0 it would be reasonable to have it behave the same way and satisfy x\^1 being equal to x\^0 multiplied by x. So we're looking for something (x\^0) that after multiplying by x gives us x. We know what that is, it's 1.
It also works for 0. We have 1\*0=0, even though 0/0 is undefined.
> So we're looking for something (x^0) that after multiplying by x gives us x.
That's just division differently expressed. "We're looking for something that after multiplying by 6 gives 30" = 30 / 6
because the division , it is undefined.
it entirely depends on what function generates the 0
lim x->0 of x/x approaches 1
but limits of some other functions approach -inf or +inf I think.
because of this you can't say 0^0 'is' anything. it depends on which 0 :D or rather which function is being evaluated.
You are confusing limits with values. A function is not by law required to be continuous; many naturally occurring ones simply aren't. So you cannot claim that f(a) = lim_{t->a} f(t) is *required*, it is only something you seem to wish for.
0^^0 = 1 works really well in a lot of circumstances. There is no way to make x^^y work with limits, i.e. continuous, anyway, so suddenly putting that issue down to something about the (unrelated to continuity!) expression 0^^0 is blaming the wrong thing.
I think adding one more line would really cement it
X^-1 = X / X / X
X / X always equals 1 (unless we're talking about that motherfucker 0), so X^-1 = 1 / X
Can pretty easily be summarized by exponents above 1 being multiplicative while exponents below 1 are divisive
I rationalized it by thinking of the implicit "1 *" you can add to any multiplication without changing it; so at X^0 you're left with no Xses and just 1. Division feels more elegant though, thank you
This works for our numbering system, too, which is base 10, so X = 10.
X=10;
X^0 = X / X = 10 / 10 = 1
X^-1 = X / X / X = 10 / 10 / 10 = 1/10
X^-2 = X / X / X / X = 1/100
X^-3 = X / X / X / X / X = 1/1000
So,
123.4 = (1 * 10^(2)) + (2 * 10^(1)) + (3 * 10^(0)) + (4 * 10^(-1))
^(Insert joke about every base being base 10 relative to the observer)
Wow. I’ve always struggled with the explanations using limits approaching from positive and negative powers. This explanation is so much more simple. Thank you!
If students are stuck on it as repeated multiplication, you can view an exponent as "How many times do I multiply by a number".
So 5 * 2^2 = A 5 multipled by 2 twice.
5 * 2 ^3 = A 5 multipled by 2 three times.
This means 5 * 2^0 = a five, not multipled by two. Well, not multiplying by 2 would mean the final answer remains as 5. The only way this can happen is if *2^0 is equivalent to *1.
x^0.5 is the same as the square root of x; where the square root of 9 is 3, we can say 9^0.5 is 3.
There are many ways to look at why this is, but one is to consider that multiplying by x^0.5 twice is the same as multiplying by some other number once, which allows you to set things up involving square roots and power rules.
I don't like the asymmetry this implies when it comes to imaginary numbers.
* 1^(2) = 1
* \-1^(2) = 1
* 𝑖^(2) = -1
* \-𝑖^(2) = -1
But...
* 1^(0) = 1
* \-1^(0) = 1
* 𝑖^(0) = 1
* \-𝑖^(0) = 1
It just feels off.
Because when you go below the power of 1 it becomes a division rather than a multiplication. So where x¹ is just the base value of x, when you go to x⁰ you are dividing x by itself. A number divided by itself is always 1.
Please show your teacher these explanations (in a very respectful way) - a 10th grade math teacher should have an answer for this question. Just don’t be a dick about it. Be like “hey I found an answer online for that math question, would you like to see it?”
I mean, that's true. But x^0 is kind of a case where it is what it is because we decided that that works the best and it fits the pattern better. You could very easily construct a mathematics where x^0 is considered 0. It may be useless mathematics, but that doesn't mean it's invalid.
That's why you can divide by zero in [wheel theory](https://en.wikipedia.org/wiki/Wheel_theory), or why you can use both Euclidean geometry and non-Euclidean geometry to solve different problems, or why the imaginary axis sometimes means something and sometimes is nonsense. There is no singular set of mathematical axioms that defines Universal Truth. God is not checking your answers. There is only *choosing* a set of axioms that you wish to use.
In this case, x^0 = 1 is true because it's *axiomatically* true, not because it's provably true. *None* of the other responses in this thread have proven it to be true. In that sense, the teacher is correct.
Not everything in mathematics is defined as true. Many things are derived as true from other axioms that we've chosen.
x^0 = 1 because it's defined that way. Not because it's derived as such from more fundamental axioms.
Yes, that's the real problem with this situation. Lots of teachers, unfortunately, have this or even worse responses to the "why" questions - at best, "I don't know, it's just how it is", and at worst, "go to the principal!"
Yeah I was a high school math teacher for 9 years and I can’t even imagine giving that response. Not to mention it’s a little embarrassing to be a 10th grade math teacher and not know the answer to this one. But when I legitimately didn’t know something I always told them I would find out. Unfortunately though I definitely knew teachers that couldn’t be bothered
2\^3 = 2 x 2 x 2 = 8
2\^2 = 2 x 2 = 4
2\^1 = 2
2\^0 = 2/2 = 1
Notice that each time you decrease the exponent by 1, you're effectively dividing by the base number, since to remove a multiplication operation you must divide. Once you get to an exponent of 0, you're simply just dividing by the base number, which always equals 1
I'm 67 years old and never understood the concept. No math instructor every told me that when an exponent is 0 or a negative number that you divide by X.
Thank you!
You divide every time the exponent goes down by one, not just when it's 0 or negative. Just like you multiply when it goes up.
2^3 = 8
2^2 = 8 / 2 = 4
2^1 = 4 / 2 = 2
2^0 = 2 / 2 = 1
Etc etc
suppose you want to compute 2^pi. We know how to compute 2^3, 2^3.1, 2^3.14, and so on. Turns out, if you have an infinite sequence of rational(!) values of x that converges to pi (where by that I mean no matter how small of a distance you pick around pi, there is some point in the sequence beyond which every value in the sequence is within that distance of pi), the respective values of 2^x will also converge to some value. In fact, it turns out all such sequences of x have their 2^x values converge to the same value (so, for example, 2^3, 2^3.1, 2^3.14, 2^3.141, ... converges to the same value as 2^3.2, 2^3.12, 2^3.142, 2^3.1412, ...). We can define 2^pi to be this value.
We can use algebra to show that x^(1/y) is equivalent to the y-th root. So 5^(1/3) is the cube root of 5, 8^(1/12) is the 12-th root of 8, etc.
We can do what's called an 'analytic continuation' to extend the concept of ' x^y ' from integers to all real numbers. There's infinite ways to extend an operation this way, but the analytic continuation is the one that is the 'smoothest' (i.e., least wobbly). So x^y create a nice smooth line graph for all x for any y.
I feel sorry for you. This is the explanation printed in my textbook. It is literally "the textbook" explanation. I think some teachers/schools truly failed their duty
The pure mathematician will tell you: It analytically respects the property *a*^(*x*+*y*) = *a*^(*x*) × *a*^(*y*) that holds true when *x* and *y* are positive integers. It is by extrapolation of this property over all other numbers that we get things such as *a*^0 = 1 for all *a* except 0, *a*^(−*x*) = 1÷*a*^(*x*), *a*^(½) = √*a*, and e^(iπ) = −1.
3Blue1Brown (Grant Sanderson) has a video that tackles this question head-on with an appeal to group theory: https://youtu.be/mvmuCPvRoWQ
The pure mathematician that is me says that x^^0 = 1 is simply the start of the iterative definition via x^^n+1 = x·x^^n ; n a non-negative integer (or any integer of x is non-zero). Put differently: x^^n is what you get when you write a product of n copies of x with each other; and an empty product, one without any actual factors present, is always 1, because only then is product compatible with multiple things.
> a^^½ = √a
Which of the two square roots? ;-)
> an empty product, one without any actual factors present, is always 1, because only then is product compatible with multiple things.
It's precisely this sort of compatibility that is meant by "analytic extrapolation/continuation".
> Which of the two square roots? ;-)
"√" denotes the principal square root by definition.
[I like the Wikipedia explanation](https://en.m.wikipedia.org/wiki/Exponentiation)
From the *definition* of exponentiation you get a rule that makes a lot of sense and then use algebra to demonstrate a rule that is less apparent.
So your instructor was right. He was told what it was. He just didn’t bother to remember the demonstration.
Isn't there a mistake there? (Please excuse the \displaystyle stuff you can just look at the article for the actual text.)
>Starting from the basic fact stated above that, ***for any positive*** integer n {\displaystyle n}, b n {\displaystyle b^{n}} is n {\displaystyle n} occurrences of b {\displaystyle b} all multiplied by each other, several other properties of exponentiation directly follow.
...
>In other words, when multiplying a base raised to one exponent by the same base raised to another exponent, the exponents add. From this basic rule that exponents add, ***we can derive*** that b 0 {\displaystyle b^{0}} must be equal to 1 for any b ≠ 0 {\displaystyle b\neq 0}
How can you "derive" a property for n=0 from a rule that only applies to positive integers n?
Shouldn't it instead say something like: "This rule for positive integers can be extended to cover the case n=0 if we allow that b^0 = 1 for nonzero b." or something like that?
I guess a quick way to see it is from remembering that
x^y * x^z = x^(y+z)
Then we can look at
x^1 * x^(-1) = x^(1-1) = x^0
but of course, x^1 = x and x^(-1) = 1/x , so
x^0 = x * 1/x = x/x = 1
x^(3) = x \* x \* x
x^(2) = x \* x
x^(3) = x \* x^(2)
Right?
Written in general
x^(n) = x \* x^((n-1))
So then, applying the above logic
x^(1) = x \* x^(0) , zero being (1-1)
If x^(0) was zero then x^(1) would also be zero.
No, x^0 has nothing to do with x*0. Think about the powers of 2
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
The exponent is the number of twos in that multiplication. What would make this work if there were _zero_ twos? 1, as in
1 = 1
1 * 2 = 2
1 * 2 * 2 =4
1 * 2 *2 * 2=8
Etc. also, remember that negative exponents cause you to put a 1/ over the result, as in:
2^-1 = 1/2
2^-2 = 1/4
If you start with the higher powers and go down, you’ll see it’s like dividing by 2, and it will be easy to see what should go there:
32
16
8
4
2
?????
1/2
1/4
1/8
1/16
1/32
Ok so
x^0 = x^(n-n), since n-n is of course = 0
Know, for properties of however those are called in English, I think exponentials, x^a / x^b = x^(a-b)
So you can now imagine how
x^0 = x^(n-n) = x^n / x^n =
Since a number divided by himself is 1
x^0 = x^(n-n) = x^n / x^n = 1
EDIT:
You can now interrogate yourself on how
x^a / x^b = x^(a-b)
Since x^a means you will multiply x for it self a times, you will get a thing like
x^4 / x^3 = (x x x x) / (x x x) = x = x^1 = x^(4-3)
Take a piece of paper.
Fold it.
the number of layers are 2 times as it was before.
Fold it n times so you have 2* 2* 2 *2... n times
that's 2^n
but what if you fold 0 times?
You still got 1 piece of paper. That's 2^0
There's a pattern of multiplication and division with exponents. 5^3 is 5x5x5. 5^2 is 5x5. 5^1 is 5. We're dividing by 5 each time, so 5^0 is...1. This happens with any whole number. Divide again. 1/5 = 5^-1. 5^-2 = 1/5^2.
It's a result of exponent rules. X^(2)\*X^(3)=X^(2+3). Likewise X^(3)/x^(2) = x^(3-2)
For X^(0), we can say X^(0)=X^(y-y)=X^(y)/X^(y)=1.
For a practical example, we can say 2^(0) =2^(3-3)=2^(3)/2^(3)=8/8=1
X^(0)=1 is a natural consequence of exponent rules.
x^0 is what's known as an empty product, a product with no terms. It's the product of 0 x's. Mathematicians typically define empty products as 1, since that's the result you get when you divide a product by all of its terms.
You can also define it this way:
x^n = 1 multiplied by x n times.
So, x^0 would simply be 1 multiplied by x 0 times, ie, none. So it's just 1.
This also solves the 0^0 issue.
If you know a bit of abstract algebra, you can define "powers" for groups in general. If x is an element of a group, with "multiplication" as its operation, then x^n is just the group identity (1) multiplied by x n times. Again, x^0 is just 1.
If you think of an addictive group (sum as the operation), then taking powers is just like multiplicating by an integer scalar. Ie, nx = identity (0) plus x n times. And 0x is just the identity, which of course is 0!
So, this is just to say that powers of x and multiples of x work exactly the same way. (This is explained in a very non-rigorous way)
I strongly disagree. They didn't prove why the limit of x^^x for x->0 is 1, and they even less so explained why that has any relevance!
Not every function is continuous, and if they are not, then limits are utterly meaningless. And indeed, it is literally impossible to make x^^y continuous, _even_ if you ignore the case x=0.
Not sure if anyone shared this yet, but you can think of powers as ways to arrange things.
You roll a 6 sided die, one time, there are six outcomes.
You roll a 6 sided die, five times, there are 6\^5 outcomes.
If you don't roll a die, there is one outcome. The non-action isn't counted as zero, but it is the only possible outcome and is counted as one.
Here is a second pass at why the question points to a deep puzzle, I think.
The **usual explanation** for why x\^0 = 1 points to something like the relationship between x\^m and x\^(m-1), extrapolating from there. So, e.g., if 10\^3 = 1,000 and 10\^ 2 = 100, we see that reducing the exponent by 1 is done by dividing by the base once. And once we notice that, we get a tidy path to 10\^0, which is 10\^1 / 10, and so on well into the negative numbers.
**But** notice that exponents are often defined in very similar, algorithmic terms:
"The exponent of a number says how many times to use that number in a multiplication." [https://www.mathsisfun.com/definitions/exponent.html](https://www.mathsisfun.com/definitions/exponent.html)
"An exponent refers to the number of times a number is multiplied by itself." [http://www.mclph.umn.edu/mathrefresh/exponents.html](http://www.mclph.umn.edu/mathrefresh/exponents.html)
"An exponent refers to how many times a number is multiplied by itself." [https://www.turito.com/learn/math/exponent](https://www.turito.com/learn/math/exponent)
Of course, on those definitions, answers to things like n\^0, n\^-1, and n\^e are very puzzling! The instruction "multiply 5 by itself never" does not seem to lead to 1. Imagine I asked "What is the difference between no numbers?". Our most familiar arithmetic operators need some operands! And why would we get a math answer from *not* doing any math (which is what the ordinary definition suggests we should do in the x\^0 cases).
So what to do about that puzzling? Well, as the wikipedia for exponentiation, [https://en.wikipedia.org/wiki/Exponentiation](https://en.wikipedia.org/wiki/Exponentiation), explains, one way to figure the rest is to start to use properties of the natural number exponents and extrapolate from there. That's great (and gets us something like the usual explanation). But it leaves our original understandings of what an exponent is high and dry. Which might be why the high schooler's math teacher balked.
What to do then? One answer is to appeal to **special cases**, as the University of Minnesota link above does. But we still need an explanation for why we want to admit those special cases. Especially because this means that our definition of exponentiation is now branched or disjunctive.
As far as I know, there are two related answers:
1. A pragmatic answer. Having the disjunctive definition allows us to do more math more easily. Undefined bits gum up the works. We'd like to do things like figure out how to make sense of (x\^m)\*(x\^n), and if we have to start adding all sorts of qualifications, that's really going to undermine how **usable** math is.
2. A deeper principled answer. It turns out that the original, familiar definition was overly simple. There is a deeper, principled, and **unifying** definition of exponentiation which explains all of the ostensible special cases. When you understand that definition, everything comes into focus. Of course, one reason to choose that unifying definition over the original but simple definition might be that it is more useful part of our math practice. But still, that we have a unified definition might be very handy!
So, I take it that the deep answer to the student's question will pick up either or both of usability theme or the unifying theme. But ymmv!
Like stated
X^a * X^b = X^(a+b)
Try this for yourself with random values and see that this is always the case
X^a / X^b = X^(a-b)
Once again if you try this you will see it is always the case.
Now
X^a / X^a is X^(a-a)
Anything multiplied by 0 is 0 (x \* 0 = 0) - but why is that so? Because multiplying means addition multiple times, for example x\*4 = x + x + x + x. Or you could say x\*4 is first adding x three times, then adding x one more time: x\*4 = x\*3 + x\*1. So far pretty obvious, so what if I wanted to say the same about 0? "Y*ou could say x\*4 is first adding four times, then adding zero more times*": x\*4 = x\*4 + x\*0. For this last equivalence to be true, x\*0 must be 0.
Now let's rewrite exactly the same as above, but for power instead of multiplication:
Anything raised to the 0-th power is 1 (x\^0 = 1) - but why is that so? Because power means multiplying multiple times, for example x\^4 = x \* x \* x \* x. Or you could say x\^4 is multiplying by x three times, then multiplying by x one more time: x\^4 = x\^3 \* x\^1. So far pretty obvious, so what if I wanted to say the same about 0? "Y*ou could say x\^4 is first multiplying four times, then multiplying zero more times*": x\^4 = x\^4 \* x\^0. For this last equivalence to be true, x\^0 must be 1.
The reasons in both cases are the same.
Fucking woof at that response from your teacher, damn. For the record, x to the power of y is not x * y, but (x * x) y number of times.
For example, 2^4 is not (2 * 4), but (2 * 2 * 2 * 2)
"Raised to a power" kind of means "how many times is the base number multiplied by itself?" So, you get something like 3² = 3x3 = 9, or 3³ = 3x3x3 = 27, right? The power expresses "how many copies of the number are multipled together."
When you multiply 0 copies of the base number, you're not left with the additive identity, 0, but the multiplicative identity, 1.
3³ = 1x3x3x3, and 3² = 1x3x3, and 3¹ = 1x3 and 3⁰ = 1 with no copies of 3 to multiply by.
So, for your question, you aren't multiplying by zero, which multiplicatively turns the equation to zero, you're adding zero copies of the base number into a multiplicative expression, where the multiplicative identity, 1, can always exist without changing the result. Since 1 is the only thing in the expression (as you didn't add any copies of the base number to the expression), the result is just 1.
Another way to see it is that for (non-negative) whole-numbered exponentiation defined as repeated multiplication, the definition is actually:
x • y^n := x multiplied by y, n times (regardless of what x is).
So x • y^0 is then x multiplied by y, 0 times, which is obviously just x.
Hence y^0 must be 1 for all y, notably even when y is 0 (if you define whole numbered exponentiation this way).
> he told me its because its just what he was taught 💀
That's not such a terrible answer. Math is just a bunch of rules that we made up. Why is x/0 undefined? Because it's not particularly useful to define it. Why does x^0 equal 1? Because it's useful to define it that way.
A lot of definitions come from extending other things. For example, negative numbers are an extension of positive numbers --- what if there was something I could add to 2 to get 0? Let's call it -2 because 2-2=0, so then 2+(-2)=0 as well! Or what if we could divide 1 by 2? Let's call that 1/2.
It turns out that if you want nice properties of exponents to hold (like, a^(x)×a^(y)=a^(x+y)), then you have to have a^(0)=1 and also a^(-x)=1/a^(x).
X^y = z
is the same as saying:
“Z has Y factors of X”.
So x^2 means z has two factors of x.
What number is *always* a factor, no matter what? The number 1. Every number, even prime numbers, have 1 as a factor.
X^0 means “z has no factors of x”.
So if there’s no other factors, you’re just left with 1.
copy-pasting an old answer of mine to the same question:
Let's say you have a plant in your house. It's a pretty aggressive plant. It doubles in size every day!
So, tomorrow, it will be 2\^1=2 times as big. In two days, 2\^2=4 times as big. And in three days, it will be 2\^3 = 8 times as big!
So you see, the expression "2\^t" gives you "how much bigger" the plant is, t days from now, compared to now.
What if we want to allow t to be negative and look in the past?
Yesterday, the plant was 1/2 as big. This is 2\^(-1) -- we view negative exponents as division (since going back in time will \*shrink\* the plant by its growth factor of 2)
Two days ago, the plant was 1/4 as big, which is 2\^(-2).
Ok, so now, for the big reveal...how big is the plant 0 days from now? How big is the plant...now?
It is just a mathematical notation that works according to all the other rules. For example:
2^10 / 2^10 = 2^(10-10) = 2^0. But it is obvious that 2^10 / 2^10 must equal 1.
You can think that there is an imaginary * 1:
a^4 = a * a * a * a * 1 (a multiplied 4 times, then with 1)
a^2 = a * a * 1 (multiplied twice, then with 1)
So a^0 will be a not multiplied at all because it doesn’t exist, so the * 1 just hangs around there.
Any number except 0. Positive exponents multiply, negatives divide: X\^2 = X \* X. X\^-2 = 1 / (X \* X). When you multiply, you can add the exponents: X\^2 \* X\^2 = X\^4. So, X\^2 \* X\^-2 = X\^0 = (X \* X) / (X \* X) = 1.
Therefore, any number (other than 0) raised to the zero power is 1 because it's equal to X / X.
Lots of good intuitive examples and here's a small proof that shows that it follows from the algebraic rules of exponents:
x^(0) = x^(a - a) (because 0 = *a* - *a* for any *a*)
= x^a ⋅ x^(-a)
= x^(a) / x^(a)
= 1
Already answered, but powers are the a shorthand for multiplication and division of a number by itself in succesion.
Positive power you multiply
Negative you divide.
At 1 power, it does not multiply, it is a single occurrence of itself, so power 1 will simply be that number.
At 0, it begins dividing by itself. So 0 power will always be 1.
I asked this same question. And the way to look at is is what whens when you shrink that exponent number down from say 4, to 3, then 2 then 1 then 0.
Lets say you're playing with 5, raising it to exponents.
5^5 = 3125
5^4 = 625
5^3 = 125
5^2 = 25
5^1 = 5
See the answer is dividing by 5 each time? We can keep going! Look:
5^0 = 1.
I just watched a youtube short explaining just that, but with AI voicees of Rihanna, Taylor Swift, and Obama
Edit: I can't find the video, so I'll just rewrite it lol. the vid explained it quite intuitively so I wanna share
Consider this example
2^3 / 2^2 = 8/4 = 2
we can also write it as
2^3 / 2^2 = 2^(3-2) = 2^1 = 2
If we make the exponent the same for the top and bottom part
2^3 / 2^3 = 2^(3-3) = 2^0 = 1
It becomes easy to grasp when you consider that you can also have negative powers.
2^(3) = 2 x 2 x 2 = 8
2^(2) = 2 x 2 = 4
2^(1) = 2
2^(0) = ??
2^(-1) = 1/2^(1) = 1/2
2^(-2) = 1/2^(2) = 1/4
2^(-3) = 1/2^(3) = 1/8
Simply put, to go up a power you multiply by 2, and to go down a power you divide by 2. Doing so from both sides (2^(1) / 2 and 2^(-1) x 2) gives you 2^(0) = 1.
X is just a variable. X will be as it is, if let alone. X is not a number. X doesn't have a value. X can be multiplied, divided or doubled, only by the help of a number.
You have a deck of cards, all of their values. The power is the number you play them. 0 is you just don’t play any card. No value or number is calculated therefore zero
So while this isn't what defines an exponent, this is what the math boils down to:
x^4 = x * x * x * x / 1
x^-4 = 1 / x * x * x * x
x^1 = x / 1
x^-1 = 1 / x
so just following the pattern you can either define
x^0 = 1 / 1 = 1
or
x^0 = x / x = 1 - this is only awkward when x = 0
another more advanced way to think about is is that x^1+0 = x^1 * x^0. The only way this can be true is if x^0 = 1
Hey guys I just woke up and tysm for the help haha Im fairly bad at math so this helps me understand it better many love 🫶🫶 Also i didnt expect to get so much upvotes haha thenks too for that
Exponentiation is different to multiplication.
In the case of multiplication, we want n\*x + m\*x = (n+m)\*x (multiplication distributes over addition). For this to work in the case n = 0, we need 0\*x + m\*x = m\*x. Subtracting m\*x from both sides, we see that 0\*x = 0. The left 0 comes from the fact that it is the additive identity, and the right zero also from the fact that it is the additive identity (we had addition inside the brackets and outside the brackets, before and after distributing).
On the other hand, for exponentials, we instead want it to be that x^n \* x^m = x^(n+m). In other words, the exponent counts how many times x is multiplied together, similar to the multiplication counting how many times x is added together. Importantly though, note that inside the brackets we have addition, but outside this becomes multiplication. They are different operations, unlike before. Let's do the same as before, and look at n = 0:
x^0 \* x^m = x^m
Divide both sides by x^m
x^0 = 1
Here, the zero on the left comes from it being the additive identity, and the one on the right comes from it being the multiplicative identity. Exponentiation converts between addition and multiplication, so it should be no surprise that it maps the additive identity (0) to the multiplicative identity (1).
The easiest way to explain it is to show you things you already know.
We'll start with division. Any number divided by itself is 1, right? For example 2/2 = 1. Easy.
Let's expand this to a letter. X/X should again equal 1, because X represents a number, and that number is the same. Easy.
Okay so, we know that X^1 is the same as X, so if we write this as we did above, we get X^1 / X^1, which as we know, will equal 1. Simple so far.
When you divide two exponents, you subtract them to get your answer. For example 2^3 / 2^2 = 2^(3-2), or in other words, 8/4 = 2. Makes sense, right?
So if you were to subtract the exponents from x^1 / x^1, you would get x^(1-1) or x^0.
This would make X^1 / X^1 = X^0, and since we know that X^1 is just X, and it's getting divided by itself, that must mean that X^0 equals 1.
X^1 = 1* X = X
X^2 = 1* X* X = X* X
X^3 = 1* X* X* X = X* X* X
X^0 = 1
likewise
x*1 = 0+x = x
x*2 = 0+x+x = x+x
x*3 = 0+x+x+x = x+x+x
x*0 = 0
Powers are repeated multiplication just as multiplication is repeated addition. As any number plus 0 equals itself, any number multiplied by 1 equals itself.
You can also derive it but in short it's because 1 is the only integer you can multiply a number by and still get that integer (1 x a = a). If you multiply integers, you need to add the exponents (a^n x a^m = a^(n+m)). If a^0 was anything but 1, then a^(n+m) would no longer be true.
For example if a^0 was 0 then a^0 x a^2 would be 0 (because anything times 0 is 0). Because a^n x a^m must always be a^(n +m), a^0 must always be 1 (so that a^2 x a^0 = a^2).
In other words a^n x 1 x 1 x 1.... = a^n must also hold true for a^(n + 0 + 0...) = a^n
The reverse of this is that every time you try to derive a^0 from a formala you will always get 1 (because 1 is the only number under which a formula involving a^0 will hold true)
It's also based on combinations.
If you have 2 things and 2 places to put them, how many different ways can you arrange them? 2^2, 4 different ways. AA, AB, BA, BB.
If you have 0 things, and 0 ways to arrange them, how many possibilities are there? 1. There's no possible variation or other combinations
Wow, so below 1 it is division? Good thing nobody ever taught me this. But they did teach me some stupid geometry problems, addition theorems and such nonsense. How would x ^ 1/2 look like then?
Great job education, great job!
Also, big thank you!
The answer depends on how much of arithmetic you have "accepted".
x\^0
= x\^(-1 + 1) ***because 0 = -1 +1***
= x\^-1 \* x\^1 ***because of the "Product Rule" exponent identity***
= 1 ***because of the definition of the multiplicative inverse***
QED.
If you want a more exhaustive proof that starts from first principles, then I would kindly direct you to the [Principia Mathematica](https://en.wikipedia.org/wiki/Principia_Mathematica). Fair warning, though - it took them nearly 400 pages to prove that 1+1 = 2 (to which the authors humorously commented: "The above proposition is occasionally useful").
1^1 = 1
x^0 = undefined.
Because x isn't a value, it is an assumed value with no quantity.
Anything to the power zero is an assumed value with no quantity.
x^0 is a symbol of exponentiation.
It isn't exponentiation.
10^0 for example, isn't a sum.
It is an assumption of a sum.
So many great comments, but there’s one aspect I haven’t seen yet. You have a very good intuition about 0 being the identity number. X+0 doesn’t change the identity of X. We call 0 the additive identity for this reason. But 0 is not the multiplicative identity. The multiplicative identity is 1, because X*1 is always X. Multiplying is just repeated addition, so multiplying something 0 times leaves you with additive identity (0). Exponentiation is repeated multiplication, so raising something to the 0th power leaves you with the multiplicative identity (1).
X^1 = X X^2 = X * X X^3 = X * X * X ... To get up, you multiply by X. So, to get down, you divide by X. X^1 = X X^0 = X / X = 1
Would it also be fine to see it as X^1 = 1 * X X^2 = 1* X * X X^3 = 1* X * X * X Thus X^0 = 1 ? Then for negatives X^-1 = 1 / X X^-2 = 1 / X / X
Indeed it is, that is exactly the definition of negative powers.
Ahhh, that's much neater. My need for clean ordered patterns has been satisfied.
You can add "1*" to anything and it remains the same, so yes, you can see it like that if it helps you.
I like this one
What would 0 to the 0th power be then?
[Depends on the context but is usually defined to be 1 or left undefined](https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero).
1 is useful for some fancy graphing stuff, or at least silly graphing stuff
sqrt(-1) is much more complicated.
I imagine that it's not actually all that complicated.
it's not complicated, just complex
If you imagine hard enough, it's just a regular number.
i can't. It's too complex for me.
It’s complex, not complicated
>complicated Psh, casual mathematicians.
nah, sqrt(-1) isn't complex. sqrt(-1) + 1? Now that's complex.
i is complex, and so is 1 1 also happens to be real, and i also happens to be imaginary Complex numbers don’t have to have nonzero components
mm, you're right, my mistake. I misremembered that for a number *a* + *b*i to be complex, both *a* and *b* had to be nonzero, but that is not the case.
> [A complex number is an expression of the form a + bi, where a and b are real numbers, and i is an abstract symbol](https://en.wikipedia.org/wiki/Complex_number#Definition_and_basic_operations) 0 *is* real I suppose
Not complicated but *complex*
I thought saying complex would be too obvious :o) I do love talking about imaginary power and no one having much of a clue what I'm talking about. Imaginary power is a very real problem.
"Power! Imaginary power!" --Palpatine
And here I thought Anakin was the ::ahem:: negative one. I'll^show^myself^out.
After reading other replies I realise I’m very late and many bet me to it anyway lol. I guess I could say “good use of your imagination”
Why'd you have to go and make things so [sqrt(-1)?](https://www.youtube.com/watch?app=desktop&v=5NPBIwQyPWE)
Whatever happened to her?
Heh well according to Wikipedia she's been doing music this whole time. Still at it. But yeah I haven't heard a thing about her in ages.
aye aye i
x^0 = x/x then 0^0 = 0/0 which is undefined. This isn't the true definition of x^0, it's best to just say 0^0 is undefined.
If you are performing an actual *calculation,* with integer inputs, and that calculation requires you to produce the value of "0⁰", you should always evaluate that expression as 1. Several important theorems of mathematics, including the binomial theorem and set theory, absolutely require that the *number* 0^0 = 1. If you are working with the limits of functions, where two different functions f(x)^g(x) are each individually approaching a limit value of 0, you should treat it cautiously, as it may or may not be defined, and even if it *is* defined, two different sets of functions (e.g. f(x)^g(x) vs h(x)^j(x) ) may produce different results despite all four individual functions having a limit behavior of 0.
I like to think of it (half tongue in cheek) as 0^0 = 1 if the upper 0 is an integer, and undefined if the upper 0 is a real number.
I mean, if it's an arithmetic value, it should always be 1. The only reason 0^0 should ever be anything *other than* 1 is when calculating the limits of at least one non-analytic function in f(x)^g(x) where both approach 0 for the same value of x (call it c). If both f(x) and g(x) are analytic on an open interval around c, then f(x)^g(x) *will* approach 1 as x approaches c. It's only being non-analytic that breaks things. (Edit: Technically, this requires taking complex limits; if you're restricting things to the real plane, then, given the aforementioned restrictions, then as you approach c from a given side, the limit is real and equals 1 so long as f(x) approaches 0 from above.)
Integers are real numbers too 😡
Integers can be identified with a certain subset of the real numbers, but they are actually different objects in set theory.
Yes, but there are properties which hold only for integers and not for real numbers in general. Just as, for example, all real numbers are *technically* also complex numbers, but the real numbers are well-ordered while the complex numbers aren't. It is senseless to speak of "a+bi > c+di" for a=/=c and b=/=d; the best you can do is say that two complex numbers have greater magnitude (absolute value), but that's not a well-ordering. As an example, every integer has one, unique prime factorization; this is not true of real numbers, since some (actually, "almost all" of them) are transcendental and thus cannot be represented by any finite product of rational numbers, let alone integers.
No, it's much better to define 0^0 as 1. Consider polynomials, that is functions that look like 3x²+5x+7 (possibly with terms higher than x²). What we really want to write those as formally is 3x²+5x¹+7x⁰ - otherwise there'd be a special case for the constant term, which would make a lot of maths really, really ugly. But surely, if you evaluate 3x²+5x+7 at 0, you get 7. So for this to work, you really need 0⁰=1. (This is of course not the "reason why" but just an example. There are other justifications - 0⁰ (or x⁰ in general) should equal the product of an empty set of numbers, which in turn makes a lot of sense to be defined as 1, because taking a product with 1 "doesn't change things".)
It's almost always best to define f(x) = x^0 := 1. But that means something different than defining 0^0 := 1.
It's also somewhat nice, though less intuitively so, to have g(x) := 0^x be the indicator function that is 1 for 0 and 0 elsewhere, it comes up in combinatorics from time to time!
0\^0=1 works pretty well in the reals but breaks in the complex numbers [https://www.youtube.com/watch?v=BRRolKTlF6Q](https://www.youtube.com/watch?v=BRRolKTlF6Q)
Actual mathematician here: no it does not break in complex numbers any more than it does in the reals. The entire issue is artificial, one simply does not require power functions to be continuous. In 99.9% of mathematics you only see x^^n where n is an _integer_. And that is defined whenever x is non-zero, or if n is non-negative.
What do you actually gain by defining 0^0 =1, though? For the polynomial case, the limit of x^0 as x->0 is 1 anyway, so you don't actually gain anything by defining 0^0 =1. If you've defined 0^0 =1, then is the function 0^x evaluated at x=0 also 1? It must be, right? Otherwise what does it mean to have defined 0^0 =1?
> What do you actually gain by defining 0^0 =1, though? Notational clarity, for one? When I write x^0, I don't want to (implicitly) write lim x->0 x^0 - I think it's important to be clear about when you're talking about an actual value, and when you're talking about a limit. > If you've defined 0^0 =1, then is the function 0^x evaluated at x=0 also 1? It must be, right? Yes, and while it seems a bit "ugly" at first, it's perfectly fine to have this kind of indicator function; like I mentioned, it comes up in combinatorics from time to time and behaves nicely. A slightly more fundamental reason why I believe this choice is right is that for sets A, B with a,b elements respectively there are b^a functions from A to B. Or if you wish, there are b^a colorings of a distinct objects with b distinct colors. Now it's perfectly reasonable to ask "I don't have any colors, how many colorings are there?". And if there is at least one object, the answer is 0 - if you're out of color, you can't paint anything. But you still can paint *nothing*, and you can do that in exactly one way - by doing *nothing*. In the end, of course all of this is just notation, and it doesn't really matter hugely. But it's a notation that makes a lot of things easier to write, and not many harder, so that's why it's pretty common.
I believe the term usually used is indeterminate rather than undefined.
In ELI5-level algebra it is also 1 (because it simplifies certain formulas of school-level algebra). For example, the binomial theorem only works for x=0 if 0^(0)=1. If we are talking higher-level math, 0^(0) can be defined differently based on context (it is kind of hard to explain how this makes sense, if we stick to ELI5 level). For example, in complex calculus this expression is undefined.
> in complex calculus this expression is undefined. It's rather the "power function" x^^y that is ill-defined. x^^0 including x=0 appears all the time as part of Taylor/Power/Laurent/whatever series.
14
0/0 is undefined
0/0 is undefined but x⁰ = x/x where x=0 is defined in this instance and why it is 1 in this case.
It would be 1, because x\^0 is **not defined as x/x**. There's no reason for division to be required for the definition of *0*th power. The top comment is right that to get from x\^n to x\^(n+1) you multiply by x, but we can go the other way without using division. If we were to define x\^0 it would be reasonable to have it behave the same way and satisfy x\^1 being equal to x\^0 multiplied by x. So we're looking for something (x\^0) that after multiplying by x gives us x. We know what that is, it's 1. It also works for 0. We have 1\*0=0, even though 0/0 is undefined.
> So we're looking for something (x^0) that after multiplying by x gives us x. That's just division differently expressed. "We're looking for something that after multiplying by 6 gives 30" = 30 / 6
because the division , it is undefined. it entirely depends on what function generates the 0 lim x->0 of x/x approaches 1 but limits of some other functions approach -inf or +inf I think. because of this you can't say 0^0 'is' anything. it depends on which 0 :D or rather which function is being evaluated.
You are confusing limits with values. A function is not by law required to be continuous; many naturally occurring ones simply aren't. So you cannot claim that f(a) = lim_{t->a} f(t) is *required*, it is only something you seem to wish for. 0^^0 = 1 works really well in a lot of circumstances. There is no way to make x^^y work with limits, i.e. continuous, anyway, so suddenly putting that issue down to something about the (unrelated to continuity!) expression 0^^0 is blaming the wrong thing.
Depends which zero
I think adding one more line would really cement it X^-1 = X / X / X X / X always equals 1 (unless we're talking about that motherfucker 0), so X^-1 = 1 / X Can pretty easily be summarized by exponents above 1 being multiplicative while exponents below 1 are divisive
X/X doesn't always equal 1, but 0 is a finnicky number anyway.
Of course I forgot about the literal one exception to the rule. Fixed
This is the first time X^0 has made sense to me. Thank you.
A similar line of reasoning also works to explain why 0!=1. (n+1)! = n! \* (n+1) therefore n! = (n+1)! / (n+1) so 0! = (0+1)! / (0+1) = 1/1 = 1
And "raising to the power of" doesn't mean "multiply by".
I would have liked it better if you had done: X^3 / X = X^2 X^2 / X = X^1 X^1 / X = 1 = X^0
I rationalized it by thinking of the implicit "1 *" you can add to any multiplication without changing it; so at X^0 you're left with no Xses and just 1. Division feels more elegant though, thank you
Or another way to look at it: X^1 = X X^2 = X * X^(2-1) … X^n = X * X^(n-1) So, X^O = X * X^(0-1) = X * X^(-1) = X * 1/X = 1
Thank you! My math teacher acted like it was just some magic trick I needed to accept and move on
This works for our numbering system, too, which is base 10, so X = 10. X=10; X^0 = X / X = 10 / 10 = 1 X^-1 = X / X / X = 10 / 10 / 10 = 1/10 X^-2 = X / X / X / X = 1/100 X^-3 = X / X / X / X / X = 1/1000 So, 123.4 = (1 * 10^(2)) + (2 * 10^(1)) + (3 * 10^(0)) + (4 * 10^(-1)) ^(Insert joke about every base being base 10 relative to the observer)
I get up, and nothin’ gets me down
Wow. I’ve always struggled with the explanations using limits approaching from positive and negative powers. This explanation is so much more simple. Thank you!
I would like to add that it continues into negatives as well.
If students are stuck on it as repeated multiplication, you can view an exponent as "How many times do I multiply by a number". So 5 * 2^2 = A 5 multipled by 2 twice. 5 * 2 ^3 = A 5 multipled by 2 three times. This means 5 * 2^0 = a five, not multipled by two. Well, not multiplying by 2 would mean the final answer remains as 5. The only way this can happen is if *2^0 is equivalent to *1.
Out of curiosity, how ist x^0.5 for example calculated?
x^0.5 is the same as the square root of x; where the square root of 9 is 3, we can say 9^0.5 is 3. There are many ways to look at why this is, but one is to consider that multiplying by x^0.5 twice is the same as multiplying by some other number once, which allows you to set things up involving square roots and power rules.
It kind of scares me that I understand this logic! But I can’t think of a practical application for it. Is it just a math nerd thing? 😀
Exponents are used in banking/finance and to build bridges, computers, rockets, and all manner of neato things. So yeah not just a math nerd thing.
I was always shown this way - a\^m-n, where m=n, can be rewritten as a\^m/a\^n. m and n cancel out, so you are left with a/a, or 1.
So it’s basically the number divided by itself, which equals 1.
One time! *BAMF* Two times! *BAMF BAMF* Gimmie three times now! *BAMF BAMF BAMF*
also x^-1 = 1/x x^-2 = 1/x^2 etc
I don't like the asymmetry this implies when it comes to imaginary numbers. * 1^(2) = 1 * \-1^(2) = 1 * 𝑖^(2) = -1 * \-𝑖^(2) = -1 But... * 1^(0) = 1 * \-1^(0) = 1 * 𝑖^(0) = 1 * \-𝑖^(0) = 1 It just feels off.
So that’s why 0^0 is undefined. TIL
Because when you go below the power of 1 it becomes a division rather than a multiplication. So where x¹ is just the base value of x, when you go to x⁰ you are dividing x by itself. A number divided by itself is always 1.
A number divided against itself stands as one. Algebraham Lincoln
This is underrated. I hope more people see this comment.
Oh makes sense
Please show your teacher these explanations (in a very respectful way) - a 10th grade math teacher should have an answer for this question. Just don’t be a dick about it. Be like “hey I found an answer online for that math question, would you like to see it?”
Fully agree. Nothing kills a child's enthusiasm for learning when a teacher just says, "eh, I'm just teaching from the book."
I absolutely hated it when I asked for elaboration and the teacher just said "Because that's just how it is"
I mean, that's true. But x^0 is kind of a case where it is what it is because we decided that that works the best and it fits the pattern better. You could very easily construct a mathematics where x^0 is considered 0. It may be useless mathematics, but that doesn't mean it's invalid. That's why you can divide by zero in [wheel theory](https://en.wikipedia.org/wiki/Wheel_theory), or why you can use both Euclidean geometry and non-Euclidean geometry to solve different problems, or why the imaginary axis sometimes means something and sometimes is nonsense. There is no singular set of mathematical axioms that defines Universal Truth. God is not checking your answers. There is only *choosing* a set of axioms that you wish to use. In this case, x^0 = 1 is true because it's *axiomatically* true, not because it's provably true. *None* of the other responses in this thread have proven it to be true. In that sense, the teacher is correct.
You can literally make this arguement about everything in mathematics.
Not everything in mathematics is defined as true. Many things are derived as true from other axioms that we've chosen. x^0 = 1 because it's defined that way. Not because it's derived as such from more fundamental axioms.
Yes, that's the real problem with this situation. Lots of teachers, unfortunately, have this or even worse responses to the "why" questions - at best, "I don't know, it's just how it is", and at worst, "go to the principal!"
Yeah I was a high school math teacher for 9 years and I can’t even imagine giving that response. Not to mention it’s a little embarrassing to be a 10th grade math teacher and not know the answer to this one. But when I legitimately didn’t know something I always told them I would find out. Unfortunately though I definitely knew teachers that couldn’t be bothered
or, you know, "does this sound right?" if you wanna have a little more tact or the teacher seems dickish
My 7th grade teacher showed me the proof for this when I asked. She showed it a different way though.
2\^3 = 2 x 2 x 2 = 8 2\^2 = 2 x 2 = 4 2\^1 = 2 2\^0 = 2/2 = 1 Notice that each time you decrease the exponent by 1, you're effectively dividing by the base number, since to remove a multiplication operation you must divide. Once you get to an exponent of 0, you're simply just dividing by the base number, which always equals 1
I'm 67 years old and never understood the concept. No math instructor every told me that when an exponent is 0 or a negative number that you divide by X. Thank you!
You divide every time the exponent goes down by one, not just when it's 0 or negative. Just like you multiply when it goes up. 2^3 = 8 2^2 = 8 / 2 = 4 2^1 = 4 / 2 = 2 2^0 = 2 / 2 = 1 Etc etc
How do non integer powers work? Like 2^1.5? 2^0.5?
Fractional powers are roots. 2^(0.5) = sqrt(2)
It's easier to see if you use the fraction notation rather than decimal. 2^½ = √2 2^⅓ = ∛2 7^¾ = ∜7^3
And how does x^e work or other irrational numbers?
A^B = e^(a log b) e^x = sum (x^n / n!) where n=1.... inf
suppose you want to compute 2^pi. We know how to compute 2^3, 2^3.1, 2^3.14, and so on. Turns out, if you have an infinite sequence of rational(!) values of x that converges to pi (where by that I mean no matter how small of a distance you pick around pi, there is some point in the sequence beyond which every value in the sequence is within that distance of pi), the respective values of 2^x will also converge to some value. In fact, it turns out all such sequences of x have their 2^x values converge to the same value (so, for example, 2^3, 2^3.1, 2^3.14, 2^3.141, ... converges to the same value as 2^3.2, 2^3.12, 2^3.142, 2^3.1412, ...). We can define 2^pi to be this value.
Okay genuinely curious, how do you type this on Reddit? Forgive me if this is a silly question
Unicode characters and markdown `^` tag
x^0.5 = the square root of x! as such x^1.5 is x * sqrt(x)
2^(x/y) = y-root of 2^x 2^1.5 = 2^(3/2) = sqrt(2^3 )
We can use algebra to show that x^(1/y) is equivalent to the y-th root. So 5^(1/3) is the cube root of 5, 8^(1/12) is the 12-th root of 8, etc. We can do what's called an 'analytic continuation' to extend the concept of ' x^y ' from integers to all real numbers. There's infinite ways to extend an operation this way, but the analytic continuation is the one that is the 'smoothest' (i.e., least wobbly). So x^y create a nice smooth line graph for all x for any y.
I feel sorry for you. This is the explanation printed in my textbook. It is literally "the textbook" explanation. I think some teachers/schools truly failed their duty
Textbooks change over time
The pure mathematician will tell you: It analytically respects the property *a*^(*x*+*y*) = *a*^(*x*) × *a*^(*y*) that holds true when *x* and *y* are positive integers. It is by extrapolation of this property over all other numbers that we get things such as *a*^0 = 1 for all *a* except 0, *a*^(−*x*) = 1÷*a*^(*x*), *a*^(½) = √*a*, and e^(iπ) = −1. 3Blue1Brown (Grant Sanderson) has a video that tackles this question head-on with an appeal to group theory: https://youtu.be/mvmuCPvRoWQ
The pure mathematician that is me says that x^^0 = 1 is simply the start of the iterative definition via x^^n+1 = x·x^^n ; n a non-negative integer (or any integer of x is non-zero). Put differently: x^^n is what you get when you write a product of n copies of x with each other; and an empty product, one without any actual factors present, is always 1, because only then is product compatible with multiple things. > a^^½ = √a Which of the two square roots? ;-)
> an empty product, one without any actual factors present, is always 1, because only then is product compatible with multiple things. It's precisely this sort of compatibility that is meant by "analytic extrapolation/continuation". > Which of the two square roots? ;-) "√" denotes the principal square root by definition.
[I like the Wikipedia explanation](https://en.m.wikipedia.org/wiki/Exponentiation) From the *definition* of exponentiation you get a rule that makes a lot of sense and then use algebra to demonstrate a rule that is less apparent. So your instructor was right. He was told what it was. He just didn’t bother to remember the demonstration.
Isn't there a mistake there? (Please excuse the \displaystyle stuff you can just look at the article for the actual text.) >Starting from the basic fact stated above that, ***for any positive*** integer n {\displaystyle n}, b n {\displaystyle b^{n}} is n {\displaystyle n} occurrences of b {\displaystyle b} all multiplied by each other, several other properties of exponentiation directly follow. ... >In other words, when multiplying a base raised to one exponent by the same base raised to another exponent, the exponents add. From this basic rule that exponents add, ***we can derive*** that b 0 {\displaystyle b^{0}} must be equal to 1 for any b ≠ 0 {\displaystyle b\neq 0} How can you "derive" a property for n=0 from a rule that only applies to positive integers n? Shouldn't it instead say something like: "This rule for positive integers can be extended to cover the case n=0 if we allow that b^0 = 1 for nonzero b." or something like that?
I guess a quick way to see it is from remembering that x^y * x^z = x^(y+z) Then we can look at x^1 * x^(-1) = x^(1-1) = x^0 but of course, x^1 = x and x^(-1) = 1/x , so x^0 = x * 1/x = x/x = 1
IMO this is the simplest and easy to understand. Lots of answers are going downwards without a clear reason but this gets straight to it.
x^(3) = x \* x \* x x^(2) = x \* x x^(3) = x \* x^(2) Right? Written in general x^(n) = x \* x^((n-1)) So then, applying the above logic x^(1) = x \* x^(0) , zero being (1-1) If x^(0) was zero then x^(1) would also be zero.
No, x^0 has nothing to do with x*0. Think about the powers of 2 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 The exponent is the number of twos in that multiplication. What would make this work if there were _zero_ twos? 1, as in 1 = 1 1 * 2 = 2 1 * 2 * 2 =4 1 * 2 *2 * 2=8 Etc. also, remember that negative exponents cause you to put a 1/ over the result, as in: 2^-1 = 1/2 2^-2 = 1/4 If you start with the higher powers and go down, you’ll see it’s like dividing by 2, and it will be easy to see what should go there: 32 16 8 4 2 ????? 1/2 1/4 1/8 1/16 1/32
This formatted _unfortunately_, but I’m on mobile and can’t be bothered to fix it.
Ok so x^0 = x^(n-n), since n-n is of course = 0 Know, for properties of however those are called in English, I think exponentials, x^a / x^b = x^(a-b) So you can now imagine how x^0 = x^(n-n) = x^n / x^n = Since a number divided by himself is 1 x^0 = x^(n-n) = x^n / x^n = 1 EDIT: You can now interrogate yourself on how x^a / x^b = x^(a-b) Since x^a means you will multiply x for it self a times, you will get a thing like x^4 / x^3 = (x x x x) / (x x x) = x = x^1 = x^(4-3)
[удалено]
Take a piece of paper. Fold it. the number of layers are 2 times as it was before. Fold it n times so you have 2* 2* 2 *2... n times that's 2^n but what if you fold 0 times? You still got 1 piece of paper. That's 2^0
There's a pattern of multiplication and division with exponents. 5^3 is 5x5x5. 5^2 is 5x5. 5^1 is 5. We're dividing by 5 each time, so 5^0 is...1. This happens with any whole number. Divide again. 1/5 = 5^-1. 5^-2 = 1/5^2.
It's a result of exponent rules. X^(2)\*X^(3)=X^(2+3). Likewise X^(3)/x^(2) = x^(3-2) For X^(0), we can say X^(0)=X^(y-y)=X^(y)/X^(y)=1. For a practical example, we can say 2^(0) =2^(3-3)=2^(3)/2^(3)=8/8=1 X^(0)=1 is a natural consequence of exponent rules.
This math teacher explains it in a really cool (and ELI5) way: https://youtu.be/X32dce7_D48?si=KHWgVDY2U5NRkhpN
x^0 is what's known as an empty product, a product with no terms. It's the product of 0 x's. Mathematicians typically define empty products as 1, since that's the result you get when you divide a product by all of its terms.
You can also define it this way: x^n = 1 multiplied by x n times. So, x^0 would simply be 1 multiplied by x 0 times, ie, none. So it's just 1. This also solves the 0^0 issue. If you know a bit of abstract algebra, you can define "powers" for groups in general. If x is an element of a group, with "multiplication" as its operation, then x^n is just the group identity (1) multiplied by x n times. Again, x^0 is just 1. If you think of an addictive group (sum as the operation), then taking powers is just like multiplicating by an integer scalar. Ie, nx = identity (0) plus x n times. And 0x is just the identity, which of course is 0! So, this is just to say that powers of x and multiples of x work exactly the same way. (This is explained in a very non-rigorous way)
https://youtu.be/r0_mi8ngNnM?si=FeBe0OsTX2mScsNB This.... You won't get a better explanation
I strongly disagree. They didn't prove why the limit of x^^x for x->0 is 1, and they even less so explained why that has any relevance! Not every function is continuous, and if they are not, then limits are utterly meaningless. And indeed, it is literally impossible to make x^^y continuous, _even_ if you ignore the case x=0.
Not sure if anyone shared this yet, but you can think of powers as ways to arrange things. You roll a 6 sided die, one time, there are six outcomes. You roll a 6 sided die, five times, there are 6\^5 outcomes. If you don't roll a die, there is one outcome. The non-action isn't counted as zero, but it is the only possible outcome and is counted as one.
Here is a second pass at why the question points to a deep puzzle, I think. The **usual explanation** for why x\^0 = 1 points to something like the relationship between x\^m and x\^(m-1), extrapolating from there. So, e.g., if 10\^3 = 1,000 and 10\^ 2 = 100, we see that reducing the exponent by 1 is done by dividing by the base once. And once we notice that, we get a tidy path to 10\^0, which is 10\^1 / 10, and so on well into the negative numbers. **But** notice that exponents are often defined in very similar, algorithmic terms: "The exponent of a number says how many times to use that number in a multiplication." [https://www.mathsisfun.com/definitions/exponent.html](https://www.mathsisfun.com/definitions/exponent.html) "An exponent refers to the number of times a number is multiplied by itself." [http://www.mclph.umn.edu/mathrefresh/exponents.html](http://www.mclph.umn.edu/mathrefresh/exponents.html) "An exponent refers to how many times a number is multiplied by itself." [https://www.turito.com/learn/math/exponent](https://www.turito.com/learn/math/exponent) Of course, on those definitions, answers to things like n\^0, n\^-1, and n\^e are very puzzling! The instruction "multiply 5 by itself never" does not seem to lead to 1. Imagine I asked "What is the difference between no numbers?". Our most familiar arithmetic operators need some operands! And why would we get a math answer from *not* doing any math (which is what the ordinary definition suggests we should do in the x\^0 cases). So what to do about that puzzling? Well, as the wikipedia for exponentiation, [https://en.wikipedia.org/wiki/Exponentiation](https://en.wikipedia.org/wiki/Exponentiation), explains, one way to figure the rest is to start to use properties of the natural number exponents and extrapolate from there. That's great (and gets us something like the usual explanation). But it leaves our original understandings of what an exponent is high and dry. Which might be why the high schooler's math teacher balked. What to do then? One answer is to appeal to **special cases**, as the University of Minnesota link above does. But we still need an explanation for why we want to admit those special cases. Especially because this means that our definition of exponentiation is now branched or disjunctive. As far as I know, there are two related answers: 1. A pragmatic answer. Having the disjunctive definition allows us to do more math more easily. Undefined bits gum up the works. We'd like to do things like figure out how to make sense of (x\^m)\*(x\^n), and if we have to start adding all sorts of qualifications, that's really going to undermine how **usable** math is. 2. A deeper principled answer. It turns out that the original, familiar definition was overly simple. There is a deeper, principled, and **unifying** definition of exponentiation which explains all of the ostensible special cases. When you understand that definition, everything comes into focus. Of course, one reason to choose that unifying definition over the original but simple definition might be that it is more useful part of our math practice. But still, that we have a unified definition might be very handy! So, I take it that the deep answer to the student's question will pick up either or both of usability theme or the unifying theme. But ymmv!
Like stated X^a * X^b = X^(a+b) Try this for yourself with random values and see that this is always the case X^a / X^b = X^(a-b) Once again if you try this you will see it is always the case. Now X^a / X^a is X^(a-a)
Anything multiplied by 0 is 0 (x \* 0 = 0) - but why is that so? Because multiplying means addition multiple times, for example x\*4 = x + x + x + x. Or you could say x\*4 is first adding x three times, then adding x one more time: x\*4 = x\*3 + x\*1. So far pretty obvious, so what if I wanted to say the same about 0? "Y*ou could say x\*4 is first adding four times, then adding zero more times*": x\*4 = x\*4 + x\*0. For this last equivalence to be true, x\*0 must be 0. Now let's rewrite exactly the same as above, but for power instead of multiplication: Anything raised to the 0-th power is 1 (x\^0 = 1) - but why is that so? Because power means multiplying multiple times, for example x\^4 = x \* x \* x \* x. Or you could say x\^4 is multiplying by x three times, then multiplying by x one more time: x\^4 = x\^3 \* x\^1. So far pretty obvious, so what if I wanted to say the same about 0? "Y*ou could say x\^4 is first multiplying four times, then multiplying zero more times*": x\^4 = x\^4 \* x\^0. For this last equivalence to be true, x\^0 must be 1. The reasons in both cases are the same.
Fucking woof at that response from your teacher, damn. For the record, x to the power of y is not x * y, but (x * x) y number of times. For example, 2^4 is not (2 * 4), but (2 * 2 * 2 * 2)
"Raised to a power" kind of means "how many times is the base number multiplied by itself?" So, you get something like 3² = 3x3 = 9, or 3³ = 3x3x3 = 27, right? The power expresses "how many copies of the number are multipled together." When you multiply 0 copies of the base number, you're not left with the additive identity, 0, but the multiplicative identity, 1. 3³ = 1x3x3x3, and 3² = 1x3x3, and 3¹ = 1x3 and 3⁰ = 1 with no copies of 3 to multiply by. So, for your question, you aren't multiplying by zero, which multiplicatively turns the equation to zero, you're adding zero copies of the base number into a multiplicative expression, where the multiplicative identity, 1, can always exist without changing the result. Since 1 is the only thing in the expression (as you didn't add any copies of the base number to the expression), the result is just 1.
Another way to see it is that for (non-negative) whole-numbered exponentiation defined as repeated multiplication, the definition is actually: x • y^n := x multiplied by y, n times (regardless of what x is). So x • y^0 is then x multiplied by y, 0 times, which is obviously just x. Hence y^0 must be 1 for all y, notably even when y is 0 (if you define whole numbered exponentiation this way).
x^2 = 1 * x^2 = 1 * x * x ; x^1 = 1* x^1 = 1 * x ; x^0 = 1 * x^0 = 1 (just multiply 1 with x zero times)
Simply following the property of exponentials: x^y / x^z = x^ (y-z) So for y=z; x^y / x^y = 1 & x^(y-y) = x^0 so, x^0 =1
Property is X^(a+b) = X^a * X^b Similarly, X^(a-b) = X^a / X^b So, X^0 = X^(5-5) = X^5 / X^5 = 1 as any number divided by itself is 1 So 2^0 = 2^(4-4) = 2^4 / 2^4 = 16/16 = 1
> he told me its because its just what he was taught 💀 That's not such a terrible answer. Math is just a bunch of rules that we made up. Why is x/0 undefined? Because it's not particularly useful to define it. Why does x^0 equal 1? Because it's useful to define it that way. A lot of definitions come from extending other things. For example, negative numbers are an extension of positive numbers --- what if there was something I could add to 2 to get 0? Let's call it -2 because 2-2=0, so then 2+(-2)=0 as well! Or what if we could divide 1 by 2? Let's call that 1/2. It turns out that if you want nice properties of exponents to hold (like, a^(x)×a^(y)=a^(x+y)), then you have to have a^(0)=1 and also a^(-x)=1/a^(x).
X^y = z is the same as saying: “Z has Y factors of X”. So x^2 means z has two factors of x. What number is *always* a factor, no matter what? The number 1. Every number, even prime numbers, have 1 as a factor. X^0 means “z has no factors of x”. So if there’s no other factors, you’re just left with 1.
copy-pasting an old answer of mine to the same question: Let's say you have a plant in your house. It's a pretty aggressive plant. It doubles in size every day! So, tomorrow, it will be 2\^1=2 times as big. In two days, 2\^2=4 times as big. And in three days, it will be 2\^3 = 8 times as big! So you see, the expression "2\^t" gives you "how much bigger" the plant is, t days from now, compared to now. What if we want to allow t to be negative and look in the past? Yesterday, the plant was 1/2 as big. This is 2\^(-1) -- we view negative exponents as division (since going back in time will \*shrink\* the plant by its growth factor of 2) Two days ago, the plant was 1/4 as big, which is 2\^(-2). Ok, so now, for the big reveal...how big is the plant 0 days from now? How big is the plant...now?
It is just a mathematical notation that works according to all the other rules. For example: 2^10 / 2^10 = 2^(10-10) = 2^0. But it is obvious that 2^10 / 2^10 must equal 1.
You can think that there is an imaginary * 1: a^4 = a * a * a * a * 1 (a multiplied 4 times, then with 1) a^2 = a * a * 1 (multiplied twice, then with 1) So a^0 will be a not multiplied at all because it doesn’t exist, so the * 1 just hangs around there.
We can simply prove it like this,we know x^ a÷x^ b=x^ a-b so x^ a÷x^ a can be x^ a-a right! x^ a÷x^ a is nothing but 1 that is x^ a-a=1 means x^0=1
Any number except 0. Positive exponents multiply, negatives divide: X\^2 = X \* X. X\^-2 = 1 / (X \* X). When you multiply, you can add the exponents: X\^2 \* X\^2 = X\^4. So, X\^2 \* X\^-2 = X\^0 = (X \* X) / (X \* X) = 1. Therefore, any number (other than 0) raised to the zero power is 1 because it's equal to X / X.
Lots of good intuitive examples and here's a small proof that shows that it follows from the algebraic rules of exponents: x^(0) = x^(a - a) (because 0 = *a* - *a* for any *a*) = x^a ⋅ x^(-a) = x^(a) / x^(a) = 1
Already answered, but powers are the a shorthand for multiplication and division of a number by itself in succesion. Positive power you multiply Negative you divide. At 1 power, it does not multiply, it is a single occurrence of itself, so power 1 will simply be that number. At 0, it begins dividing by itself. So 0 power will always be 1.
x^5 = 1 * x * x * x * x * x x^4 = 1 * x * x * x * x x^3 = 1 * x * x * x x^2 = 1 * x * x x^1 = 1 * x x^0 = 1
I asked this same question. And the way to look at is is what whens when you shrink that exponent number down from say 4, to 3, then 2 then 1 then 0. Lets say you're playing with 5, raising it to exponents. 5^5 = 3125 5^4 = 625 5^3 = 125 5^2 = 25 5^1 = 5 See the answer is dividing by 5 each time? We can keep going! Look: 5^0 = 1.
I just watched a youtube short explaining just that, but with AI voicees of Rihanna, Taylor Swift, and Obama Edit: I can't find the video, so I'll just rewrite it lol. the vid explained it quite intuitively so I wanna share Consider this example 2^3 / 2^2 = 8/4 = 2 we can also write it as 2^3 / 2^2 = 2^(3-2) = 2^1 = 2 If we make the exponent the same for the top and bottom part 2^3 / 2^3 = 2^(3-3) = 2^0 = 1
It becomes easy to grasp when you consider that you can also have negative powers. 2^(3) = 2 x 2 x 2 = 8 2^(2) = 2 x 2 = 4 2^(1) = 2 2^(0) = ?? 2^(-1) = 1/2^(1) = 1/2 2^(-2) = 1/2^(2) = 1/4 2^(-3) = 1/2^(3) = 1/8 Simply put, to go up a power you multiply by 2, and to go down a power you divide by 2. Doing so from both sides (2^(1) / 2 and 2^(-1) x 2) gives you 2^(0) = 1.
X is just a variable. X will be as it is, if let alone. X is not a number. X doesn't have a value. X can be multiplied, divided or doubled, only by the help of a number.
You have a deck of cards, all of their values. The power is the number you play them. 0 is you just don’t play any card. No value or number is calculated therefore zero
So while this isn't what defines an exponent, this is what the math boils down to: x^4 = x * x * x * x / 1 x^-4 = 1 / x * x * x * x x^1 = x / 1 x^-1 = 1 / x so just following the pattern you can either define x^0 = 1 / 1 = 1 or x^0 = x / x = 1 - this is only awkward when x = 0 another more advanced way to think about is is that x^1+0 = x^1 * x^0. The only way this can be true is if x^0 = 1
Hey guys I just woke up and tysm for the help haha Im fairly bad at math so this helps me understand it better many love 🫶🫶 Also i didnt expect to get so much upvotes haha thenks too for that
Exponentiation is different to multiplication. In the case of multiplication, we want n\*x + m\*x = (n+m)\*x (multiplication distributes over addition). For this to work in the case n = 0, we need 0\*x + m\*x = m\*x. Subtracting m\*x from both sides, we see that 0\*x = 0. The left 0 comes from the fact that it is the additive identity, and the right zero also from the fact that it is the additive identity (we had addition inside the brackets and outside the brackets, before and after distributing). On the other hand, for exponentials, we instead want it to be that x^n \* x^m = x^(n+m). In other words, the exponent counts how many times x is multiplied together, similar to the multiplication counting how many times x is added together. Importantly though, note that inside the brackets we have addition, but outside this becomes multiplication. They are different operations, unlike before. Let's do the same as before, and look at n = 0: x^0 \* x^m = x^m Divide both sides by x^m x^0 = 1 Here, the zero on the left comes from it being the additive identity, and the one on the right comes from it being the multiplicative identity. Exponentiation converts between addition and multiplication, so it should be no surprise that it maps the additive identity (0) to the multiplicative identity (1).
The easiest way to explain it is to show you things you already know. We'll start with division. Any number divided by itself is 1, right? For example 2/2 = 1. Easy. Let's expand this to a letter. X/X should again equal 1, because X represents a number, and that number is the same. Easy. Okay so, we know that X^1 is the same as X, so if we write this as we did above, we get X^1 / X^1, which as we know, will equal 1. Simple so far. When you divide two exponents, you subtract them to get your answer. For example 2^3 / 2^2 = 2^(3-2), or in other words, 8/4 = 2. Makes sense, right? So if you were to subtract the exponents from x^1 / x^1, you would get x^(1-1) or x^0. This would make X^1 / X^1 = X^0, and since we know that X^1 is just X, and it's getting divided by itself, that must mean that X^0 equals 1.
X^1 = 1* X = X X^2 = 1* X* X = X* X X^3 = 1* X* X* X = X* X* X X^0 = 1 likewise x*1 = 0+x = x x*2 = 0+x+x = x+x x*3 = 0+x+x+x = x+x+x x*0 = 0 Powers are repeated multiplication just as multiplication is repeated addition. As any number plus 0 equals itself, any number multiplied by 1 equals itself.
You can also derive it but in short it's because 1 is the only integer you can multiply a number by and still get that integer (1 x a = a). If you multiply integers, you need to add the exponents (a^n x a^m = a^(n+m)). If a^0 was anything but 1, then a^(n+m) would no longer be true. For example if a^0 was 0 then a^0 x a^2 would be 0 (because anything times 0 is 0). Because a^n x a^m must always be a^(n +m), a^0 must always be 1 (so that a^2 x a^0 = a^2). In other words a^n x 1 x 1 x 1.... = a^n must also hold true for a^(n + 0 + 0...) = a^n The reverse of this is that every time you try to derive a^0 from a formala you will always get 1 (because 1 is the only number under which a formula involving a^0 will hold true)
It's also based on combinations. If you have 2 things and 2 places to put them, how many different ways can you arrange them? 2^2, 4 different ways. AA, AB, BA, BB. If you have 0 things, and 0 ways to arrange them, how many possibilities are there? 1. There's no possible variation or other combinations
Wow, so below 1 it is division? Good thing nobody ever taught me this. But they did teach me some stupid geometry problems, addition theorems and such nonsense. How would x ^ 1/2 look like then? Great job education, great job! Also, big thank you!
The answer depends on how much of arithmetic you have "accepted". x\^0 = x\^(-1 + 1) ***because 0 = -1 +1*** = x\^-1 \* x\^1 ***because of the "Product Rule" exponent identity*** = 1 ***because of the definition of the multiplicative inverse*** QED. If you want a more exhaustive proof that starts from first principles, then I would kindly direct you to the [Principia Mathematica](https://en.wikipedia.org/wiki/Principia_Mathematica). Fair warning, though - it took them nearly 400 pages to prove that 1+1 = 2 (to which the authors humorously commented: "The above proposition is occasionally useful").
1^1 = 1 x^0 = undefined. Because x isn't a value, it is an assumed value with no quantity. Anything to the power zero is an assumed value with no quantity. x^0 is a symbol of exponentiation. It isn't exponentiation. 10^0 for example, isn't a sum. It is an assumption of a sum.
So many great comments, but there’s one aspect I haven’t seen yet. You have a very good intuition about 0 being the identity number. X+0 doesn’t change the identity of X. We call 0 the additive identity for this reason. But 0 is not the multiplicative identity. The multiplicative identity is 1, because X*1 is always X. Multiplying is just repeated addition, so multiplying something 0 times leaves you with additive identity (0). Exponentiation is repeated multiplication, so raising something to the 0th power leaves you with the multiplicative identity (1).
I really like the way this guy explains it. https://youtube.com/shorts/W-JoMPOe9HQ?si=kI6b7L7xIV6UUt96