Why does any nonzero number to the zeroth power = 1? [duplicate]

I can't properly wrap my head around this odd concept, though I feel I'm almost there.

A non-zero base raised to the power of 0 always results in 1. After some thinking, I figured this is the proof: $\frac{{x}^{2}}{{x}^{2}}= x^{2-2}=x^{0}=1$

Assuming that's true, would it be correct to assume that anything raised to 0 is a "whole" (1)? Because if $\frac{{x}^{2}}{{x}^{2}}=1$, then no matter what x is, it will always result in 1.

I would like to understand this concept intuitively and deeply, rather than just memorizing that $x^{0}=1$

EDIT: Thank you all for the answers. Each and everyone of them have been insightful and I've now gained a deeper understanding. This is a new account, so it seems I can't upvote, but if I could I would upvote each and everyone of you. Thanks :)

marked as duplicate by Hayden, JMoravitz, S.C.B., Hans Lundmark, J. M. is a poor mathematicianJan 31 '17 at 8:39

• I have a few good intuitive ways to think about this, if you'd like me to post them. – The Count Jan 31 '17 at 2:26
• @TheCount Yes, it would be very much appreciated. Thank you. – J. Doerty Jan 31 '17 at 2:30
• Well, smaude and I had the same idea, but there you go! – The Count Jan 31 '17 at 2:33
• Careful: Make sure you take note that your reasoning a la $\frac{x^2}{x^2}$ cannot be used for $x=0$, which is an intuition as to why, contrary to popular claims, $0^0$ is undefined and not equal to $1$. See the links above for further details on this. – MM8 Jan 31 '17 at 2:36
• @J.Doerty You can probably upvote now. 29 rep. – The Count Jan 31 '17 at 15:18

Here's a good heuristic that helped me feel better about it back in the day:

Consider any number $x$ to any power. We will choose, say, $5^3$.

$5^3 = 125$. Divide both sides by $5$. You get $5^2=25$, which we know. Again. Then you get $5^1=5$. We usually leave exponents of $1$ off but we'll keep it here. Do you see what happens? The power goes down by $1$ each time. We can do it again. Then following our pattern, $5^0=1$. But $5$ wasn't special!

You can do this with any real number. Of course, for irrationals it can get a little fuzzy.

By the way, we can keep going. Divide by $5$ again. You get $5^{-1}=\frac{1}{5}$. I hope this helps!

• Thank you for the answer, very nicely put. I have one more related question: Does $x^{0}$ always just mean 1? So for whatever equation I have, I can just replace 1 with $x^{0}$ and nothing would change? – J. Doerty Jan 31 '17 at 2:45
• Yes, for the most part. $0^0$ is undefined in most situations. There is a link to all sorts of good info on that in the comments. Also, be sure to stick to real numbers for this trick. You can do it with complex but it's... muddy, to say the least. – The Count Jan 31 '17 at 2:47
• Also, the other answers are all very good and useful. Be sure to look at them all, and upvote everyone else who helped you! – The Count Jan 31 '17 at 2:48

Clearly the value of $x^0$ is a matter of definition. A simple way to think about this that only involves non-negative integer powers is to notice that for positive integers $a$ and $b$ you want $$x^{a+b}=x^a \, x^b.$$ If you were to define $x^0$ you would likely want to have that hold for $b=0$, that is $$x^{a+0}=x^a \, x^0 \rightarrow x^{a}=x^a \, x^0 \rightarrow x^0=1.$$ Now there are other ways of getting to this that involves the definition of the exponential function (as in infinite polynomial series) but the above is the simplest.

If you would think of it backwards, you would see that, in the case of 2, for example,

$2^4$ =16

$2^3$ = 8

$2^2$ = 4

$2^1$ = 2

So, it is logical to continue dividing by two,and reach the conclusion that $2^0$ = 1. This is true for negative powers, and is so because it is the most logical way to define powers outside of the natural numbers.

• I posted my answer at the same time as you, but I guess it's good to have two examples. Have an upvote! – The Count Jan 31 '17 at 2:32
• haha thanks, I'll upvote you too. – smaude Jan 31 '17 at 2:34
• fwiw I think it's clearer when written like this in a stack – MichaelChirico Jan 31 '17 at 6:13
• @MichaelChirico I can see that. I'm just a verbose person. – The Count Jan 31 '17 at 14:40
• @TheCount it's nice to see written out too, which is why I upvoted yours as well :) – MichaelChirico Jan 31 '17 at 14:51

because $x^0=e^{0log(x)}=e^0=1, x> 0$.

• who said $e^0 =1$? – MichaelChirico Jan 31 '17 at 6:13
• @MichaelChirico One way to see this is via a power series. – Henricus V. Jan 31 '17 at 6:17
• afaik power series involve the zeroth power for the constant term, so it's still circular – MichaelChirico Jan 31 '17 at 6:18
• @MichaelChirico The exponential function can be defined by $\exp'=\exp$ and $\exp(0)=0$. In that sense, $e^0=1$ is a convention, but of course the only useful one ... :) – Hagen von Eitzen Jan 31 '17 at 7:48
• @MichaelChirico You can explicitly move the first term out and have the series start at $1$. – Henricus V. Jan 31 '17 at 15:33

Let me start off by saying that I am not a mathematician, and that I will be using some pseudo-mathematical terms in the interest of writing something more akin to simplified English rather than accurate mathematical jargon.

The question as stated is:

Why does any non-zero number to the zeroth power equal one?

To answer this question lets first talk about what is meant by "zeroth" power.

"zeroth power" refers to exponentiation. To understand why the zeroth power works the way it does it's important that we properly define exponentiation.

Exponentiation is the act of raising a number to the power of another number. That's actually not too helpful because now you need to know what "raising to the power" means.

...

But first, lets talk about multiplication.

Multiplication is the act of adding a number ($a$) some other number ($b$) of times ($a \times b$).

$$2 + 2 + 2 = 2 \times 3$$

This is all well and good, but when we talk about mulitplying by $0$ we need to know what number to put on the left hand side:

$$? = 2 \times 0$$

The base for addition is $0$. It's the additive identity. Every addition equation may be implicitly started with $0$. This means that above, two times three is actually:

$$0 + 2 + 2 + 2 = 2 \times 3$$

In this form certain behaviors become quite clear:

$$0 + 2 + 2 + 2 = 2 \times 3$$ $$0 + 2 + 2 = 2 \times 2$$ $$0 + 2 = 2 \times 1$$ $$0 = 2 \times 0$$

Negatives also make sense, because instead of adding numbers, you do the opposite, you un-add (often called "subtraction"):

$$0 = 2 \times 0$$ $$0 - 2 = 2 \times -1$$ $$0 - 2 - 2 = 2 \times -2$$

...Ok, with all that in mind, now it's time to look at exponentiation.

Exponentiation is the act of multiplying a number ($a$) some other number ($b$) of times ($a ^ b$)

$$2 \times 2 \times 2 = 2 ^ 3$$

This is all well and good but when we talk about "raising to the power of 0" we need to know what number to put on the left hand side:

$$? = 2 ^ 0$$

The base for multiplication is $1$. It's the multiplicative identity. Every multiplication equation may be implicitly started with $1$. This means that above, two to the power of three is actually:

$$1 \times 2 \times 2 \times 2 = 2 ^ 3$$

In this form certain behaviors become quite clear:

$$1 \times 2 \times 2 \times 2 = 2 ^ 3$$ $$1 \times 2 \times 2 = 2 ^ 2$$ $$1 \times 2 = 2 ^ 1$$ $$1 = 2 ^ 0$$

Likewise, negatives also make sense, because instead of multiplying numbers, you do the opposite, you un-multiply (often called "division"):

$$1 = 2 ^ 0$$ $$1 \div 2 = 2^{-1}$$ $$1 \div 2 \div 2 = 2^{-2}$$

Note that these patterns hold regardless of the base:

$$n \times n \times n \times 1 = n^3$$ $$n \times n \times 1 = n^2$$ $$n \times 1 = n^1$$ $$1 = n^0$$ $$1 \div n = n^{-1}$$ $$1 \div n \div n = n^{-2}$$ $$1 \div n \div n \div n = n^{-3}$$