# Thread: number to the zeroth exponent

1. ## number to the zeroth exponent

I'm rereading my precalculus material and i realized my book completely skimmed on this on.
why a number that is not equal to zero to the zeroth power equals 1?

2. ## Re: number to the zeroth exponent Originally Posted by mohsentux I'm rereading my precalculus material and i realized my book completely skimmed on this on.
why a number that is not equal to zero to the zeroth power equals 1?
That is a simple one. To compute say x^9/x^6 you get x^(9-6)= x^3.

Consider x^3/x^3. On one hand, x^3/x^3=1 since when you divide the same things (and they are not 0's, since you said x is not 0) you always get 1. On the other hand, x^3/x^3= x^(3-3) = x^0. Equating both results we get x^0 = 1. This will be true if we replace the 3's with any one number.

3. ## Re: number to the zeroth exponent

I think that you should have known about the Index Law from your high school mathematics

Actually x^0 is intentionally defined to be zero for the nonzero x. Recall from the Index Law that x^m/x^n = x^(m-n). So the motive for this definition is to render the Index Law valid when m=n. Thus when m=n we get x^m/x^m = x^(m-m) = x^0 = 1. We also have to be caution that we call the Index Law a "law", but not a "Theorem". Note that m and n ought to be natural numbers and we would not offered proof for the Index Law

4. ## Re: number to the zeroth exponent Originally Posted by KLHON I think that you should have known about the Index Law from your high school mathematics

Actually x^0 is intentionally defined to be zero for the nonzero x.
Of course you mean "defined to be 1", not 0.

Recall from the Index Law that x^m/x^n = x^(m-n). So the motive for this definition is to render the Index Law valid when m=n. Thus when m=n we get x^m/x^m = x^(m-m) = x^0 = 1. We also have to be caution that we call the Index Law a "law", but not a "Theorem". Note that m and n ought to be natural numbers and we would not offered proof for the Index Law
Yes, defining $\displaystyle x^0$ is a definition that makes that 'index law' true for x= 0. We can extend that idea. The reason we define $\displaystyle x^{-n}$ to be $\displaystyle \frac{1}{x^n}$ is that, that way, we have $\displaystyle x^nx^m= x^{m+n}$ still true even for negative n and m. And the reason we define $\displaystyle x^{1/n}= \sqrt[n]{x}$ is so that $\displaystyle (x^n)^m= x^{nm}$ will still be true for m and n rational numbers.

5. ## Re: number to the zeroth exponent

I thought the OP was talking about 0^0 ?

6. ## Re: number to the zeroth exponent Originally Posted by rodders I thought the OP was talking about 0^0 ?
Here is the OP: Originally Posted by mohsentux I'm rereading my precalculus material and i realized my book completely skimmed on this on.
why a number that is not equal to zero to the zeroth power equals 1?

7. ## Re: number to the zeroth exponent

Ah yes, my bad

8. ## Re: number to the zeroth exponent

Speaking of $0^0$, if you want the binomial expansion $(1+x)^n = \sum_{k=0}^n \binom n k x^k$ to work for all $x$, you need to define $0^0 = 1$. Look at the first term when $x=0$.

9. ## Re: number to the zeroth exponent Originally Posted by Walagaster Speaking of $0^0$, if you want the binomial expansion $(1+x)^n = \sum_{k=0}^n \binom n k x^k$ to work for all $x$, you need to define $0^0 = 1$. Look at the first term when $x=0$.
Actually there are a number of reasons to call $\displaystyle 0^0$ either 1 or 0. So it is labeled "undefined."

I have noted a number of high powered Physicists have a tendency to define $\displaystyle 0^0 = 1$. Because it works in certain situations we use it. (It's part of what I call "Physics Math" which sometimes only has a vague relationship to actual Mathematics.)

-Dan