# Math question

1. May 22, 2006

### Universe_Man

I always learned this in math, but never really questioned it.

why does $$n^0=1$$

2. May 22, 2006

### LeonhardEuler

Alright, first you know $x=x^1$ and $x^1x^1=x^{1+1}=x^2$, right? And in general, $x^ax^b=x^{a+b}$ when a and b are greater than or equal to one. So why not define $x^0$ so that this is true even if a or b is zero? If this is the case then
$$x^0x^a=x^{a+0}=x^a$$
We have x0xa=xa, so as long as x is not zero we can divide by xa to come out with $x^0=1$.

More generally, powers are defined so that $x^ax^b=x^{a+b}$ is true even when a and b are not integers and even when they are not real.

Last edited: May 22, 2006
3. May 22, 2006

### Markjdb

A simple proof:

Because $$(n^a)*(n^b) = n^{a+b}$$,
it can be said that $$n^0$$ is equivalent to $$n^1 * n^{-1}$$.
$$n * (1/n) = 1$$,
therefore, $$n^0 = 1$$

Last edited: May 22, 2006
4. May 22, 2006

### Universe_Man

OH OH ok I get it, thanks alot guys.

5. May 25, 2006

### HallsofIvy

Staff Emeritus
In other words, if we want aman= an+m to be true for 0 as well as positive integer value of m and n, we must define a0= 1.
It is also true that if we want aman= an+m to be true for negative integer powers, then we must have ana-n= a0= 1. In other words, we must define $a^{-n}= \frac{1}{n}$.

To see how we must define $a^{\frac{1}{n}}$, look at the law
(an)m= amn.