I know that x^0 = 1 and x^1 = x where x is any integer. Why is that the case? How can I best understand this concept of zero and one powers without getting too technical?

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- Dec 10th 2013, 07:30 PMnycmathx^0 and x^1
I know that x^0 = 1 and x^1 = x where x is any integer. Why is that the case? How can I best understand this concept of zero and one powers without getting too technical?

- Dec 10th 2013, 07:54 PMromsekRe: x^0 and x^1
This is discussed in countless places on the web. Here for example

- Dec 10th 2013, 08:19 PMnycmathRe: x^0 and x^1
I did not realize that such a basic idea has a complicated definition.

- Dec 10th 2013, 08:39 PMromsekRe: x^0 and x^1
- Dec 11th 2013, 01:26 AMMelody2Re: x^0 and x^1
Hi Nycmath,

I have been trying to learn Latex and I used your post for practice inspiration.

I finally got it to look almost how I wanted so this is what I wanted to say.

Unfortunately I don't know how to transfer the program that I built in TEX to this site. Maybe someone can help me with that, so I had to snip the output and attach it as a thumbnail.

It might be what you were after. - Dec 11th 2013, 05:48 AMnycmathRe: x^0 and x^1
- Dec 28th 2013, 04:15 PMHallsofIvyRe: x^0 and x^1
You don't have to use a "complicated definition". For this question, it is sufficient to use the definition "$\displaystyle a^n$, for n positive integer, is a multiplied by itself n times" From that, it follows immediately that $\displaystyle x^1= x$. To get $\displaystyle x^0$, we have to extend that original definition to 0, which is not a "positive integer". We are, of course, free to

**define**something however we want, but from that previous definition, it is easy to show that $\displaystyle a^{n+ m}= a^na^m$. That is such a nice property that we would like to define $\displaystyle a^0$ in such a way as to have that still true. That is, we want $\displaystyle a^{n+ 0}= a^na^0$. On the left, n+ 0= n so $\displaystyle a^{n+0}= a^n$. That mean that, to have $\displaystyle a^{n+0}= a^n= a^na^0$, we must define $\displaystyle a^0= 1$. (Part of showing that $\displaystyle a^{n+m}= a^na^m$ requires that a not be equal to 0. So we really have "$\displaystyle a^0= 1$ as long as a is not 0" and $\displaystyle 0^0$ is not defined.) - Dec 28th 2013, 05:34 PMProve ItRe: x^0 and x^1
$\displaystyle \displaystyle \begin{align*} x^0 \end{align*}$ is not $\displaystyle \displaystyle \begin{align*} 1 \end{align*}$ for any integer. $\displaystyle \displaystyle \begin{align*} x^0 = 1 \end{align*}$ for all $\displaystyle \displaystyle \begin{align*} x \neq 0 \end{align*}$. $\displaystyle \displaystyle \begin{align*} 0^0 \end{align*}$ is undefined.

- Dec 28th 2013, 07:22 PMsakonpure6Re: x^0 and x^1
Simply: $\displaystyle x^3 / x^3 = 27 / 27 = 1 $ Now $\displaystyle x^3/x^3 = x^(3-3) (according to exponential law) = x^0 = 1$

- Dec 28th 2013, 07:43 PMMINOANMANRe: x^0 and x^1
- Dec 28th 2013, 09:12 PMsakonpure6Re: x^0 and x^1
- Dec 29th 2013, 04:16 AMnycmathRe: x^0 and x^1
Thank you everyone for your help.