# Math Help - Exponent Rules: (x)^1 and (x)^0

1. ## Exponent Rules: (x)^1 and (x)^0

We know that anything raised to the first power equals itself.
For example, (x)^1 = x. We also know that anything raised to the zero power equals 1. For example, (x)^0 = 1. Why is that true for both cases?

2. Originally Posted by RTC1996
We know that anything raised to the first power equals itself.
For example, (x)^1 = x. We also know that anything raised to the zero power equals 1. For example, (x)^0 = 1. Why is that true for both cases?
Well, $x^1 = x$ makes sense, does it not? If $x^2 = x*x$ it also follows that $x^1 = x$. One x multiplied together (so to speak) is just x.

Similar logic applies to $x^0$. Zero x's multiplied together is obviously 0. [EDIT: Wow, I've handled x^0 enough times you'd think I wouldn't mess this up! x^0 = 1 as explained below. Let that be a lesson to me!]

Do note that 0^0 is a little tougher to settle. Why? An explanation can be found here:
sci.math FAQ: What is 0^0?

3. $\displaystyle x^0=x^{a-a}=\frac{x^a}{x^a}=1$

4. Originally Posted by Grep
Well, $x^1 = x$ makes sense, does it not? If $x^2 = x*x$ it also follows that $x^1 = x$. One x multiplied together (so to speak) is just x.

Similar logic applies to $x^0$. Zero x's multiplied together is obviously 0.

Do note that 0^0 is undefined, though. Why? An explanation can be found here:
sci.math FAQ: What is 0^0?
$\displaystyle x^0\neq 0$

5. $\frac{a^x}{a^x} = 1$...we know this.

$\frac{a^x}{a^y} = a^{x-y}$

$\frac{a^x}{a^x} = 1 = a^{x-x} = a^0$

We have shown one statement.

$\frac{a^{x+1}}{a^x} = a = a^1$

This is the other statement.

6. Originally Posted by dwsmith
$\displaystyle x^0\neq 0$
Sorry, wrote too quickly there, didn't I? My apologies.

7. Originally Posted by RTC1996
We know that anything raised to the first power equals itself.
For example, (x)^1 = x. We also know that anything raised to the zero power equals 1. For example, (x)^0 = 1. Why is that true for both cases?
$\displaystyle\frac{x^5}{x^3}=\frac{(x)(x)(x)(x)(x) }{(x)(x)(x)}=\frac{(x)(x)(x)}{(x)(x)(x)}\;(x)(x)=( 1)x^2$

We can get this result by forming the new index from 5-3.
Basically 3 of the x's in the denominator "cancel" 3 of the x's in the numerator to give 1.

Hence we can write the "general" way as shown in earlier posts.

Therefore, in index form... $\displaystyle\frac{x^5}{x^5}=\frac{x^n}{x^n}=1=x^0$ if we subtract the indices.

$x^0$ is 1 because it is an "index" way to write a number divided by itself.

$\displaystyle\frac{x^5}{x^4}=\frac{x^3}{x^2}=\frac {x^{n+1}}{x^n}=x^1=x$

Following on from this, you will see why negative indices represent a power of a number divided by a higher power of the same number.