# Thread: any no raised to zero

1. ## any no raised to zero

Can anyone explain me -- Why any number raised to (Power of) zero is always equal to one.

Regards,

2. Consider Numerical Approach:

$\displaystyle 5^2 = 25$
$\displaystyle 5^1 = 5$
$\displaystyle 5^0= 1$
$\displaystyle 5^{-1}= \frac{1}{5}$

See, We are dividing by 5 each of the power values to get the one below. At any number to the power of zero, we will get one. It's like that for any number.

For example, 9:
$\displaystyle 9^2 = 81$
$\displaystyle 9^1 = 9$
$\displaystyle 9^0= 1$
$\displaystyle 9^{-1}= \frac{1}{9}$

3. Originally Posted by avinash
Can anyone explain me -- Why any number raised to (Power of) zero is always equal to one.

Regards,
Well $\displaystyle 0^0$ is undefined, but other than that it is so that $\displaystyle a^xa^y=a^{x+y}$ works for any real $\displaystyle x$ and $\displaystyle y$ and $\displaystyle a >0$ . Also for other consistency reasons.

RonL

4. Originally Posted by avinash
Can anyone explain me -- Why any number raised to (Power of) zero is always equal to one.

Well that is not always true $\displaystyle 0^0$ is not defined
.
However, if $\displaystyle x \ne 0 \Rightarrow \quad x^0 = 1$.
Every nonzero number has a multiplicative inverse denoted by $\displaystyle x^{-1}$.
Use the law of exponents: $\displaystyle 1 = \left( x \right)\left( {x^{ - 1} } \right) = x^{1 - 1} = x^0$

Alternatively: $\displaystyle 1 = \frac{{x^k }}{{x^k }} = x^{k - k} = x^0$