raise to power zero?

Why is it that anything raise to power 0 is 1?

#Algebra #AlgebraicManipulation

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

$0 = k-k$ , Thus $n^0 = n^{k-k}$ (say k is non-zero as we want to prove for 0) = $\frac{n^k}{n^k} = 1$

But this won't be true for $n=0$ as the denominator becomes $0^k = 0$ for non zero k , and hence the value is not defined (this is why we say that $0^0$ is not defined . ) Thus ANYTHING raised to 0 is not 1 .... any "non-zero" thing raised to 0 is 1....
;)

Aditya Raut - 7 years ago

as u said that , 0 power 0 is not defined.... bt as far i know.. 0 to the 0 is e whose value is 2.17828(approx.)

Aman Agrawal - 7 years ago

Let a number $a$ be raised to $x$ .

We have $a^x$ .

Let's divide it by itself-

$\frac{a^x}{a^x}$ .

Surely, these two quantities are the same-

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

Let's apply the law of indices to L.H.S & normal division to the R.H.S. Now, we have,

$a^{x-x} = 1$

$\Rightarrow \boxed{a^{0} = 1}$ .

Ameya Salankar - 7 years ago

Take example as 2^0. It can be expessed as 2^(x-x) = (2^x)/(2^x)=1

Kushagra Jaiswal - 7 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$