Why does $n^0 = 1$?

Question

Why is it that $n^0 = 1$?
I understand how $n^2 = n*n$ and how $n^1 = n$ but I can't understand why $n^0 = 1$.

Answer 1

Exponentiation satisfies the laws of exponents: $a^{b+c} = a^ba^c$ . If we want this law to still be satisfied when we extend to the case $b=0$, we need to have $a^{0+c} = a^0a^c$ , and therefore we need to have $a^0 = 1$.

Answer 2

HINT:

Using Exponent Laws(1) , (2)

$$n^{a+b}=n^a\cdot n^b$$

Setting $a=0,$ $$n^b=n^b\cdot n^0$$

Check when $n^b$ can be cancelled

Answer 3

You can think of the rule that $\frac{x}{x}=1$ for any nonzero number $x$.

Then, using for example $n^2$, we get:

$$1=\frac{n^2}{n^2}=n^{2-2}=n^0$$

Hence it makes sense to have $n^0=1$.

Answer 4

\begin{align*} n^m &= n\times \ldots \times n\; (\textrm{for}\; m \in \mathbb{Z}_{>0})\\ &\vdots\\ n^2 &= n\times n\\ n^1 &= n\\ n^0 &= 1\\ n^{-1} &= \frac{1}{n}\\ n^{-2} &= \frac{1}{n\times n}\\ &\vdots\end{align*}

To go from $n^{m}$ to $n^{m+1}$, you multiply by $n$, and to go from $n^{m}$ to $n^{m-1}$, you divide by $n$.

Answer 5

When you do $m\cdot n$, you start from $0$ and keep adding $n$ as many times as $m$ says: so if $m=0$, you just have $0$ because you have nothing else to do; if $m=1$, you get $0+n$; if $m=2$ you do $0+n+n$. And so on.

Exponentiation is exactly the same, but with multiplication replacing addition and $1$ replacing $0$: for doing $n^m$ you start from $1$ and keep multiplying.

If $m=0$ you have $1$ and do nothing else; if $m=1$ you have $1\cdot n$; if $m=2$ you have $1\cdot n\cdot n$. And so on.

This is usually formalized in a recursive definition: $$ 0n=0,\qquad (m+1)n = mn+n $$ for multiplication, that becomes $$ n^0=1,\qquad n^{m+1}=n^m\cdot n. $$ So, for instance, $$ n^3=n^2\cdot n=(n^1\cdot n)\cdot n=((n^0\cdot n)\cdot n)\cdot n =((1\cdot n)\cdot n)\cdot n $$ and parentheses can be omitted.

Of course this has been recognized much time later than when exponentiation was introduced as a shorthand for “repeated multiplication“. The law of exponents $$ n^{a+b}=n^a\cdot n^b $$ is (almost) obvious when $a,b>1$; but it can extended to the exponents $0$ and $1$ so that it still holds by defining $n^1=n$ and $n^0=1$.

Note that nowhere it's assumed that $n>0$. It can be any “number”, as long as a neutral element form multiplication exists and multiplication is associative; indeed this can be done in any semigroup with identity (also called monoid).

The symmetry with multiplication could be seen better if we wrote $^mn$, but, alas, people started to write it the other way around.