Why ##a^0=1##?

[Mike_bb]

Hello!

I have problem in understanding why $a^0=1$.

I represent $a^x$ as $1*a*a*a*a*...*a$ (x times)

Why $a^0=1$? Why is $a^0$ not equal to $0$?

Thanks.

 

[Ibix]

It's consistent with $a^{n+1}=a\cdot a^n$

 

[Ibix]

Or, more generally, $a^{n+m}=a^na^m$ implies $a^0=1$, because otherwise $a^{n+0}\neq a^n$.

 

[martinbn]

Hello!

I have problem in understanding why $a^0=1$.

I represent $a^x$ as $1*a*a*a*a*...*a$ (x times)

Why $a^0=1$? Why is $a^0$ not equal to $0$?

Thanks.

Well if your $x$ is zero, then you will have no $a$'s, so just the $1$ in the begining.

 

[jedishrfu]

Another way to look at it

$a^0 = 1$

$a * a^0 = a * 1$

$a^1 = a$

Of course you can reverse the steps and conclude that $a^0 = 1$

 

[Hill]

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

 

[Hornbein]

Exponents are about multiplication, and 1 is the multiplicative identity.

If exponents represented repeated addition the zeroeth power would be 0, the additive identity.

 

[Gavran]

Very similar to post #6.
$$ a^0=a^{x-x}=\frac{a^x}{a^x}=1 $$

 

[mathwonk]

This has several excellent answers and I cannot improve on them. I just have this persistent nagging voice wanting to express this abstractly, if perhaps not actually usefully for a beginner.
Namely, all answers point out, one way or another, that exponentiation is a function that changes addition into multiplication. (It also has positivity and continuity hypotheses.)

I.e. a>0, then f(x) = a^x, is the only homomorphism from (Q,+) to (R>0,*). I.e. the only function from rationals to positive reals, such that, for all real x,y, f(x+y) = f(x)*f(y).
As a consequence, f carries the additive identity, namely 0, to the multiplicative identity, namely 1.

If you want such a uniqueness statement also for the exponentiation function defined on all reals, i.e. for the function f(x) = a^x, from (R,+) to (R>0,*), you need a continuity hypothesis on f, but this is unnecessary for the desired statement about 0 and 1.