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, hence changes zero into 1. (It also has positivity and continuity hypotheses.)

[edit: I have added a necessary condition for uniqueness that I originally forgot to mention.]

I.e. if a>0, then f(x) = a^x, is a homomorphism from (Q,+) to (R>0,*). I.e. a 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.

Indeed it is the only such homomorphism such that f(1) = a.
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; i.e. the value of the function at zero is always uniquely determined, no matter what the value is at 1.
For the same reason, i.e. because logs change multiplication into addition, the value of a logarithm function at 1 is always zero, no matter what (positive) base is chosen for the log. And a continuous log function L, from (R>0,*) to (R,+), is determined by a choice of positive a, (i.e. the base), such that L(a) = 1.