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.

 

[Ssnow]

$a^0$ can't be $0$ because if $a^0=a\cdot 0 =0$ for every $a$, then $0^0=0$, that is not true in mathematics ... In addiction $a^0=a^{0+0}=a^0\cdot a^0$, now $a^0=\left(a^0\right)^2$ that is only if $a^0=1$ or $a^0=0$. By the nosense of $a^0=0$ follow that $a^0=1$.
Ssnow

 

[wrobel]

$$a^x:=\exp(x\ln a),\quad \exp(x):=1+\sum_{k=1}^\infty\frac{x^k}{k!},\quad a>0$$

 

[symbolipoint]

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

Nearly the best answer!

 

[symbolipoint]

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

Also nearly the best answer!

 

[sophiecentaur]

I think this is one of those things that we 'learn' first - as a bit of trickery and then follow it up with a believable proof.

 

[DaveC426913]

I think this is one of those things that we 'learn' first - as a bit of trickery and then follow it up with a believable proof.

I agree with that, because that's the way I learned it:

a^0 is defined as 1.

Only now do I see why it is so.

 

[symbolipoint]

I agree with that, because that's the way I learned it:

a^0 is defined as 1.

Only now do I see why it is so.

So much like how some of us can refer to Laws Of Exponents and what happens when applying the Laws when you have a number to power of 1 in numerator and same number to power of 1 in denominator --- Boom ! Exponent 0; already understanding the ration is 1.

 

[Hill]

In arithmetic, $x^k$ is defined recursively. The recursive step is $x^k=x^{k-1}x$. What is the base case then, $x^0$? For the recursion to work it just has to be 1.

 

[Gavran]

Also nearly the best answer!

@symbolipoint, thank you.

In any of the earlier posts, nobody mentioned the solution for $0^0$. Using post #8 for defining $0^0$ will generate the result $0^0 =0^{x-x}=0^x/0^x=0/0$, which is undefined.

The mathematical expression $0^0$ has different interpretations, and it is defined depending on the context.

It can be proved that $$ \lim_{x\to0^+}x^x=1 $$, but it cannot be proved that $0^0=1$. To support the continuity of function $x^x$ on the interval $\left[0,\infty\right)$, $0^0$ can be defined as equal to $1$.

One more illustration is the power series $$ e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!} $$, where $e^0=1$ holds only for $0^0=1$.

There are other examples with their own interpretations of $0^0$, and all these examples have one thing in common, and it is that they are all based on convenience. There is no example that assigns the value to $0^0$ other than $1$ except in scenarios where $0^0$ is undefined.