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If $16^2 = 16 \times 16$ and if $16^1 = 16$ why doesn't $16^0 = 0$? I don't understand the rationale. Does $16^0$ express the idea that $16$ is multiplied by nothing and therefore is $1$?

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Why should it be $0$? –  Samrat Mukhopadhyay Jul 29 '13 at 7:35
What is $16^m\cdot16^n$? Then put $n=-m$. –  Andrea Mori Jul 29 '13 at 7:36
This is an axiom in a sense. –  Samrat Mukhopadhyay Jul 29 '13 at 7:37
$16^0$ doesn't express that $16$ is multiplied by nothing but that 16 is multiplied by itself $0$ times, so actually there's nothing multiplied and it doesn't depend on the $16$ at all. –  user87690 Jul 29 '13 at 7:43
If $a > 0$, then $a^0$ is defined to be $1$, and we chose to define it this way so that the rule $a^{m+n} = a^m a^n$ will be true not just when $m$ and $n$ are positive integers, but also when either $m$ or $n$ is equal to $0$. –  littleO Jul 29 '13 at 8:06
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marked as duplicate by Mark Bennet, Jared, ShreevatsaR, Davide Giraudo, mau Jul 29 '13 at 8:21

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5 Answers

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$16^0$ is the product of zero factors with value $16$ each. So the question reduces to the more general question of what is the product of no numbers, that is, an empty product.

Now of course that's a matter of definition, so the actual question is: What is the most reasonable definition of an empty product?

Now if you look at the definition of the product of more than two numbers, you notice that you can do the multiplication step by step. So if you want to do the product of four numbers, you first multiply three of them, and then multiply the result with the fourth. To calculate the product of three numbers, you first multiply two numbers, and multiply this result with the third.

Now the product of two numbers is elementary (we need to have that defined in order to perform the recursion from the previous paragraph!). However we can still ask what the product of exactly one factor should be. Now the obvious choice is to make it exactly that number. If we do so, we again find that we find the product of two numbers by first determining the product of one of the numbers (which of course is the number itself), and then multiplying that with the other one.

So now we get back to the question of how we should define the product of no numbers. Well, given the observations above, it is clear that we want it to define so that we get the product of one number by taking the product of no numbers, and multiplying it with that number. Since the product of exactly one number is the number itself, the product of no number should be a number which, when multiplied with an arbitrary other number, gives that number.

Now it turns out that such a number indeed exists, and moreover is unique. It is the number $1$. Therefore there's only one reasonable choice for the value of the empty product, and that is $1$.

Since $a^0$ is just the product of zero factors with value $a$ each, that is, the empty product, it also gets the value $1$. This is of course also true if $a$ has the value $16$.

We can also verify that this is a reasonable definition by checking that with this definition all the power laws also continue to hold for exponent $0$: $$\begin{align} a^m = a^m \cdot 1 = &&\color{red}{a^m\cdot a^0} &\color{red}{= a^{m+0}} &&= a^m\\ 1 = a^0 = &&\color{red}{a^{0\cdot m}} &\color{red}{= (a^0)^m} &&= 1^m = 1\\ 1 = a^0 = &&\color{red}{a^{m\cdot 0}} &\color{red}{= (a^m)^0} &&= 1 \end{align}$$ Note that the first one would fail with the definition $a^0=0$.

Note however that the empty product argument is stronger than the demand that the power laws should hold because from the empty power argument you also immediately get that it is reasonable to define $0^0=1$, while the power law rules work both with $0^0=0$ and $0^0=1$. Also, we know that eventually the power laws do partially break down (as soon as negative bases and non-integer exponents are considered), so the power laws by themselves make a rather weak argument.

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took me a couple of days to get the concept in full but now your answer really makes sense to me now. Thanks again. –  Jessica M. Jul 30 '13 at 17:30
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The power law is that $a^{n+1}=a\cdot a^n$

Apply with $n=0$ to give $a=a^1=a\cdot a^0$

So either $a=0$ or divide by $a$ to give $a^0=1$

Set $a=16$.

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Perhaps this is what you're looking for ($x \neq 0$ of corse):

$$x^0 = x^{(-1+1)} = x^{(-1)} \cdot x^1 = \frac 1 x \cdot x = 1$$

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And if you punch in $0^0$ in your calculator, you get an error because of the second last step. A bonus –  sidht Jul 29 '13 at 7:39
$0^{-1}$ is a different problem, so thanks for your hint –  ulead86 Jul 29 '13 at 7:41
Although you can argue that $0^0 = 1$ even without using $0^{-1}$. –  user87690 Jul 29 '13 at 7:43
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Also note that $16^3=16\cdot 16^2$, $16^2=16\cdot 16^1$, so you'd expect that $16^1=16\cdot 16^0$, which requires $16^0=1$.

You can also view $n^k$ as the number of ways to colour $k$ distinct objects with $n$ available colours. If there is no object ($k=0$) there is exactly ne way to do this: do nothing. (Remarkibly, this also gives $0^0=1$ whereas $0^k=0$ for $k>0$).

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Were $16^0=0$ then any number involving this, like $16^7 = 16^7 \times 16^0$ would also be zero. $a^0$ is the identity element of multiplication, and is thus $a^0=1$ for all $a$.

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