Why is (virtually) anything raised to $0$th power equal to $1$?

Question

Taking a shot in the dark:

$10^3 = 10 * 10 * 10 = 1 (10 * 10 * 10)$

$10^2 = 1 (10 * 10)$

$10^1 = 1 (10)$

$10^0 = 1$

This isn't quite proofy enough, but seems to be reasonably convincing to a non-mathematician like myself. Is there a deeper reason why this is true?

Answer 1

The simplest explanation is the following for me: For any $x>0$ (to avoid branching), $$x^a=x^{a+0}=x^a\cdot x^0$$$$1=x^0$$

In other words, exponentiation converts addition to multiplication, so we expect the additive identity to map to the multiplicative identity.

Answer 2

Another way to think about this could be, $$\begin{gathered} {a^x} = 1 + x\left( {\ln a} \right) + \frac{{{x^2}}}{{2!}}{\left( {\ln a} \right)^2} + \frac{{{x^3}}}{{3!}}{\left( {\ln a} \right)^3} + \frac{{{x^4}}}{{4!}}{\left( {\ln a} \right)^4} + \cdots \\ \end{gathered}$$ where you put $x = 0$. It's a different issue to prove that this series absolutely converges, etc. - but given all that, I think this is a nice way to look at why $a^0 = 1$ when $a \ne 0$.

P.S. This is the Taylor series of $f(x) = a^x$ centered at $x= 0$ (hence also called the Maclaurin series).

Answer 3

There is a rule for exponents: $$ a^{n+m} = a^n a^m \tag1 $$ true for $n,m = 1,2,3,4,\dots$ If we (by convention) extend this to $a^0$, then we get $$ a = a^1 = a^{1+0} = a^1 a^0 = a\;a^0 $$ so as long as $a\ne 0$, we divide by $a$ to get $$ 1 = a^0 $$

---

Extending the convention even more to negative integer exponents: $$ 1 = a^0 = a^{n+(-n)} = a^n\;a^{-n} $$ Divide to get $$ a^{-n} = \frac{1}{a^n} $$

Answer 4

Martin R has linked a number of posts, but they are cluttered with many wrong explanations, with few exceptions (such as Henning Makholm's and Gadi A's). Even in this thread, the other posts do not really explain why we can choose to have $x^0 = 1$, even though they give motivation for why we want it so.

Let me emphasize that mathematics is not about wishful thinking. Wishful thinking can lead to ridiculous things like the liar's paradox (for allowing circular reasoning), Russell's paradox (for naive set theory), Quine's paradox (for assuming that a certain statement has a well-defined truth-value without justifying it), and so on.

Similarly here, before we even start asking whether $x^0 = 1$, we have to first define what "$x^0$" means. Hence Henning's answer is the true answer. I want to elaborate a bit more. If you think carefully, you would actually realize that you are unable to define $x^k$ for general $k∈ℕ$ without an inductive/recursive definition, so either way you have to pick an arbitrary base case in your definition. Then clearly the question is whether it is better to start from $x^0 = 1$ or $x^1 = x$. It turns out that the structure of exponentiation is much nicer and simpler with the former choice, which is why things like the binomial theorem, taylor series, combinatorial intepretations and so on all work better if we have $x^0 = 1$.