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Math and Arithmetic

Why does any number to the zero power equal one?

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December 04, 2010 4:07PM

Here is why any number to the zero power equals one.

Consider this.

a^b. it is natural to restrict a > 0, but we'll only assume that number b is any real number.

We'll use the natural exponential function defined by the derivative of the exponential function.

Now we have a^r=e^rln(a). And we know that e^rln(a)=e^((ln(a))^r), where a >0 and r is in the domain of all real numbers negative infinity to infinity.

We can apply this definition to any number a to any power r.

Particularly, a^0. By the provided definition, a^0=e^(0*ln(a))=e^0=1.

Furthermore, a^1=e^(1*ln(a))=e(ln(a))=a.

And a^2=(e^(ln(a))^2)=a^2.

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Here is a simpler approach:

In general, a^n/a^m = a^(n-m) and a/a = 1. We can use these facts to prove that x^0 = 1 so long as x isn't 0.

First, state the obvious: 1 = 1

Next, since any non-zero number divided by itself is one: 1 = a^n/a^n

(It doesn't change how the equation looks, but for the sake of being thorough, you could subsitute (a^n/a^n) in place of 1 in the original equation.)

Then, since dividing like bases requires that you subtract their exponents:

a^n/a^n = a^(n-n) = a^0

Substitute (a^0) in for (a^n/a^n) and you obtain: a^0 = 1

There are two reasons "a" cannot be 0 in this proof: firstly, raising 0 to non-zero powers would still result in zero, so "a" being 0 would cause division by zero in the initial theorems we used, and secondly, 0^0 is considered undefined in itself.