Asked in Math and Arithmetic
Why does any number to the zero power equal one?
December 04, 2010 4:07PM
Here is why any number to the zero power equals one.
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.
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.