Definition: If a ≠ 0 and a is a real number, then a^{0} = 1.

In layman's terms, a real number raised to the zero power is always 1. One way to understand why this is so is to observe the pattern of the results of the following series of equations:

2^{5} = 32
2^{4} = 16
2^{3} = 8
2^{2} = 4
2^{1} = 2
2^{0} = 1

Starting at the top, you can divide 32 by 2, to arrive at the result of the equation below it. By dividing any result in this series of equations by 2, you arrive at the result of the equation below. Put in generic terms, by dividing the result of a^{x} by a, the result is always a^{x-1}.

Following this pattern in the sample series of equations provided, you can see that by dividing the result of 2^{1} by 2, the result is 1. This provides a concrete intuitive example of why a^{0} = 1.

Extending the series of equations shows that the same progression makes sense with negative exponents.

2^{-1} = 1/2
2^{-2} = 1/4
2^{-3} = 1/8
2^{-4} = 1/16

This also holds true if the base is negative

-2^{0} = 1

or if the base is a non-integer

-2.32423^{0} = 1

To prove this definition rigorously requires application of one of the laws of exponents

r^{x}
--- = r^{(x-y)}
r^{y}

for all r, x, and y. For example:

2^{6}
--- = 2^{(6-4)} = 2^{2} = 4
2^{4}

Consider the following fraction:

2^{4}/2^{4}

This fraction equals 1 because the numerator and the denominator are equal, so 2^{4}/2^{4} = 1. Applying the law of exponents, observe that

2^{4}
--- = 2^{(4-4)} = 2^{0}
2^{4}

Therefore, 2^{0} = 1

Substituting x for 2 and y for 4 gives us the following:

x^{y}
--- = x^{(y-y)} = x^{0} = 1
x^{y}