Why 0! = 1

in #math6 years ago

Most of you should be familiar with the factorial function where n factorial (written n!) is the product of the first n natural numbers, e.g. 5! = 1 * 2 * 3 * 4 * 5 = 120. This is a straightforward definition for positive numbers but in maths you will sometimes see zero factorial. It feels like this should come out to zero because zero will appear as a factor but actually the common definition is 0! = 1. Why is that?

If we again look at the definition n! = n * (n-1) * (n-2) * … * 3 * 2 * 1 it becomes clear that n! = n * (n-1)! since (n-1)! = (n-1) * (n-2) * ... * 3 * 2 * 1. The interesting thing about this is that we can rearrange the equation to yield (n-1)! = (n!)/n. Because we divide by n this will not hold true for n=0 but it will do so for n=1. This yields (1-1)! = (1!)/1 which is 0! = 1/1 =1.

There are actually ways to extend the factorial even further so it is also defined for fractions and negative numbers, always keeping the fundamental property x! = x * (x-1)! but they are not as easy to understand as this.

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