Anyone who has taken high school math is familiar with the constant .

Today we are going to prove that **e is in fact irrational**! We will go through Joseph Fourier‘s famous proof by contradiction. The maths background we need is to know the power series expansion: . The proof is slightly tricky so stay focussed!

(Reference: http://en.wikipedia.org/wiki/Proof_that_e_is_irrational)

Suppose to the contrary that e is a rational number, so .

Using the power series formula mentioned above, we have

Multiply both sides by ,

Now, we split the sum into two parts:

Rearranging,

Now, denote . is an integer since both and are integers and their difference (which is x) will be an integer.

We now prove that . For all terms with we have the upper estimate

This inequality is strict for every . Changing the index of summation to and using the formula for the infinite geometric progression , we obtain:

We have that is an integer but . This is a contradiction (since there is no integer strictly between 0 and 1), and so must be irrational. (QED)

Interesting? 🙂

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Did you know the constant *e* is sometimes called Euler’s number?

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Watch this video for another proof that *e* is irrational!