To prove that e is irrational

We define

Now each term of a_n is positive, so a_n > 0.

The terms of b_n are a geometric series, so

Moreover in comparing each term of a_n with the corresponding term of b_n we have

etc, so it follows that a_n < b_n.

Hence we have .

By definition,

.

Now suppose e is rational, ie that

Then we have

We can choose n so that q divides n (eg n = q).

Then the expression enclosed by inequality signs is an integer.

An integer cannot lie between 0 and .

That is a contradiction, so the assumption (that e is rational) can only be false.

Hence finally e is irrational.

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