We have already proved that is irrational (see the example in this post!)

The fact that is irrational, i.e. there are no integers such that was discovered a couple of hundreds of years ago and there are several proofs of this result; I like Ivan Niven’s proof better and that’s what I’m going to explain here.

Throughout this post, is the -th derivative of a function

**Problem 1**. Let be a polynomial of degree Show that

**Solution**. The proof is by induction over If then is a constant and so

Now suppose the equality in the problem holds for polynomials of degree and let be a polynomial of degree Then using integration by parts with and gives

Now, in , we use integration by parts again, this time with and to get

But since is a polynomial of degree we can use our induction hypothesis to write and so becomes

**Problem 2**. Let be integers with and Let and Show that and are integers for all

**Solution**. Since is a polynomial of degree we have for all and and so there is nothing to prove for

Now, since is a polynomial of degree we have

On the other hand, using the binomial theorem, it is clear that

where are some integers.

So and together give for and for Thus, since is an integer for we have proved that is an integer for all

Finally, since we have for all and thus So is an integer because we have already proved that is an integer.

**Problem 3**. Show that is irrational.

**Solution** (*Ivan Niven*). Suppose, to the contrary, that is rational and put where are integers. Let be an integer and put

Then by Problem 1 and 2, must be an integer. But we are now going to prove that if is large enough, then is not an integer and this contradiction proves that our assumption that is rational is false.

First note that, on the interval we have

and thus

(Note that because is not identically zero on ).

But for every real number we have because the series is convergent (to ). So Hence, if is large enough, we will have and thus, by Therefore is not an integer if is large enough.

**Remark**. Another way to say that a real number is irrational is to say that is not the root of a polynomial of degree one with integer coefficients. If is not a root of any polynomial with integer coefficients, then is called *transcendental*. Clearly every transcendental number is irrational. It is known that both and are in fact transcendental.

**Exercise 1**. Let be a polynomial of degree Show that

**Exercise 2**. Show that the inequality in the solution of Problem 3 can be improved by proving that