What does it mean to say that a number \(x\) is *irrational*?

It means that we cannot write \(x=m/n\) where \(m\) and \(n\) are integers with \(n\ne0\).

Prove by contradiction statements A and B below, where \(p\) and \(q\) are real numbers.

**A:** If \(pq\) is irrational, then at least one of \(p\) and \(q\) is irrational.

**B:** If \(p+q\) is irrational, then at least one of \(p\) and \(q\) is irrational.

Proof by contradiction works like this:

We want to prove a statement S.

We assume S to be untrue, then show a contradiction follows from this.

We cannot have a contradiction in mathematics!

This means S must be true.

We first prove statement A.

Assume instead that \(pq\) is irrational, but both \(p\) and \(q\) are rational.

But then \(pq\) is the product of two rational numbers, so is rational.

This contradicts the assumption that \(pq\) is irrational. So statement A is true.

For statement B we argue similarly.

Assume instead that \(p+q\) is irrational, but both \(p\) and \(q\) are rational.

But then \(p+q\) is the sum of two rational numbers, so is rational.

This contradicts the assumption that \(p+q\) is irrational. So statement B is true.

Disprove by means of a counterexample statement C below, where \(p\) and \(q\) are real numbers.

**C:** If \(p\) and \(q\) are irrational, then \(p+q\) is irrational.

To disprove a statement by means of a counterexample, we only need to find ONE example for which the statement is untrue.

One counterexample is \(p=\sqrt{2}\), \(q=-\sqrt{2}\). Here \(p + q = 0\), which is rational.

If the numbers \(e\), \(\pi\), \(\pi^2\), \(e^2\) and \(e\pi\) are irrational, prove that at most one of the numbers \(\pi+e\), \(\pi-e\), \(\pi^2-e^2\), \(\pi^2+e^2\) is rational.

We’ve been given lots of suggestions from statements \(A\) and \(B\), along with a warning from statement \(C\). The question is, how to use these?

We can work using proof by contradiction again.

We want to prove \(S\), the statement ‘no pair of these numbers is rational.’

Suppose some pair is rational. There are six possible pairs, and we will check them one by one.

If \(\pi + e\) and \(\pi -e\) are both rational, the their sum is, but this is \(2\pi\) which is irrational. Contradiction!

Now \(\pi^2-e^2 = (\pi-e)(\pi+e)\), so if \(\pi^2-e^2\) and \(\pi-e\) are both rational, then \(\pi + e\) is too, which we have seen is impossible.

Similarly if \(\pi^2-e^2\) and \(\pi+e\) are both rational, then \(\pi - e\) is too, which is impossible.

If \(\pi + e\) is rational and so is \(\pi^2 + e^2\), then \((\pi + e)^2\) is rational, which means that \(\pi^2 + e^2 + 2e\pi\) is, which means that \(e\pi\) is, which is a contradiction.

Similarly it’s impossible for \(\pi - e\) and \(\pi^2 + e^2\) to both be rational.

Finally if \(\pi^2 + e^2\) and \(\pi^2 -e^2\) are both rational, then their sum is too, but this is \(2\pi^2\) which we are told is irrational. Contradiction!

So no pair of these four numbers can be rational, and thus at most one can be rational.

Mathematicians have shown that \(e\), \(\pi\), \(\pi^2\) and \(e^2\) are irrational, and that at most one of \(\pi+e\), \(\pi-e\) and \(e\pi\) is rational.

It’s currently unknown whether any of \(e\pi\), \(\pi\pm e\) or \(\pi^2\pm e^2\) is rational.