**Fact:** The square root of \(2\), \(\sqrt{2}\), is irrational.

Sort the statements below into a proof of this fact. Can you explain how the ordered statements prove this result?

Suppose, for a contradiction, that \(\sqrt{2}\) is rational.

What’s the difference between a rational and irrational number?

That is, we can write \(\sqrt{2} = \frac{m}{n}\) where \(m\) and \(n\) are integers and where \(n \neq 0\).

Squaring, we have \(2 = \frac{m^2}{n^2}\).

Multiply across to get \(2n^2 = m^2\).

In the prime factorisations of \(m^2\) and \(n^2\), \(2\) occurs to an even power.

Remember that any number can be written in terms of its prime factors. For example \(24 = 2^3 \times 3\). If \(n\) is even, which powers of \(2\) could be in the prime factorisation? What happens when \(n\) is then squared?

What about for odd numbers? If \(n\) is odd, which powers of \(2\) could appear in the prime factorisation? What about \(n^2\)?

Is zero even or odd?

In the prime factorisation of \(2n^2\), \(2\) occurs to an odd power.

How does this statement follow on from the previous card? Could these cards appear in the opposite order?

But prime factorisations are unique, so \(2\) should appear to the same power in both \(2n^2\) and \(m^2\).

This is a contradiction, so our original assumption that \(\sqrt{2}\) is rational must be wrong.

So \(\sqrt{2}\) is irrational.

Can you adapt this argument to show that \(\sqrt{3}\) is irrational?

Suppose, for a contradiction, that \(\sqrt{3}\) is rational.

That is, we have \(\sqrt{3} = \frac{m}{n}\) for integers \(m\) and \(n\), where \(n \neq 0\).

Squaring, we have \(3 = \frac{m^2}{n^2}\).

Multiply across to get \(3n^2 = m^2\).

In the prime factorisations of \(m^2\) and \(n^2\), \(3\) occurs to an even power.

In the prime factorisation of \(3n^2\), \(3\) occurs to an odd power.

But prime factorisations are unique, so \(3\) should appear to the same power in both \(3n^2\) and \(m^2\).

This is a contradiction, so our original assumption that \(\sqrt{3}\) is rational must be wrong.

So \(\sqrt{3}\) is irrational.

For which positive integers \(d\) is \(\sqrt{d}\) irrational? Justify your answer.

If \(d\) is a square, say \(d = n^2\) where \(n\) is an integer, then clearly \(\sqrt{d} = n\) is rational.

If \(d\) is not a square, then \(\sqrt{d}\) is irrational. We can prove this using a very similar argument to that used for \(2\) and \(3\) above.

Here’s a special case to show how we can adapt the argument to work even when \(d\) is not prime.

Let’s consider \(\sqrt{10}\).

Suppose, for a contradiction, that \(\sqrt{10}\) is rational, so there are integers \(a\) and \(b\) with \(b \neq 0\) and \(\sqrt{10} = \frac{a}{b}\).

Then, squaring and multiplying across, we have \(10b^2 = a^2\).

In the prime factorisations of \(a^2\) and \(b^2\), \(2\) occurs to an even power. In the prime factorisation of \(10b^2\), \(2\) occurs to an odd power.

But this is not possible, so we have a contradiction. So \(\sqrt{10}\) is irrational.

Notice that if we try to run the same argument on a number that is divisible by a square, such as \(12 = 2^2 \times 3\), then we have to be slightly careful. One way round this is to notice that \(\sqrt{12} = 2\sqrt{3}\), and this is irrational if and only if \(\sqrt{3}\) is irrational. So it suffices to show that if \(d\) is *square-free* (is not divisible by a square) then \(\sqrt{d}\) is irrational.

To check your understanding, see what happens when you run the usual contradiction argument on a square value of \(d\) such as \(4\) or \(9\). Why do we not get a contradiction in this case?

Can you show that \(1 + \sqrt{2}\) is irrational? Is \(\sqrt{2} + \sqrt{3}\) rational or irrational?

Suppose for a contradiction that \(1 + \sqrt{2}\) is rational.

The sum of two rational numbers is rational, so we see that \((1 + \sqrt{2}) + (-1) = \sqrt{2}\) is rational.

But we showed above that \(\sqrt{2}\) is irrational, so this is a contradiction. So \(1 + \sqrt{2}\) is irrational.

Suppose for a contradiction that \(\sqrt{2} + \sqrt{3}\) is rational.

Then \(\sqrt{2} + \sqrt{3} = \frac{a}{b}\) for some integers \(a\) and \(b\) where \(b \neq 0\).

Squaring both sides gives \(2 + 2\sqrt{6} + 3 = \frac{a^2}{b^2}\).

Rearranging then gives \(2\sqrt{6} = \frac{a^2}{b^2} - 5\).

In particular, that tells us that \(\sqrt{6}\) is rational.

Which of the arguments above tells us that \(\sqrt{6}\) is irrational? This gives us a contradiction. So \(\sqrt{2} + \sqrt{3}\) is irrational.