# $\sqrt{2}$ is irrational Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

## Solution

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.