Why the square root of 2 is irrational

What we want to prove

A rational number can be written as $\dfrac{a}{b}$, where $a$ and $b$ are whole numbers and $b \neq 0$. An irrational number cannot. Is $\sqrt{2}$ rational?

A calculator shows $\sqrt{2} = 1.41421356\ldots$ It shows only a few digits. It cannot tell us whether the digits start to repeat much later. We need a proof, and it was found more than two thousand years ago.

The proof, step by step

Theorem. $\sqrt{2}$ is irrational.

Proof.

1. Suppose the opposite. Suppose $\sqrt{2} = \dfrac{a}{b}$, where $a$ and $b$ are whole numbers with no common factor other than $1$. Every fraction can be cancelled down to lowest terms, so this is allowed.
2. Square both sides. $2 = \dfrac{a^2}{b^2}$, so $a^2 = 2b^2$.
3. So $a$ is even. $a^2$ is a multiple of $2$. Since $2$ is a prime, $2$ divides $a$, by the key fact of the last lesson. Write $a = 2c$.
4. Put it back. $(2c)^2 = 2b^2$ gives $4c^2 = 2b^2$, so $b^2 = 2c^2$.
5. So $b$ is even. $b^2$ is a multiple of $2$, so $2$ divides $b$.
6. The contradiction. Now $2$ divides both $a$ and $b$. But we chose them with no common factor. This is impossible.

So the supposition is false, and $\sqrt{2}$ is irrational.

The picture shows the same chain. Each step follows from the one before it. The only step that can be wrong is the first one, the supposition, so it must be wrong.

Fractions come close, but never arrive

The best fractions, such as $\dfrac{7}{5}$, $\dfrac{17}{12}$ and $\dfrac{41}{29}$, give $p^2 - 2q^2$ equal to $1$ or $-1$. They come very close to $\sqrt{2}$. If $p^2 - 2q^2$ were ever $0$, the square of $\dfrac{p}{q}$ would be exactly $2$, and the proof says that this never happens.

The same proof for other roots

Take $\sqrt{3}$. Suppose $\sqrt{3} = \dfrac{a}{b}$ in lowest terms. Then $a^2 = 3b^2$. The prime $3$ divides $a^2$, so $3$ divides $a$. Write $a = 3c$. Then $9c^2 = 3b^2$, so $b^2 = 3c^2$, and $3$ divides $b$. Both are multiples of $3$, which is a contradiction.

The same steps work for $\sqrt{5}$, $\sqrt{7}$, $\sqrt{11}$ and the square root of any prime $p$. Only the prime changes.

Why it fails for the square root of 4

Try the steps with $\sqrt{4} = \dfrac{a}{b}$. Then $a^2 = 4b^2$, so $a = 2b$ and $\dfrac{a}{b} = 2$. There is no contradiction, because $\sqrt{4} = 2$ really is a fraction, $\dfrac{2}{1}$. Notice that $4$ is not a prime, so the key fact does not apply to it.

In general, $\sqrt{n}$ is irrational for every whole number $n$ that is not a perfect square. A perfect square, such as $4$, $9$ or $16$, has a whole number as its square root.

Practise with new numbers

Try it in Python

The program searches for a fraction $\dfrac{a}{b}$ with $a^2 = 2b^2$, for every bottom $b$ up to $100\,000$. Then it does the same for $\sqrt{3}$.

A search can only check finitely many fractions. The proof covers all of them at once.

Common mistakes

• Writing $\sqrt{2} = 1.414$. $1.414$ is the fraction $\dfrac{1414}{1000}$, and its square is $1.999396$, not $2$. It is only close.
• Forgetting the words "lowest terms". The whole contradiction comes from them.
• Skipping a reason. From $a^2 = 2b^2$ we may say that $2$ divides $a$ only because $2$ is a prime.
• Thinking every square root is irrational. $\sqrt{9} = 3$ and $\sqrt{16} = 4$.

Remember these

• $\sqrt{p}$ is irrational for every prime $p$. $\sqrt{n}$ is irrational when $n$ is not a perfect square.
• The proof: suppose lowest terms, square both sides, use the key fact twice, and find a common factor.
• Decimals from a calculator, or a computer search, cannot prove that a number is irrational.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. In the proof for $\sqrt{2}$, why may we start with $a$ and $b$ that have no common factor?

2. In the proof for $\sqrt{7}$, we reach $a^2 = 7b^2$ and write $a = 7c$. What equation do we get for $b$?

3. Is $\sqrt{11}$ rational or irrational?

4. $\dfrac{7}{5} = 1.4$ is close to $\sqrt{2}$. Work out $7^2 - 2 \times 5^2$. What does it tell you?

5. Which of $\sqrt{13}$, $\sqrt{36}$ and $\sqrt{40}$ are irrational?