Sums and products with irrational numbers

Two rules, and their proofs

In Class 9 you met two rules. Now you can prove them, by contradiction.

Rule 1. If $r$ is rational and $x$ is irrational, then $r + x$ is irrational.

Proof. Suppose $r + x = s$, where $s$ is rational. Then $x = s - r$. A rational number minus a rational number is rational, so $x$ would be rational. That contradicts the fact that $x$ is irrational.

Rule 2. If $r$ is rational and not $0$, and $x$ is irrational, then $rx$ is irrational.

Proof. Suppose $rx = s$, where $s$ is rational. Then $x = \dfrac{s}{r}$, which is rational, since $r \neq 0$. Again $x$ would be rational, a contradiction.

Why does rule 2 need $r \neq 0$? Because $0 \times \sqrt{2} = 0$, which is rational.

Proving that a number is irrational

Example 1. Show that $3 + 2\sqrt{5}$ is irrational.

Suppose $3 + 2\sqrt{5} = s$, where $s$ is rational. Then

The right side is rational, because $s$ and $3$ are rational. So $\sqrt{5}$ would be rational. But $5$ is a prime, so $\sqrt{5}$ is irrational, as the last lesson proved. This contradiction shows that $3 + 2\sqrt{5}$ is irrational.

Example 2. Show that $\dfrac{1}{\sqrt{3}}$ is irrational.

Rationalise: $\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3} = \dfrac{1}{3} \times \sqrt{3}$. That is a non-zero rational times an irrational number, so it is irrational by rule 2.

The plan is always the same. Suppose that the number is rational, get the root alone on one side, and reach a contradiction.

Two irrational numbers

With two irrational numbers, anything can happen.

• A rational sum: $(4 + \sqrt{2}) + (6 - \sqrt{2}) = 10$.
• An irrational sum: $\sqrt{2} + \sqrt{3}$, as you will see below.
• A rational product: $\sqrt{8} \times \sqrt{2} = \sqrt{16} = 4$.
• An irrational product: $\sqrt{2} \times \sqrt{3} = \sqrt{6}$.

So never say that two irrational numbers must give an irrational answer. Check each case.

A harder one

Example 3. Show that $\sqrt{2} + \sqrt{3}$ is irrational.

Suppose $\sqrt{2} + \sqrt{3} = s$, where $s$ is rational. Then $\sqrt{3} = s - \sqrt{2}$. Square both sides:

Here $s$ is not $0$, since $\sqrt{2} + \sqrt{3}$ is positive. The right side is rational, so $\sqrt{2}$ would be rational. That is a contradiction, so $\sqrt{2} + \sqrt{3}$ is irrational.

Practise with new numbers

Try it in Python

A number $a + b\sqrt{2}$, with whole numbers $a$ and $b$, can be kept as the pair (a, b). It is rational exactly when $b = 0$. The program adds and multiplies two such numbers.

Common mistakes

• "Irrational plus irrational is irrational." Not always: $(4 + \sqrt{2}) + (6 - \sqrt{2}) = 10$.
• Forgetting the zero in rule 2. $0 \times \sqrt{5} = 0$ is rational.
• Proving with decimals. $3 + 2\sqrt{5} = 7.472\ldots$ does not show that the digits never repeat.
• Not naming the contradiction. Say which known fact breaks, such as "$\sqrt{5}$ is irrational".

Remember these

• Rational plus irrational is irrational. A non-zero rational times an irrational number is irrational.
• Two irrational numbers can give a rational or an irrational sum or product.
• To prove that $a + b\sqrt{p}$ is irrational, with $b \neq 0$ and $p$ a prime: suppose it is rational, get $\sqrt{p}$ alone, and reach a contradiction.

Check yourself

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

1. Show that $7 - \sqrt{3}$ is irrational.

2. Is $\dfrac{\sqrt{12}}{\sqrt{3}}$ rational or irrational?

3. Give two irrational numbers whose product is $10$.

4. $0 \times \sqrt{7} = 0$ is rational. Why does this not break rule 2?

5. Is $(\sqrt{5} - 1)(\sqrt{5} + 1)$ rational or irrational?