Operations on real numbers

Adding and multiplying real numbers

The real numbers are made of the rational numbers and the irrational numbers together. Every point on the number line is a real number. What kind of number do we get when we add or multiply them?

Rational with rational. The sum, difference and product of rational numbers are rational. The quotient is rational too, if the divisor is not zero.

Rational with irrational.

• The sum or difference of a rational and an irrational number is irrational. Example: $2 + \sqrt{3}$, $5 - \sqrt{7}$.
• The product or quotient of a non-zero rational and an irrational number is irrational. Example: $3\sqrt{5}$, $\dfrac{\sqrt{2}}{4}$.

Why is $2 + \sqrt{3}$ irrational? Suppose it were rational, equal to $r$. Then $\sqrt{3} = r - 2$, which is a rational number minus a rational number, so it would be rational. But $\sqrt{3}$ is irrational. This contradiction shows that $2 + \sqrt{3}$ is irrational.

Irrational with irrational. Here the answer can be either.

• $\sqrt{2} + \sqrt{3}$ is irrational.
• $(2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$ is rational.
• $\sqrt{2} \times \sqrt{2} = 2$ is rational.
• $\sqrt{2} \times \sqrt{3} = \sqrt{6}$ is irrational.

So the sum or product of two irrational numbers is not always irrational.

Rules for square roots

For positive numbers $a$ and $b$:

Examples: $\sqrt{3} \times \sqrt{12} = \sqrt{36} = 6$, and $\dfrac{\sqrt{18}}{\sqrt{2}} = \sqrt{9} = 3$.

We cannot add square roots like this. In general, $\sqrt{a} + \sqrt{b} \ne \sqrt{a + b}$. For example, $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$, but $\sqrt{9 + 16} = \sqrt{25} = 5$.

Set $a = 0$. Now blue and red match, because $\sqrt{0} = 0$. For any other pair, √a + √b is taller than √(a + b), while green and amber always match.

Simplifying a square root

We take out any perfect square factor.

• $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$.
• $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$.
• $\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$.

Square roots with the same irrational part can be added, in the same way as $3x + 2x = 5x$.

If the irrational parts are different, the terms cannot be combined: $\sqrt{2} + \sqrt{3}$ stays as it is.

Useful identities

For positive numbers $a$ and $b$:

1. $(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$
2. $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$
3. $(\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab}$
4. $(\sqrt{a} - \sqrt{b})^2 = a + b - 2\sqrt{ab}$

Examples:

• $(\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2}) = 5 - 2 = 3$.
• $(\sqrt{3} + \sqrt{2})^2 = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6}$.
• $(3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7$.
• $(2\sqrt{3} + 1)^2 = 12 + 1 + 4\sqrt{3} = 13 + 4\sqrt{3}$.

These identities are the same as the ones that you know for $(x + y)(x - y)$ and $(x + y)^2$, with $x = \sqrt{a}$ and $y = \sqrt{b}$.

Practise with new numbers

Try it in Python

This program checks some results. A calculator rounds irrational numbers, so we round the answers.

Common mistakes

• Writing $\sqrt{a} + \sqrt{b} = \sqrt{a + b}$. This is false.
• Saying that the sum of two irrational numbers is irrational. It can be rational.
• Adding unlike roots. $\sqrt{2} + \sqrt{3}$ cannot be combined.
• Forgetting the middle term. $(\sqrt{3} + \sqrt{2})^2$ is not just $3 + 2$.

Remember these

• Rational plus or times irrational (non-zero) is irrational.
• Two irrational numbers can give a rational sum or product, such as $\sqrt{2} \times \sqrt{2} = 2$.
• $\sqrt{a}\sqrt{b} = \sqrt{ab}$, and $(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$.
• $\sqrt{a} + \sqrt{b}$ is not $\sqrt{a + b}$.

Check yourself

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

1. Is $3 + \sqrt{2}$ rational or irrational?

2. Simplify $\sqrt{8} \times \sqrt{2}$.

3. Simplify $(\sqrt{7} + 2)(\sqrt{7} - 2)$.

4. Simplify $\sqrt{12} + \sqrt{27}$.