You can now sort single numbers into rational and irrational. What happens when you add or multiply them? Some answers are always the same, and some depend on the numbers.
Start with what is certain. Rational numbers are closed under the four operations: add, subtract, multiply or divide (not by zero) two fractions and you get another fraction. For example, $\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$ and $\frac{2}{3} \cdot \frac{9}{4} = \frac{3}{2}$.
Why it works
Suppose $r$ is rational and $x$ is irrational, and imagine $r + x$ were some rational number $s$. Then $x = s - r$, a difference of two rationals, so $x$ would be rational. That is impossible, so rational + irrational is always irrational. The same argument works for a product: if $r \ne 0$ and $r \cdot x = s$ were rational, then $x = s \div r$ would be rational. So a nonzero rational times an irrational is irrational. (If $r = 0$, the product is $0$, which is rational.)
| Combination | Result | Example |
|---|---|---|
| rational $\pm$ rational | always rational | $\frac{1}{2} - 3 = -\frac{5}{2}$ |
| rational $\pm$ irrational | always irrational | $4 + \sqrt{3}$ |
| nonzero rational $\times$ irrational | always irrational | $5\pi$ |
| irrational $\pm$ irrational | it depends | $\sqrt{2} + \sqrt{2} = 2\sqrt{2}$, but $\sqrt{2} + (-\sqrt{2}) = 0$ |
| irrational $\times$ irrational | it depends | $\sqrt{2}\cdot\sqrt{3} = \sqrt{6}$, but $\sqrt{2}\cdot\sqrt{8} = 4$ |
Worked examples
Worked example: Rational plus irrational
Is $\frac{3}{4} + \sqrt{7}$ rational or irrational?
- $\frac{3}{4}$ is rational. $7$ is not a perfect square, so $\sqrt{7}$ is irrational.
- Rational + irrational is always irrational.
Answer: Irrational.
Worked example: Two irrationals whose product is rational
Is $\sqrt{3} \cdot \sqrt{12}$ rational or irrational?
- Each factor is irrational, so no rule decides it. Compute.
- $\sqrt{3}\cdot\sqrt{12} = \sqrt{36}$.
- $\sqrt{36} = 6$, a rational number.
Answer: Rational.
Worked example: Irrational parts that cancel
Is $(5 + \sqrt{2}) + (1 - \sqrt{2})$ rational or irrational?
- Group like parts: $(5 + 1) + (\sqrt{2} - \sqrt{2})$.
- $\sqrt{2} - \sqrt{2} = 0$, so the sum is $6$.
Answer: Rational, even though both numbers being added are irrational.
Common mistake
Assuming two irrational numbers must give an irrational answer. Irrational parts can cancel or combine into a perfect square, as in $\sqrt{2} + (-\sqrt{2}) = 0$. When both numbers are irrational, simplify and look at the result; when one is rational, the rules in the table decide it.
Key idea
Rational with rational stays rational. Rational $\pm$ irrational, and nonzero rational $\times$ irrational, are always irrational. Irrational with irrational can go either way, so simplify and see.