What are irrational numbers? An irrational number is a real number that cannot be written as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $q \ne 0$. Its decimal expansion is non-terminating and non-repeating.
Examples: $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$, $\pi$, $e$. In contrast, $\frac{1}{2} = 0.5$ (terminating) and $\frac{2}{3} = 0.\overline{6}$ (repeating) are rational.
A key supporting result. If a prime $p$ divides $a^{2}$, then $p$ divides $a$. This follows from the Fundamental Theorem of Arithmetic: the prime $p$ must appear in the prime factorisation of $a^{2}$, and since the primes of $a^{2}$ are exactly the primes of $a$ (with doubled powers), $p$ must already be a factor of $a$. This lemma is the heart of every irrationality proof below.
Proof that $\sqrt{2}$ is irrational (by contradiction).
- Assume the opposite: $\sqrt{2} = \frac{p}{q}$ in lowest terms (so $p$ and $q$ have no common factor).
- Square both sides: $2 = \frac{p^{2}}{q^{2}}$, hence $p^{2} = 2q^{2}$.
- So $2 \mid p^{2}$, which means $2 \mid p$. Write $p = 2k$.
- Then $(2k)^{2} = 2q^{2} \Rightarrow 4k^{2} = 2q^{2} \Rightarrow q^{2} = 2k^{2}$, so $2 \mid q^{2}$ and $2 \mid q$.
- Now 2 divides both $p$ and $q$, contradicting "lowest terms". The assumption is false, so $\sqrt{2}$ is irrational.
The same argument with the prime 3 proves $\sqrt{3}$ irrational, and with the prime 5 proves $\sqrt{5}$ irrational.
Combinations with rationals. A few standard results follow easily, each by assuming the result is rational and rearranging to make a known irrational equal to a rational — a contradiction:
- The sum or difference of a rational and an irrational (e.g. $5 + \sqrt{3}$, $5 - \sqrt{3}$) is irrational.
- The product or quotient of a non-zero rational and an irrational (e.g. $3\sqrt{2}$, $\frac{1}{\sqrt{2}}$) is irrational.
- Hence numbers like $3 + 2\sqrt{5}$ and $6 + \sqrt{2}$ are irrational.
A useful caution about combining two irrationals. The sum, difference, product, or quotient of two irrationals need not be irrational. For instance $\sqrt{2} \times \sqrt{2} = 2$ (rational) and $\sqrt{2} \div \sqrt{2} = 1$ (rational). So always reason carefully rather than assuming the result is irrational.
Common mistakes to avoid:
- Calling $\sqrt{9}$ or $\sqrt{16}$ irrational — perfect-square roots like $\sqrt{9}=3$ are rational.
- Forgetting the "lowest terms" or "non-zero rational" condition, which is exactly what creates the contradiction.
- Assuming any expression containing a root is irrational (e.g. $2\sqrt{3} \div 3\sqrt{3} = \frac{2}{3}$ is rational because the roots cancel).
- Trying to prove irrationality without contradiction — the standard Class 10 method is always "assume rational, derive a contradiction".
Classify $0.101001000100001\ldots$ (one extra zero each time) as rational or irrational.
Solution- Step 1: Look at the pattern: the blocks of zeros keep growing, so no fixed block of digits repeats.
- Step 2: Terminating or repeating decimals are rational; this decimal is neither.
- Step 3: It is non-terminating and non-repeating.
Answer: It is irrational
Prove that $\sqrt{2}$ is irrational.
Solution- Step 1: Assume $\sqrt{2} = \frac{p}{q}$ in lowest terms, $q \ne 0$.
- Step 2: Square: $2 = \frac{p^{2}}{q^{2}}$, so $p^{2} = 2q^{2}$, giving $2 \mid p$. Write $p = 2k$.
- Step 3: Then $4k^{2} = 2q^{2} \Rightarrow q^{2} = 2k^{2}$, so $2 \mid q$.
- Step 4: 2 divides both $p$ and $q$, contradicting lowest terms.
Answer: $\sqrt{2}$ is irrational
Prove that $\sqrt{5}$ is irrational.
Solution- Step 1: Assume $\sqrt{5} = \frac{p}{q}$ in lowest terms.
- Step 2: Square: $p^{2} = 5q^{2}$, so $5 \mid p^{2}$, hence $5 \mid p$. Write $p = 5k$.
- Step 3: Then $25k^{2} = 5q^{2} \Rightarrow q^{2} = 5k^{2}$, so $5 \mid q$.
- Step 4: 5 divides both $p$ and $q$ — contradiction.
Answer: $\sqrt{5}$ is irrational
Prove that $5 + \sqrt{3}$ is irrational, given that $\sqrt{3}$ is irrational.
Solution- Step 1: Assume $5 + \sqrt{3} = \frac{p}{q}$ is rational, $q \ne 0$.
- Step 2: Then $\sqrt{3} = \frac{p}{q} - 5 = \frac{p - 5q}{q}$.
- Step 3: The right side is a ratio of integers, hence rational.
- Step 4: But $\sqrt{3}$ is irrational — contradiction.
Answer: $5 + \sqrt{3}$ is irrational
Show that $3\sqrt{2}$ is irrational, given that $\sqrt{2}$ is irrational.
Solution- Step 1: Assume $3\sqrt{2} = \frac{p}{q}$ is rational, $q \ne 0$.
- Step 2: Then $\sqrt{2} = \frac{p}{3q}$, a ratio of integers, hence rational.
- Step 3: This contradicts the irrationality of $\sqrt{2}$.
Answer: $3\sqrt{2}$ is irrational
Prove that $3 + 2\sqrt{5}$ is irrational, given that $\sqrt{5}$ is irrational.
Solution- Step 1: Assume $3 + 2\sqrt{5} = \frac{p}{q}$ is rational, $q \ne 0$.
- Step 2: Rearrange: $2\sqrt{5} = \frac{p}{q} - 3 = \frac{p - 3q}{q}$.
- Step 3: So $\sqrt{5} = \frac{p - 3q}{2q}$, a ratio of integers, hence rational.
- Step 4: This contradicts $\sqrt{5}$ being irrational.
Answer: $3 + 2\sqrt{5}$ is irrational
Show that $\frac{1}{\sqrt{2}}$ is irrational.
Solution- Step 1: Rationalise: $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.
- Step 2: Assume $\frac{\sqrt{2}}{2} = \frac{p}{q}$ is rational; then $\sqrt{2} = \frac{2p}{q}$, which is rational.
- Step 3: This contradicts the irrationality of $\sqrt{2}$.
Answer: $\frac{1}{\sqrt{2}}$ is irrational
Prove that $6 + \sqrt{2}$ is irrational.
Solution- Step 1: Assume $6 + \sqrt{2} = \frac{p}{q}$ is rational, $q \ne 0$.
- Step 2: Then $\sqrt{2} = \frac{p}{q} - 6 = \frac{p - 6q}{q}$, a rational number.
- Step 3: But $\sqrt{2}$ is irrational — contradiction.
Answer: $6 + \sqrt{2}$ is irrational
Determine whether $\frac{2\sqrt{3}}{3\sqrt{3}}$ is rational or irrational.
Solution- Step 1: Cancel the common $\sqrt{3}$: $\frac{2\sqrt{3}}{3\sqrt{3}} = \frac{2}{3}$.
- Step 2: $\frac{2}{3}$ is a ratio of integers with non-zero denominator.
- Step 3: Its decimal $0.\overline{6}$ is repeating, so it is rational.
Answer: $\frac{2}{3}$ is rational
Show that $\sqrt{2} + \sqrt{3}$ is irrational.
Solution- Step 1: Assume $\sqrt{2} + \sqrt{3} = r$ is rational.
- Step 2: Square: $2 + 2\sqrt{6} + 3 = r^{2}$, so $2\sqrt{6} = r^{2} - 5$.
- Step 3: Then $\sqrt{6} = \frac{r^{2} - 5}{2}$ would be rational.
- Step 4: But $\sqrt{6}$ is irrational (6 is not a perfect square) — contradiction.
Answer: $\sqrt{2} + \sqrt{3}$ is irrational
Classify each of $\sqrt{16}$, $\sqrt{9}$ and $\sqrt{7}$ as rational or irrational.
Solution- Step 1: $\sqrt{16} = 4$ — a perfect square, so rational.
- Step 2: $\sqrt{9} = 3$ — a perfect square, so rational.
- Step 3: $7$ is not a perfect square, so $\sqrt{7}$ is irrational (same proof style as $\sqrt{2}$).
Answer: $\sqrt{16}, \sqrt{9}$ rational; $\sqrt{7}$ irrational
Prove that $7 - 3\sqrt{2}$ is irrational, given that $\sqrt{2}$ is irrational.
Solution- Step 1: Assume $7 - 3\sqrt{2} = \frac{p}{q}$ is rational, $q \ne 0$.
- Step 2: Rearrange: $3\sqrt{2} = 7 - \frac{p}{q} = \frac{7q - p}{q}$.
- Step 3: So $\sqrt{2} = \frac{7q - p}{3q}$, a ratio of integers, hence rational.
- Step 4: This contradicts the irrationality of $\sqrt{2}$.
Answer: $7 - 3\sqrt{2}$ is irrational