Irrational numbers — Class 10 CBSE Mathematics MCQs with Solutions

Solution: An irrational number is precisely one that cannot be written in the form $\frac{p}{q}$ where $p$ and $q$ are integers with $q \neq 0$. This is the defining characteristic. Note that 'not an integer' is too broad (includes fractions like $\frac{1}{2}$, which are rational), and negative numbers can be rational or irrational.

Solution: $\sqrt{2}$ is irrational. We can prove this: assume $\sqrt{2} = \frac{p}{q}$ for some integers $p$ and $q$. Then $2q^2 = p^2$. By prime factorization, the left side has an even power of 2, but the right side has an odd power, a contradiction. Therefore, $\sqrt{2}$ cannot be rational; it must be irrational.

Solution: $\sqrt{3}$ is irrational (similar proof to $\sqrt{2}$: assuming $\sqrt{3} = \frac{p}{q}$ leads to a contradiction using prime factorization). $\frac{22}{7}$ is rational by definition. $0.333... = \frac{1}{3}$ is rational (repeating decimal). $\sqrt{16} = 4$ is an integer, hence rational.

Solution: Suppose $r$ is rational and $i$ is irrational, and assume $r + i = q$ is rational. Then $i = q - r$, which would be the difference of two rationals, hence rational. But this contradicts $i$ being irrational. Therefore, $r + i$ must be irrational. For example, $2 + \sqrt{2}$ is irrational because 2 is rational and $\sqrt{2}$ is irrational.

Solution: Suppose $r \neq 0$ is rational and $i$ is irrational, and assume their product $r \times i = q$ is rational. Then $i = \frac{q}{r}$, the quotient of two rationals (with $r \neq 0$), so $i$ would be rational. This contradicts $i$ being irrational. Therefore, $r \times i$ must be irrational. Example: $3 \times \sqrt{2} = 3\sqrt{2}$ is irrational.

Solution: For any prime $p$: assume $\sqrt{p} = \frac{a}{b}$ where $\gcd(a, b) = 1$. Squaring gives $pb^2 = a^2$. By the Fundamental Theorem of Arithmetic, the prime $p$ appears an odd number of times (exactly once) on the left, but an even number of times on the right (as part of $a^2$). This is impossible, so $\sqrt{p}$ is irrational. This applies to every prime.

Solution: While the product of a rational and irrational is always irrational, the product of two irrationals can be either rational or irrational. The key example: $\sqrt{2} \times \sqrt{2} = 2$, which is rational. Another example: $\sqrt{2} \times \sqrt{3} = \sqrt{6}$, which is irrational. So there is no universal rule — it depends on the specific irrationals.