Irrational numbers — Class 10 CBSE Mathematics MCQs with Solutions

Solution: $\sqrt{4} = 2$ and $\frac{22}{7}$ are rational, and $0.\overline{3} = \frac{1}{3}$ is rational because a non-terminating recurring decimal always represents a rational number. Since $7$ is not a perfect square, $\sqrt{7}$ cannot be written as $\frac{p}{q}$ and is irrational.

Solution: Let $r$ be rational and $s$ irrational. If $r + s$ were rational, then $s = (r + s) - r$ would be a difference of two rational numbers and hence rational — a contradiction. So $r + s$ is always irrational; for example, $2 + \sqrt{3}$ is irrational.

Solution: From $\sqrt{5} = \frac{p}{q}$ we get $p^2 = 5q^2$, so $5$ divides $p^2$ and hence $5$ divides $p$. Writing $p = 5k$ gives $25k^2 = 5q^2$, i.e. $q^2 = 5k^2$, so $5$ also divides $q$. Thus $5$ is a common factor of $p$ and $q$, contradicting the assumption that they are coprime.

Solution: Using the identity $(a + b)(a - b) = a^2 - b^2$: $(3 + \sqrt{5})(3 - \sqrt{5}) = 9 - 5 = 4$. This also shows that the product of two irrational numbers can be rational.

Solution: Counterexamples defeat both statements: $\sqrt{2} + (-\sqrt{2}) = 0$ is rational, so (I) fails, and $\sqrt{2} \times \sqrt{2} = 2$ is rational, so (II) fails. The sum or product of two irrationals is only sometimes irrational, e.g. $\sqrt{2} + \sqrt{3}$ and $\sqrt{2} \times \sqrt{3} = \sqrt{6}$ are irrational.

Solution: Simplify the denominator: $\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$. Therefore $\frac{7\sqrt{3}}{\sqrt{75}} = \frac{7\sqrt{3}}{5\sqrt{3}} = \frac{7}{5}$, a rational number. This shows a quotient of two irrational numbers need not be irrational.

Solution: If $r$ is rational, then $\frac{5 - r}{2}$ is also rational, since subtracting rationals and dividing by $2$ keeps the result rational. But this expression equals $\sqrt{3}$, which is irrational — a contradiction. Hence the assumption was wrong and $5 - 2\sqrt{3}$ is irrational.