Irrational numbers

The governing concept is simplification of a radical by separating a perfect-square factor. Since 18 = 9 × 2 and 9 is a perfect square, √18 = √(9 × 2) = √9 × √2 = 3√2. The factor 2 remains inside the radical because it is not a perfect square and has no square factor greater than 1. Thus option B is correct. Option A leaves 9 unchanged instead of replacing √9 by 3. Option C is the simplified form of √12, not √18, while option D is wrong because 6² = 36 rather than 18. A direct verification is (3√2)² = 9 × 2 = 18. This confirms both the numerical equivalence and the fact that 3√2 is already in simplest radical form.

(2) is rational and ( \sqrt{5} ) is irrational, so the sum is irrational. Adding rational and irrational gives irrational.

Adding like terms gives ( \sqrt{2}+\sqrt{2}=2\sqrt{2} ). Add like radicals like algebraic terms.

Since \(49=7\times7\), \(\sqrt{49}=7=7/1\), so it is rational. A radical sign alone does not make a number irrational. In exams, first check whether the radicand is a perfect square.

The governing concept is extraction of perfect-square factors from a square root. Rewrite 12 as 4 × 3, where 4 is a perfect square. Then √12 = √(4 × 3) = √4 × √3 = 2√3. The factor 3 remains under the radical because it has no perfect-square factor other than 1, so 2√3 is fully simplified. Option B, 3√2, squares to 18 and therefore represents √18, not √12. Option C is incorrect because 6² = 36, and option D squares to 48, which is too large. Squaring the selected expression gives (2√3)² = 4 × 3 = 12, providing a direct check that option A is the only correct answer.

(48=16\times3), so ( \sqrt{48}=4\sqrt{3} ). Taking out the largest perfect square gives the simplest answer.

( \sqrt{3}\times\sqrt{12}=\sqrt{36}=6 ), so it is rational. The product of two irrational numbers is not always irrational.

(72=36\times2), so ( \sqrt{72}=6\sqrt{2} ). Take the perfect square (36) out of the radical.

( \sqrt{11} ) is irrational, and its negative remains irrational. Changing the sign does not change rational or irrational type.

A repeating decimal is rational. Here \(0.2727\ldots=\frac{27}{99}=\frac{3}{11}\), since 27 repeats. Non-termination alone does not make it irrational. Exam tip: first check for repetition.

(1.01001000100001\ldots) has no fixed repetition and does not terminate. Such a decimal is irrational.

( \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} ), so it is irrational. If the denominator is irrational, first think of the simplified form.

Subtracting irrational ( \sqrt{6} ) from rational (4) gives an irrational number. The sum or difference of rational and irrational is generally irrational.

(108=36\times3), so ( \sqrt{108}=6\sqrt{3} ). While simplifying radicals, take out the largest perfect square.

Since (9<13<16), (3<\sqrt{13}<4), and (13) is not a perfect square. Therefore, ( \sqrt{13} ) is irrational.

( \sqrt{18}=3\sqrt{2} ), so ( \frac{\sqrt{18}}{3}=\sqrt{2} ). First simplify the radical and then cancel common factors.

An irrational number cannot be expressed as \\(\frac{p}{q}\\), where \\(p\\) and \\(q\\) are integers and \\(q\neq0\\). Its decimal expansion is non-terminating and non-repeating. Option A defines a rational number; integers and terminating decimals are also rational. Exam tip: identify an irrational number by a decimal that neither terminates nor repeats.

An irrational number has a decimal expansion that never ends and never repeats in a fixed pattern. A non-terminating recurring decimal is rational. Exam tip: remember “non-terminating and non-recurring” for irrational numbers.

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but their sum is \(0\), which is rational. Hence, the word “always” makes Riya’s statement false. Exam tip: a single counterexample is enough to disprove an “always” statement.

The governing concept is the distinction between perfect-square and non-perfect-square roots. Since 7 is not a perfect square of any integer, √7 cannot be expressed as p/q where p and q are integers and q is non-zero. Therefore √7 is irrational, making option B correct. It is not an integer or a natural number because squaring an integer never gives 7. Option A is also false because rational square roots of integers occur when the integer is a perfect square, such as √9 = 3. The correct test is to check whether the radicand is a perfect square before classifying its square root.