Irrational numbers

\(0.333\ldots=\frac{1}{3}\), so it is a rational number even though its decimal expansion never ends; it repeats. In contrast, \(\sqrt{2}\) has a non-terminating, non-repeating decimal. Exam tip: check whether the decimal repeats.

The direct answer is option C, √18. A decimal expansion is non-terminating and non-recurring when the number is irrational. First simplify √18: since 18 = 9 × 2, √18 = √9 × √2 = 3√2. The number √2 is irrational, and multiplying it by the non-zero rational number 3 keeps it irrational. Therefore √18 has a decimal expansion that never ends and never repeats in a fixed pattern. Option A, 5/6, is rational; its decimal is 0.8333..., which is non-terminating but recurring. Option B, 0.875, is already a terminating decimal and equals 7/8. Option C, √18, is irrational and therefore has the required non-terminating, non-recurring expansion. Option D, 2, is an integer, and every integer has a terminating decimal representation, such as 2.0. The key distinction is that a non-terminating decimal may be recurring or non-recurring; only an irrational number gives a non-terminating non-recurring decimal. Remember: rational numbers have terminating or recurring decimals, while irrational numbers have non-terminating non-recurring decimals.

( \sqrt{14} ) is irrational, and every irrational number is also real. Real numbers include both rational and irrational numbers.

(27) is not a perfect square, so ( \sqrt{27} ) is irrational. Distinguish perfect squares from non-perfect squares.

An irrational number cannot be expressed as
p/q
, where
q \ne 0
. Hence, its decimal expansion is non-terminating and non-repeating; a repeating decimal is rational. Exam tip: look for “non-repeating.”

The direct answer is option B: √32 is irrational. To understand this, simplify the radical: 32 = 16 × 2, so √32 = √16 × √2 = 4√2. The square root of 2 is irrational, meaning it cannot be written as a fraction of two integers. Therefore 4√2 is also irrational. Another quick test is that 32 is not a perfect square. The perfect squares near it are 25 and 36; their square roots are 5 and 6. Since 32 lies between them, √32 lies between 5 and 6 and is not an integer. Option A, rational, is wrong because √32 cannot be expressed as a terminating or recurring rational decimal. Option B is correct because the simplified form still contains √2. Option C, 16, is wrong because the square root of 32 is not 16; in fact, 16² = 256. Option D, 8, is wrong because 8² = 64, not 32. The exact value is approximately 5.657, not 8 or 16. Memory cue: first remove any square factor, then check whether a non-square radical remains; if it does, the result is irrational.

( \sqrt{26} ) is the square root of a non-perfect square, so it is irrational. ( \frac{22}{7} ) is rational, but ( \pi ) is not.

The decimal expansion of an irrational number is non-terminating non-recurring. This is a key exam identification.

The governing concept is classification through exact simplification. Write 0.09 as 9/100. Then √0.09 = √(9/100) = √9/√100 = 3/10 = 0.3. Since 3/10 is a ratio of integers with a non-zero denominator, it is rational; its decimal form also terminates. The value is positive and real, so it is neither non-real nor an irrational number. It is also not a non-terminating, non-recurring decimal. This example shows that a square root of a decimal is not automatically irrational: when the decimal is the square of a rational number, its square root remains rational. Therefore option B is correct.

An irrational number has a decimal expansion that neither ends nor repeats a fixed pattern, so A is correct. Terminating or repeating decimals are rational. Exam tip: remember “non-terminating, non-repeating.”

( \sqrt{4}=2 ) is rational and ( \sqrt{7} ) is irrational. Check the type of both numbers in a pair.

( \sqrt{5} ) lies approximately between (2) and (3) and is irrational. A square root of a non-perfect square is a good example.

Adding a rational number to an irrational number gives an irrational number. (3) is rational and ( \sqrt{2} ) is irrational.

Multiplying non-zero rational (5) by irrational ( \sqrt{3} ) gives an irrational number. The zero case is different.

A rational number is a number that can be written as a fraction of two integers with a nonzero denominator. Every integer is rational, so if the square root of a number is an integer, it is automatically rational. A perfect square has an integer square root, such as 1, 4, 9, 16, or 64. This makes the first choice always rational under its stated condition.

For example, \(\sqrt{64}=8=\frac{8}{1}\), so it is rational. By contrast, the square root of a non-perfect square is irrational, such as \(\sqrt{2}\). A multiple of \(\pi\) is not automatically rational, and a non-terminating, non-recurring decimal is irrational by definition. Therefore the square root of a perfect square, option A, is the correct choice.

( \sqrt{20} ) is irrational and all irrational numbers are real. Real but not rational means irrational.

( \sqrt{3} ) is irrational, so its decimal is non-terminating non-recurring. The decimal of an irrational number is never recurring.

(6), (10), and (11) are not perfect squares, so all three square roots are irrational. Check each number separately.

( \sqrt{2}\times\sqrt{2}=2 ), so it is rational. The product of two irrational numbers is not always irrational.

The governing rule is √(ab) = √a × √b, together with the extraction of perfect-square factors from a radical. Factor 28 as 4 × 7, where 4 is a perfect square. Therefore √28 = √(4 × 7) = √4 × √7 = 2√7. The factor 4 can leave the radical because its square root is the integer 2, whereas 7 remains under the radical because it has no square factor greater than 1. Option A is therefore the simplified form. Option B is not equivalent because (7√2)^2 = 98, option C has square 112, and option D has square 196. Squaring 2√7 gives 4 × 7 = 28, which directly verifies the answer and shows why the other choices are invalid.

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.