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SubjectsMathematics
अपरिमेय संख्याएँ
In this Class 9 Mathematics topic from the Number Systems chapter, students learn that irrational numbers cannot be written in the form p/q, where p and q are integers and q is not zero. They explore familiar examples such as √2 and π, understand their non-terminating, non-repeating decimal expansions, and distinguish them from rational numbers. The topic also develops skills for representing irrational numbers on the number line and understanding their place within the real number system.
Correct answer: A. \(0.333\ldots\)
Explanation: \(0.333\ldots=\frac{1}{3}\), 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.
Correct answer: C. ( \sqrt{18} )
Explanation: 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.
Correct answer: A. ( \sqrt{14} )
Explanation: ( \sqrt{14} ) is irrational, and every irrational number is also real. Real numbers include both rational and irrational numbers.
Correct answer: C. ( \sqrt{27} )
Explanation: (27) is not a perfect square, so ( \sqrt{27} ) is irrational. Distinguish perfect squares from non-perfect squares.
Correct answer: A. Its decimal expansion is non-terminating and non-repeating
Explanation: 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.”
Correct answer: B. It is irrational
Explanation: 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.
Correct answer: C. ( \sqrt{26} )
Explanation: ( \sqrt{26} ) is the square root of a non-perfect square, so it is irrational. ( \frac{22}{7} ) is rational, but ( \pi ) is not.
Correct answer: C. Non-terminating non-recurring
Explanation: The decimal expansion of an irrational number is non-terminating non-recurring. This is a key exam identification.
Correct answer: B. Rational
Explanation: 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.
Correct answer: A. Non-terminating and non-repeating
Explanation: 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.”
Correct answer: A. ( \sqrt{4} ) and ( \sqrt{7} )
Explanation: ( \sqrt{4}=2 ) is rational and ( \sqrt{7} ) is irrational. Check the type of both numbers in a pair.
Correct answer: A. ( \sqrt{5} )
Explanation: ( \sqrt{5} ) lies approximately between (2) and (3) and is irrational. A square root of a non-perfect square is a good example.
Correct answer: B. Irrational
Explanation: Adding a rational number to an irrational number gives an irrational number. (3) is rational and ( \sqrt{2} ) is irrational.
Correct answer: C. Irrational
Explanation: Multiplying non-zero rational (5) by irrational ( \sqrt{3} ) gives an irrational number. The zero case is different.
Correct answer: A. Square root of a perfect square
Explanation: 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.
Correct answer: A. ( \sqrt{20} )
Explanation: ( \sqrt{20} ) is irrational and all irrational numbers are real. Real but not rational means irrational.
Correct answer: C. Non-terminating non-recurring
Explanation: ( \sqrt{3} ) is irrational, so its decimal is non-terminating non-recurring. The decimal of an irrational number is never recurring.
Correct answer: A. ( \sqrt{6} ), ( \sqrt{10} ), ( \sqrt{11} )
Explanation: (6), (10), and (11) are not perfect squares, so all three square roots are irrational. Check each number separately.
Correct answer: B. Rational
Explanation: ( \sqrt{2}\times\sqrt{2}=2 ), so it is rational. The product of two irrational numbers is not always irrational.
Correct answer: A. 2√7
Explanation: 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.
Correct answer: B. 3√2
Explanation: 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.
Correct answer: B. It is irrational
Explanation: (2) is rational and ( \sqrt{5} ) is irrational, so the sum is irrational. Adding rational and irrational gives irrational.
Correct answer: A. (2\sqrt{2})
Explanation: Adding like terms gives ( \sqrt{2}+\sqrt{2}=2\sqrt{2} ). Add like radicals like algebraic terms.
Correct answer: A. \(\sqrt{49}\) is rational because \(\sqrt{49}=7\)
Explanation: 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.
Correct answer: A. 2√3
Explanation: 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.
