Irrational numbers

(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.

Direct answer: option B, 6√2. First factor 72 into a perfect square and another factor: 72 = 36 × 2. Since √36 = 6, we get √72 = √(36 × 2) = √36 × √2 = 6√2. The purpose of simplification is to take every perfect-square factor outside the radical. Option A, 8√2, would equal √128, not √72, so it is too large. Option B is correct because 6 is the square root of 36. Option C, 3√8, has not been fully simplified; in fact 3√8 = 3 × 2√2 = 6√2, so it is numerically equal but is not the standard simplest form. Option D, 12, is wrong because 12² = 144, not 72. Exam cue: search for the largest perfect-square factor inside 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.

Direct answer: option D, √13. To lie between 3 and 4, a number must have a value greater than 3 and less than 4. Squaring positive numbers preserves order, so square the endpoints: 3² = 9 and 4² = 16. Because 9 < 13 < 16, taking square roots gives 3 < √13 < 4. Also, 13 is not a perfect square, so √13 cannot be written as a terminating or repeating fraction; it is irrational. Option A, 3.5, is between 3 and 4 but is rational because it equals 7/2. Option B, 7/2, is also rational, although it lies in the interval. Option C, √16 = 4, is rational and is not strictly between 3 and 4 because it is an endpoint. Option D satisfies both requirements: it is between 3 and 4 and irrational. Exam cue: for √n, compare n with consecutive perfect squares.

( \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.

The relevant concept is the classification of real numbers by their decimal representation and fractional form. π is irrational: it cannot be expressed exactly as p/q for integers p and q with q ≠ 0, and its decimal expansion is non-terminating and non-repeating. Thus option B is correct. The fraction 22/7 is a useful approximation to π, but it is not exactly equal to π, so option C is false. Since every integer is rational, π cannot be an integer either; option D is therefore false. Option A reverses the correct classification. Always distinguish an approximation symbol from exact equality in numerical questions.

A rational number has a terminating decimal when, in lowest terms, its denominator contains only the prime factors 2 and 5; otherwise its decimal is repeating. The numbers 1/4 = 0.25, 0.75 and 2.5 all have terminating decimal expansions, so they are rational. In contrast, 3 is not a perfect square, and √3 is irrational. Its decimal expansion therefore continues indefinitely without a repeating block. Hence option C is correct. The question tests a standard property of irrational numbers: their decimal expansions are non-terminating and non-repeating. Do not confuse a long decimal with an irrational one; the essential feature is the absence of termination and repetition.

The three dots indicate that the digit 3 repeats indefinitely: 0.333… = 0.333333… . Every repeating decimal represents a rational number because it can be converted into a fraction. For example, let x = 0.333…; then 10x = 3.333…, and subtracting gives 9x = 3, so x = 3/9 = 1/3. Therefore option B is correct. It is not irrational because its decimal digits repeat, and it is not an integer or natural number because 1/3 lies between 0 and 1. The important distinction is that non-terminating decimals may be rational when they repeat, or irrational when they do not repeat.

(5+\sqrt{2}) is irrational because adding a rational and an irrational gives an irrational. In such questions watch the irrational part.

(3\sqrt{5}) is irrational because multiplying by a non-zero rational keeps it irrational. Remember the multiplier must not be (0).

Both \(\sqrt{7}\) and \(-\sqrt{7}\) are irrational, but their sum is \(0\), which is rational. Hence the statement is false. Exam tip: one counterexample is enough to disprove an “always” statement.

The correct answer is option C: \(\sqrt{10}\) is irrational. A rational number can be written as \(p/q\), where p and q are integers and \(q\ne0\). The square root of a perfect square is an integer, so it is rational: \(\sqrt9=3\), \(\sqrt{25}=5\), and \(\sqrt{36}=6\). The number 10 is not a perfect square; it lies between \(3^2=9\) and \(4^2=16\). Therefore \(\sqrt{10}\) cannot be simplified to an integer or a fraction of the required kind and is irrational. Option A is wrong because \(\sqrt9=3\). Option B is wrong because \(\sqrt{25}=5\). Option C is correct. Option D is wrong because \(\sqrt{36}=6\). A common mistake is thinking every square root is irrational. Only the square root of a non-perfect square is irrational, when the number is a whole number. Memory cue: check whether the number inside the root is a perfect square.

\(49\) is a perfect square, and \(\sqrt{49}=7\), which is a rational integer. In contrast, 6, 15, and 30 are not perfect squares, so their square roots are irrational. Exam tip: the square root of a perfect square is rational.

An irrational number cannot be written as a ratio of two integers, and its decimal expansion is non-terminating and non-repeating. To identify the answer, each option must be checked both for its location and for its type. The number \(\sqrt{2}\) is approximately 1.414, so it lies strictly between 1 and 2.

Also, 2 is not a perfect square, so its square root is irrational. In contrast, \(\sqrt{4}=2\) is rational, while \(3/2\) and 1.5 are both equal to 1.5 and are rational. Therefore only option B satisfies both requirements: it lies between 1 and 2 and is irrational. The supplied answer and explanation are correct.

Direct answer: option C, irrational. The decimal 0.101001000100001… continues forever, so it is non-terminating. More importantly, its digits do not settle into one fixed block that repeats forever. The gaps between the 1s keep changing: after the first 1 there is one zero, then two zeros, then three, then four, and so on. A decimal is rational if it terminates or eventually repeats a fixed pattern. This decimal does neither, so it is irrational. Option A, terminating rational, is wrong because the decimal does not end. Option B, repeating rational, is wrong because the repeating block is not fixed; the zeros increase. Option C is correct. Option D, integer, is wrong because the number is a positive decimal less than 1, not a whole number. Be careful: seeing repeated 1s does not automatically mean the decimal is repeating. Check the complete pattern.