Irrational numbers

Correct answer: C. 5

Explanation: The governing concept is solving an inequality over the integers and then counting the resulting set. Since x²<5, we have -√5<x<√5, so the possible integers are -2,-1,0,1,2. Their squares are 4,1,0,1,4, all less than 5. The next integers, -3 and 3, have square 9 and are excluded. Thus V has five elements, so C is correct.

Correct answer: A. \(\frac{4\sqrt{5}}{5}\)

Explanation: To remove the square root from the denominator, multiply both numerator and denominator by \(\sqrt{5}\): \(\frac{4}{\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}=\frac{4\sqrt{5}}{5}\), since \(\sqrt{5}\times\sqrt{5}=5\). Therefore, option A is correct. Option B is the reciprocal of the original fraction, while C and D are not equivalent to it. Exam tip: when a denominator contains a single square root, multiply the numerator and denominator by that same square root.

Correct answer: B. 6 and 7

Explanation: Since \(7^2=49<53<64=8^2\), it follows that \(7<\sqrt{53}<8\). Therefore, \(\sqrt{53}\) lies between 7 and 8, so option B is correct. Option C is incorrect because \(\sqrt{53}\) is less than 8, as \(8^2=64\). Exam tip: compare the number with the nearest consecutive perfect squares to locate its square root.

Correct answer: B. Irrational real number

Explanation: A real number is rational if its decimal expansion terminates or eventually repeats. The given expansion continues indefinitely and, as stated, has no fixed repeating block: the runs of digits keep changing. Therefore it cannot be represented as p/q with integers p and q, so it is irrational. Option B is correct; an integer is rational, and a terminating decimal also represents a rational number.

Correct answer: C. Irrational real number

Explanation: The governing concept is decimal representation of rational and irrational numbers. A rational number, when written in decimal form, either terminates or eventually repeats a fixed block of digits. For example, 0.333... repeats 3 and is rational. The given decimal 0.040040004... is stated to continue without termination and without any fixed repeating pattern. Therefore it cannot be expressed as a ratio of two integers and is irrational. It is still a real number because irrational numbers are included in the real-number system. It is not a whole or natural number, since those are non-negative integers. Hence option C is correct.

Correct answer: B. Irrational real number

Explanation: \(\sqrt{7}\times\sqrt{14}=\sqrt{98}=\sqrt{49\times2}=7\sqrt{2}\). Since \(\sqrt{2}\) is irrational, multiplying it by the nonzero rational number \(7\) still gives an irrational number. Therefore, the expression is an irrational real number, not a rational number, integer, or terminating decimal. Exam tip: combine the square roots first and extract perfect-square factors.

Correct answer: A. \(\frac{6\sqrt{7}}{7}\)

Explanation: To rationalise the denominator, multiply both the numerator and denominator by \(\sqrt{7}\): \(\frac{6}{\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}=\frac{6\sqrt{7}}{7}\), since \(\sqrt{7}\times\sqrt{7}=7\). Hence, option A is correct. Remember that the numerator must also be multiplied by \(\sqrt{7}\); changing only the denominator to 7 is not valid.

Correct answer: B. Irrational real number

Explanation: \(\sqrt{3}\times\sqrt{21}=\sqrt{63}=3\sqrt{7}\). Since 7 is not a perfect square, \(\sqrt{7}\) is irrational, and multiplying it by the non-zero rational number 3 remains irrational. Therefore, the given expression is an irrational real number. It is neither an integer, a rational number, nor a terminating decimal. Exam tip: the square root of a positive non-perfect square is irrational.

Correct answer: B. Irrational real number

Explanation: The governing classification theorem states that a real number is rational if and only if its decimal expansion terminates or eventually repeats. A decimal that continues forever without settling into any repeating block is non-terminating and non-repeating, so it cannot be written as p/q for integers p and q with q ≠ 0. Therefore the given number is irrational; because it is represented by a real decimal, it is an irrational real number. Option B is correct. Option A is ruled out by the absence of repetition, option C is ruled out because digits continue indefinitely, and option D is impossible because an integer has a terminating decimal representation such as 4.000.... The stated condition is decisive even if only an initial pattern is displayed.

Correct answer: A. 6√2

Explanation: The governing concept is simplification of like surds before addition. Rewrite each radical using 2 as the square-free factor: √2 remains √2, √8 = √(4×2) = 2√2, and √18 = √(9×2) = 3√2. These are now like surds, so their coefficients can be added: √2 + 2√2 + 3√2 = (1 + 2 + 3)√2 = 6√2. Thus option A is correct. Option B omits one coefficient, while option C adds incorrectly. Option D, √28, is actually equal to 2√7 and is not equal to the given sum. Radicals should be simplified before combining; unlike surds cannot be added by simply adding their radicands.

Correct answer: A. Irrational

Explanation: The number 3 is rational because it can be written as 3/1, while √2 is irrational; its decimal expansion is non-terminating and non-repeating. A fundamental property is that the sum of a rational number and an irrational number is irrational. To see why, suppose 3+√2 were rational. Subtracting the rational number 3 would then make √2 rational, contradicting the known irrationality of √2. Therefore 3+√2 is irrational, so option A is correct. It cannot be a rational number, integer, or whole number. The fact that the expression contains a radical does not by itself prove the result; the decisive reasoning is the rational-plus-irrational property and the contradiction argument. Thus the number is real, but among the listed choices its most specific correct classification is irrational.

Correct answer: A. Non-terminating and non-repeating

Explanation: An irrational number cannot be written as \(p/q\), so its decimal expansion is non-terminating and non-repeating. A repeating decimal is rational. Exam tip: look for both features together.

Correct answer: C. (5\sqrt{2})

Explanation: (\sqrt{18}=3\sqrt{2}) and (\sqrt{8}=2\sqrt{2}), so the sum is (5\sqrt{2}). Simplify radicals first.

Correct answer: A. Its decimal expansion is non-terminating and non-repeating.

Explanation: An irrational number has a decimal expansion that never ends and never repeats a block of digits. Option C describes a rational number. Exam tip: non-terminating, non-repeating decimals are irrational.

Correct answer: D. (2-\sqrt{3})

Explanation: Rationalising the denominator gives (\frac{1}{2+\sqrt{3}}=2-\sqrt{3}). Remember to multiply by the conjugate.

Correct answer: A. It is irrational because its decimal expansion is non-terminating and non-repeating.

Explanation: Writing a number in decimal form does not make it rational. The blocks of 1s grow as 1, 2, 3, ..., so no fixed repeating block exists. A non-terminating, non-repeating decimal is irrational. Exam tip: check for repetition.

Correct answer: C. (4\sqrt{2})

Explanation: To simplify a radical, separate any perfect-square factor from the number inside the root. Since 50 = 25 × 2 and \(\sqrt{25}=5\), we get \(\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}\). The original expression then becomes \(5\sqrt{2}-\sqrt{2}\).

These are like radicals because both terms contain the same \(\sqrt{2}\). Subtract their coefficients: \(5\sqrt{2}-1\sqrt{2}=(5-1)\sqrt{2}=4\sqrt{2}\). No further simplification is possible because 2 has no factor that is a perfect square greater than 1. Therefore option C is correct. The answer is not 5 or \(\sqrt{48}\); those forms do not represent the simplified subtraction as directly.

Correct answer: D. (6\sqrt{3})

Explanation: (\sqrt{12}=2\sqrt{3}) and (\sqrt{27}=3\sqrt{3}), so the total is (6\sqrt{3}). First convert all terms into like radicals.

Correct answer: B. Irrational

Explanation: It has no fixed repeating block, so it is irrational. Avoid treating a non-repeating decimal as repeating.

Correct answer: A. 3

Explanation: For the quotient of square roots, use \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\). Thus, \(\frac{\sqrt{27}}{\sqrt{3}}=\sqrt{\frac{27}{3}}=\sqrt{9}=3\), so option A is correct. The closest distractor, \(\sqrt{3}\), results from failing to simplify the quotient correctly. Exam tip: first divide the radicands when the denominator is nonzero, then simplify the resulting square root.

Correct answer: D. (5\sqrt{5})

Explanation: (\sqrt{45}=3\sqrt{5}) and (\sqrt{20}=2\sqrt{5}), so the sum is (5\sqrt{5}). First take out perfect-square factors.

Correct answer: C. √5

Explanation: The governing concept is that the square-root function preserves order for non-negative numbers. Since 2 < 5 < 8, taking square roots gives √2 < √5 < √8, so option C is correct without needing decimal approximations. A numerical check confirms this: √2 is about 1.41, √5 is about 2.24, and √8 is about 2.83. Option A is 1, which is less than √2 because 1² < 2. Options B and D are greater than √8: 3² = 9 > 8 and 4² = 16 > 8. The important reasoning is to compare the radicands 2, 5, and 8, since all are non-negative; their square roots retain the same order.

Correct answer: A. it is irrational

Explanation: The side of the square is \(\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}\,\text{m}\). Since \(\sqrt{2}\) is irrational, \(3\sqrt{2}\) is also irrational. Option B has the correct length but the wrong classification. Exam tip: first extract perfect-square factors from a square root.

Correct answer: A. 0.101001000100001...

Explanation: An irrational number has a decimal expansion that is non-terminating and non-repeating. In 0.101001000100001..., the blocks of zeros grow and no fixed digit pattern repeats indefinitely, so the expansion is non-terminating and non-recurring. Thus option A is irrational. Option B is recurring, option C terminates, and 11/13 is rational because it is a ratio of two integers.

Correct answer: B. Riya is incorrect, because this decimal is non-terminating and non-repeating.

Explanation: In \(0.1010010001\ldots\), the zeros between consecutive 1s keep increasing, so no fixed block repeats. Hence it is irrational. Exam tip: identify whether an endless decimal has a repeating pattern.