Irrational numbers and real numbers

Correct answer: C. A non-terminating recurring decimal is rational.

Explanation: In \(0.272727\ldots\), the block 27 repeats, so the number is rational. Let \(x=0.272727\ldots\); then \(100x-x=27\), giving \(x=\frac{27}{99}=\frac{3}{11}\). Exam tip: recurring decimals are rational, unlike non-recurring non-terminating ones.

Correct answer: A. Rational number

Explanation: A terminating decimal can be expressed as a ratio of two integers by writing it over an appropriate power of 10 (for example 0.75 = 75/100 = 3/4). Therefore such numbers are rational. Option B (irrational) is incorrect because irrational numbers have non-terminating, non-repeating decimal expansions (e.g. √2). Option C (non‑real) is wrong since decimal expansions represent real numbers. Option D (natural number only) is incorrect because terminating decimals include non-integer rationals (e.g. 0.5). Exam tip: To convert a terminating decimal to a fraction, write the decimal without the point as the numerator and use 10^n as denominator where n is number of decimal places, then simplify (e.g. 0.125 = 125/1000 = 1/8).

Correct answer: A. Rational number

Explanation: The block (27) repeats, so this is a recurring (repeating) decimal. Every repeating decimal is rational because it can be written as a fraction. For example let \\(x=3.2727\ldots\\). Then \\(100x=327.2727\ldots\\) and subtracting gives \\(99x=324\\), so \\(x=\frac{324}{99}=\frac{36}{11}\\). Thus the number is rational. The closest distractor, "irrational", is wrong because irrational numbers are non-terminating and non-repeating decimals. "Integer" is wrong since the value is not an integer, and "non-real" is wrong because this is a real number. Exam tip: convert repeating decimals to fractions by multiplying to align repeats and subtracting to eliminate the repeating part.

Correct answer: A. Irrational number

Explanation: A decimal expansion that is non-terminating and non-repeating represents an irrational number. Rational numbers always have decimal expansions that either terminate or eventually repeat a fixed block. The given expansion 5.123123312333... shows no fixed repeating block, so it is irrational. The closest distractor, 'rational number', is incorrect because a rational decimal must exhibit a repeating pattern if it does not terminate. Exam tip: to test quickly, look for a consistent repeating block—if none exists, the number is irrational.

Correct answer: A. Rational number

Explanation: \(\sqrt{121}=11\). The number 11 is an integer and can be written as the fraction \(11/1\), so it is rational. Option B (irrational) is incorrect because irrational numbers cannot be expressed as a ratio of integers; 11 can. Option C (non‑real) is wrong because 11 is a real number. Option D (zero) is wrong because the value is 11, not 0. Exam tip: the square root of a perfect square integer is an integer and thus rational.

Correct answer: A. It is irrational

Explanation: The square root of a whole number is rational only when the number under the root is a perfect square, such as \(1,4,9,16,25,36\), or another square of an integer. The number 37 lies between \(36=6^2\) and \(49=7^2\), so it is not a perfect square. Therefore \(\sqrt{37}\) cannot be written as an integer or as a terminating or repeating rational number; it is irrational.

Thus option A is correct. It is not a perfect square itself, because 37 is the radicand rather than a square number. Also, \(\sqrt{37}\) is slightly greater than 6, so it cannot equal 6; squaring 6 gives 36, not 37. It is not an integer for the same reason. The key test is that the square root of a positive integer is irrational when that integer is not a perfect square.

Correct answer: A. (4), (-2), (0.75)

Explanation: Integers and terminating decimals are rational. Options containing irrational roots or (\pi) are not only rational.

Correct answer: A. They include both rational and irrational numbers

Explanation: Real numbers form the large set of rational and irrational numbers. They can be represented on the number line.

Correct answer: A. Rational and real number

Explanation: You can write (-8) as \( -8=\frac{-8}{1}\), so it is a ratio of two integers and therefore rational. Every integer is also a real number, so (-8) is real as well.
Why other choices are wrong: B is incorrect because irrational numbers cannot be expressed as a ratio of integers (e.g. \(\sqrt{2}\)). C is wrong because natural numbers are positive integers (1,2,3,...); -8 is not natural. D is wrong because non-real (complex) numbers have a nonzero imaginary part (e.g. \(3+2i\)); -8 has no imaginary part.
Exam tip: To decide if a number is rational, try to express it as \(\frac{p}{q}\) with integers \(p,q\) and \(q\neq0\).

Correct answer: A. Irrational number

Explanation: Adding irrational (\sqrt{11}) to rational (6) gives an irrational number. Do not mistake the root for a perfect square.

Correct answer: A. Irrational number

Explanation: Subtracting an irrational number from a rational number gives an irrational result. (\sqrt{6}) is irrational because (6) is not a perfect square.

Correct answer: A. (9)

Explanation: Direct answer: Option A, 9. Use the square-root product rule for non-negative numbers: \(\sqrt{a}\sqrt{b}=\sqrt{ab}\). Thus \(\sqrt{3}\times\sqrt{27}=\sqrt{81}=9\), because the principal square root of 81 is the positive number 9. Equivalently, \(\sqrt{27}=3\sqrt{3}\), so the product is \(3\sqrt{3}\times\sqrt{3}=3\times3=9\). Option A is correct. Option B, \(3\sqrt{3}\), is only the value of \(\sqrt{27}\), not the whole product. Option C incorrectly adds the radicands, giving \(\sqrt{30}\); multiplication requires multiplying 3 and 27. Option D, 27, forgets the factor \(\sqrt{3}\). The result being rational is not surprising: two irrational factors can have a rational product. Memory cue: multiply inside one radical, then simplify.

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

Explanation: \(\sqrt{32}=\sqrt{16\times2}=\sqrt{16}\cdot\sqrt{2}=4\sqrt{2}\). Hence the simplified form is \(4\sqrt{2}\). Option B (\(8\sqrt{2}\)) is much larger and incorrect; option C (\(4\)) equals \(\sqrt{16}\), not \(\sqrt{32}\); option D is the unsimplified radical. Exam tip: always factor out the largest perfect square from under the root.

Correct answer: A. \(5\sqrt{3}\)

Explanation: Since \(75=25\times3\), and \(25\) is a perfect square, \(\sqrt{75}=\sqrt{25}\times\sqrt{3}=5\sqrt{3}\). \(3\sqrt{5}\) is incorrect because \((3\sqrt{5})^2=45\), not 75. Exam tip: To simplify a surd, identify and factor out the greatest perfect-square factor.

Correct answer: A. 6\sqrt{3}

Explanation: \(\sqrt{48}=\sqrt{16\times3}=4\sqrt{3}\) and \(\sqrt{12}=\sqrt{4\times3}=2\sqrt{3}\). Adding like terms gives \(4\sqrt{3}+2\sqrt{3}=6\sqrt{3}\). The distractor \(2\sqrt{15}\) comes from wrongly treating the sum as \(\sqrt{48+12}=\sqrt{60}=2\sqrt{15}\); you cannot combine square roots across addition that way. Exam tip: simplify each radical into simplest surd form first, then add or subtract only like surds.

Correct answer: A. \sqrt{5}

Explanation: \sqrt{45}=\sqrt{9\cdot5}=3\sqrt{5} and \sqrt{20}=\sqrt{4\cdot5}=2\sqrt{5}. Subtracting gives 3\sqrt{5}-2\sqrt{5}=\sqrt{5}. The closest distractor D (2\sqrt{5}) is incorrect because it equals \sqrt{20}, not the difference. Exam tip: factor common square factors (like \sqrt{5}) to simplify radicals before adding or subtracting.

Correct answer: A. \(\sqrt{19}\)

Explanation: \(\sqrt{19}\) is correct because \(4^2=16\) and \(5^2=25\), and since \(16<19<25\), we have \(4<\sqrt{19}<5\). Also 19 is not a perfect square, so \(\sqrt{19}\) is irrational. The closest distractor \(\frac{9}{2}=4.5\) lies between 4 and 5 but is rational. \(\sqrt{16}=4\) and \(\sqrt{25}=5\) are the end points, not inside the open interval (4,5). Exam tip: to check whether a square root is between two integers, compare the radicand with the squares of those integers and check if it's a perfect square.

Correct answer: A. Non terminating and non repeating

Explanation: The decimal never terminates and has no fixed repeating block: the number of zeros between successive 4s increases (1, 2, 3, ...), so there is no periodic pattern. Therefore it is non-terminating and non-repeating (an irrational-type decimal). Option C is incorrect because a repeating decimal must have a fixed repeating cycle, which this expansion lacks. Exam tip: a decimal is rational only if it terminates or eventually repeats; check for a constant periodic block to decide.

Correct answer: A. Rational number

Explanation: The decimal is repeating ('18' repeats), so it must be rational because repeating decimals can be written as fractions. For example, let \(x=2.\overline{18}\). Then \(100x=218.\overline{18}\), subtracting gives \(99x=216\), so \(x=\dfrac{216}{99}=\dfrac{24}{11}\). Thus the number is rational. Option B (irrational) is wrong because irrational numbers have non‑repeating, non‑terminating decimals; here the decimal repeats. Options C and D are also incorrect: the number is real (not non‑real) and it is not an integer. Exam tip: convert repeating decimals to fractions by multiplying by an appropriate power of 10 equal to the repeating block length and subtracting.

Correct answer: A. 13

Explanation: Since \((\sqrt{13})^2=13\), squaring the square root returns the original nonnegative number. Option B is just the original \(x\), not \(x^2\). Option C (169) results from incorrectly squaring 13 itself. Option D is wrong because \(x^2\) cannot be negative here. Exam tip: whenever \(x=\sqrt{a}\) for \(a\ge0\), you can directly replace \(x^2\) with \(a\).

Correct answer: A. (\sqrt{3})

Explanation: (\frac{3}{\sqrt{3}}=\sqrt{3}) because (3=\sqrt{3}\times\sqrt{3}). Practice simplifying denominators with roots.

Correct answer: A. (0) which is rational

Explanation: Multiplying any real number — whether rational or irrational — by zero always yields zero (0). Zero is rational because it can be written as \(0=0/1\). Hence the correct choice is (0), which is rational. The closest distractor is B (Always irrational), which is incorrect since zero is not irrational. Exam tip: Whenever you see "×0" in a question, the product is immediately 0, regardless of the other factor's type.

Correct answer: A. \(\sqrt{5}+(-\sqrt{5})\)

Explanation: Both \(\sqrt{5}\) and \(-\sqrt{5}\) are irrational, but their sum \(\sqrt{5}+(-\sqrt{5})=0\) is rational, so A is correct. Closest distractor B, \(\sqrt{2}+\sqrt{7}\), remains irrational (sums of distinct square‑free roots are generally irrational). C gives \(\sqrt{3}+\sqrt{3}=2\sqrt{3}\), still irrational. D is a sum of a rational and an irrational, which is irrational. Exam tip: look for additive inverses — if both terms cancel each other, the sum is rational (often 0).

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

Explanation: (\sqrt{2}\times\sqrt{5}=\sqrt{10}), which is irrational. Multiplying equal roots can often give a rational number.

Correct answer: A. 0.3

Explanation: Since \(0.09=\tfrac{9}{100}\), we have \(\sqrt{0.09}=\sqrt{\tfrac{9}{100}}=\tfrac{3}{10}=0.3\). The square root symbol denotes the principal (non‑negative) root, so \(-0.3\) is not the value of \(\sqrt{0.09}\). Exam tip: convert decimals to fractions (e.g. \(0.09=9/100\)) to make square roots easier to compute.