Irrational numbers and real numbers

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.

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).

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.

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.

\(\sqrt{121}=11\). 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.

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.

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

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

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 \(q\neq0\).

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

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

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.

\(\sqrt{32}=\sqrt{16\times2}=\sqrt{16}\cdot\sqrt{2}=4\sqrt{2}\). 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.

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.

\(\sqrt{48}=\sqrt{16\times3}=4\sqrt{3}\) 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.

\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.

\(\sqrt{19}\) 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.

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.

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.

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 \(a\).

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

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.

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).

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

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.