Irrational numbers and real numbers

Correct answer: A. (n) is a perfect square

Explanation: The square root of a positive integer is rational only when the number is a perfect square. This is a direct exam rule.

Correct answer: A. Rational number

Explanation: Let \(x = 2.\overline{45}\). The repeating block has length 2, so multiply by 100: \(100x = 245.\overline{45}\). Subtracting gives \(99x = 243\), hence \(x = \frac{243}{99} = \frac{27}{11}\). This is a fraction, so the number is rational. Option B (irrational) is incorrect because irrational numbers do not have terminating or repeating decimal expansions; here the decimal repeats. Option C (non-real) is wrong because the number is real, and option D (whole number) is wrong because there is a nonzero fractional part. Exam tip: An overline means a repeating block — use multiplication by \(10^n\) (where n is block length) to convert to a fraction quickly.

Correct answer: A. \(9\sqrt{2}\)

Explanation: \(\sqrt{162}=\sqrt{81\times2}=\sqrt{81}\cdot\sqrt{2}=9\sqrt{2}\). Therefore the correct simplified form is \(9\sqrt{2}\). Option B (\(6\sqrt{3}\)) is incorrect because \(6\sqrt{3}=\sqrt{36\times3}=\sqrt{108}\), which is not equal to \(\sqrt{162}\). Options C and D are also numerically different (\(18=\sqrt{324}\), and \(18\sqrt{2}\) is much larger). Exam tip: factor out the largest perfect square from under the root to simplify quickly.

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

Explanation: Simplify first: \(\sqrt{27}=\sqrt{9\cdot3}=3\sqrt{3}\). Thus \(4\sqrt{3}-\sqrt{27}=4\sqrt{3}-3\sqrt{3}=\sqrt{3}\), so option A is correct. Option B (1) is wrong because \(\sqrt{3}\approx1.732\), not 1. Option C (\(3\sqrt{3}\)) is incorrect because it treats the second term as if it should be added or left unsimplified; the correct simplification gives subtraction of like terms. Exam tip: always simplify radicals first and then combine only like surd terms (same radical part).

Correct answer: A. 0

Explanation: Simplify each radical: \(\sqrt{20}=2\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\). So \(\sqrt{5}+\sqrt{20}-\sqrt{45}=\sqrt{5}+2\sqrt{5}-3\sqrt{5}=(1+2-3)\sqrt{5}=0\). The closest distractor \(-\sqrt{5}\) would arise only from an arithmetic/sign error when combining coefficients; correct combination gives zero. Exam tip: factor out the common \(\sqrt{5}\) first to add coefficients quickly.

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

Explanation: The correct answer is \(\sqrt{2}\). Reason: \(\frac{2}{\sqrt{2}}=\frac{2\sqrt{2}}{\sqrt{2}\sqrt{2}}=\frac{2\sqrt{2}}{2}=\sqrt{2}\). Option C, \(\dfrac{\sqrt{2}}{2}\), equals \(\dfrac{1}{\sqrt{2}}\) and is the reciprocal form, so it is incorrect; options B (1) and D (2) are also not equal to the given expression. Exam tip: multiply numerator and denominator by the square root in the denominator to remove the root and simplify quickly.

Correct answer: A. Irrational number

Explanation: \(a-1=\sqrt{6}\). Since 6 is not a perfect square, \(\sqrt{6}\) cannot be written as a ratio \(p/q\) of integers, so it is irrational. Option B (rational) is incorrect because \(\sqrt{6}\) is not expressible as a fraction of integers; options C (integer) and D (zero) are incorrect because \(\sqrt{6}\) is neither an integer nor zero. Exam tip: simplify the expression first and check whether the square root is of a perfect square — if not, it is irrational.

Correct answer: A. (1) and (2)

Explanation: Since \(1^2=1<2<4=2^2\), it follows that \(1<\sqrt{2}<2\). Therefore \(\sqrt{2}\) lies between the integers 1 and 2. The other choices are incorrect because they give integers that are too small (0 and 1) or too large (2 and 3; 3 and 4). Exam tip: to locate \(\sqrt{x}\) between integers compare squares of consecutive integers or use a quick decimal estimate (\(\sqrt{2}\approx1.414\)).

Correct answer: A. (\sqrt{72}>\sqrt{50})

Explanation: For positive numbers a larger number inside the root gives a larger square root. Since (72>50), (\sqrt{72}>\sqrt{50}).

Correct answer: A. 0.9

Explanation: The square root symbol denotes the principal (non-negative) root. Since \(0.81=\frac{81}{100}\), \(\sqrt{0.81}=\sqrt{\frac{81}{100}}=\frac{9}{10}=0.9\). Option B (−0.9) is a common trap because \((-0.9)^2=0.81\) as well, but the principal square root is positive. Options C and D are incorrect: \(0.09^2=0.0081\), and 0.81 is the original number, not its square root. Exam tip: convert decimals to fractions to simplify square-root calculations (e.g., \(0.81=81/100\)).

Correct answer: A. 23

Explanation: These are conjugates, so use the difference of squares identity: \((a+b)(a-b)=a^2-b^2\). With \(a=5\) and \(b=\sqrt{2}\), \((5+\sqrt{2})(5-\sqrt{2})=5^2-(\sqrt{2})^2=25-2=23\). Option B (27) is incorrect — it confuses adding instead of subtracting the square; options C and D give irrational forms that are not the product. Exam tip: recognize conjugates and apply the difference of squares to compute such products quickly and reliably.

Correct answer: A. Rational number

Explanation: \(\sqrt[3]{64}=4\) because 64 is a perfect cube (\(4^3\)). The result 4 is an integer, and every integer is rational (can be written as a fraction, e.g. \(4=4/1\)). Option B is incorrect since irrational numbers have non-terminating, non-repeating decimals, which does not apply to 4. Option D is incorrect because a non-repeating decimal implies irrationality, whereas 4 is a terminating decimal. Option C is wrong because non-real numbers involve imaginary parts; \(\sqrt[3]{64}\) is a real number. Exam tip: when evaluating nth roots, first check whether the radicand is a perfect power—if it is, the root is an integer (hence rational).

Correct answer: A. It is irrational

Explanation: Prime factorization gives 16 = 2^4. A number is a perfect cube only if every prime exponent is a multiple of 3. Since 4 is not divisible by 3, \(\sqrt[3]{16}\) is not a perfect cube and therefore is irrational. Options B and C are incorrect because 4^3 = 64 and 8^3 = 512, not 16. Option D is wrong because an irrational number cannot be expressed as a terminating decimal. Exam tip: to test whether \(\sqrt[k]{N}\) is rational, factor N and check if every prime exponent is divisible by k.

Correct answer: A. (18)

Explanation: The key idea is that the square root of a positive integer is rational when the integer is a perfect square, and irrational when it is not a perfect square. Adding the rational number 2 to an irrational number keeps the result irrational. Therefore, we must find the option whose square root cannot be written as a fraction or whole number.

For 18,
\(\sqrt{18}=3\sqrt{2}\), which is irrational because 2 is not a perfect square. In contrast, \(\sqrt{16}=4\), \(\sqrt{25}=5\), and \(\sqrt{36}=6\) are rational. Hence \(2+\sqrt{18}\) is irrational, so option A is correct. The other choices produce rational sums.

Correct answer: A. (\sqrt{225}) is rational and (\sqrt{226}) is irrational

Explanation: The direct answer is A: \(\sqrt{225}\) is rational and \(\sqrt{226}\) is irrational. A perfect square has an integer square root, while a positive integer that is not a perfect square has an irrational square root. Since \(225=15^2\), we get \(\sqrt{225}=15\), an integer and hence a rational number. The consecutive squares around 226 are \(15^2=225\) and \(16^2=256\). Therefore 226 is not a perfect square, so \(\sqrt{226}\) is irrational. Option A is correct because it states exactly these results. Option B is wrong: the second root cannot be rational because 226 is not a perfect square. Option C is wrong: the first root is 15, not irrational. Option D assigns the two properties in reverse order. Do not decide only from the fact that both expressions contain a radical; a radical can have a rational value when its radicand is a perfect square. The useful exam method is to compare the radicand with nearby squares.

Correct answer: A. It is an irrational number

Explanation: The digit '2' appears at positions 1, 3, 6, 10, ... which are triangular numbers with general term \(T_n=\tfrac{n(n+1)}{2}\). The gaps between successive '2's increase, so there is no fixed repeating block (period). Any rational number's decimal expansion is either terminating or eventually periodic; this decimal is neither, therefore it is irrational. Exam tip: to test rationality, check whether the decimal eventually repeats or terminates — if not, it is irrational.

Correct answer: A. Rational number

Explanation: Direct answer: Option A, a rational number. A decimal that stops is called terminating; a decimal that continues in a repeating pattern is called recurring or repeating. Every terminating decimal can be converted into a fraction by using a denominator of 10, 100, 1000 and so on. Every repeating decimal also represents a fraction, although its conversion may require an algebraic step. Therefore both kinds are rational numbers. Option A is correct. Option B is wrong because irrational decimals are non-terminating and non-repeating. Option C is wrong because these decimals are real numbers; they are not non-real. Option D is too narrow: rational numbers include fractions, negative numbers and many non-natural numbers, not only natural numbers. For example, 0.5=1/2 is rational, although it is not a natural number. Memory cue: terminating or repeating decimal means rational; non-terminating and non-repeating means irrational.

Correct answer: A. -2

Explanation: Use the conjugate-product identity:

\((1+\sqrt{3})(1-\sqrt{3})=1^2-(\sqrt{3})^2=1-3=-2.\)

This follows from \((a+b)(a-b)=a^2-b^2\). The tempting wrong choice 4 comes from incorrectly adding 1 and 3 instead of subtracting. Exam tip: apply the conjugate formula and watch the signs carefully when multiplying surds.

Correct answer: A. (4\sqrt{3})

Explanation: (\sqrt{12}=2\sqrt{3}), (\sqrt{75}=5\sqrt{3}), and (\sqrt{27}=3\sqrt{3}). The result is (4\sqrt{3}).

Correct answer: A. Irrational number

Explanation: \(\frac{1}{2}\) is rational and \(\sqrt{2}\) is irrational. If their sum were rational, then \(\sqrt{2}=(\frac{1}{2}+\sqrt{2})-\frac{1}{2}\) would be a difference of rationals and thus rational, contradicting that \(\sqrt{2}\) is irrational. Hence \(\frac{1}{2}+\sqrt{2}\) is irrational. The closest distractor "rational" is incorrect for the reason above; "integer" and "terminating decimal" are specific types of rationals so they are also incorrect. Exam tip: memorize that rational + irrational = irrational (contradiction proof is quick and useful in exams).

Correct answer: A. 10\sqrt{3}

Explanation: \(\sqrt{300}=\sqrt{100\times3}=\sqrt{100}\cdot\sqrt{3}=10\sqrt{3}\). Hence option A is correct. The other choices are not equal to \(\sqrt{300}\): \(5\sqrt{6}\approx12.25\) (not 17.32), \(3\sqrt{100}=30\), and \(6\sqrt{50}=6\cdot5\sqrt{2}=30\sqrt{2}\). Exam tip: always extract the largest perfect square factor when simplifying square roots (here 100).

Correct answer: A. (2\sqrt{7})

Explanation: On subtracting, the (4) terms cancel and (\sqrt{7}-(-\sqrt{7})=2\sqrt{7}). Watch the signs carefully.

Correct answer: A. \(6\sqrt{5}\)

Explanation: \(\sqrt{180}=\sqrt{36\times5}=\sqrt{36}\,\sqrt{5}=6\sqrt{5}\), so the simplified form is \(6\sqrt{5}\). Why other options are wrong: \(9\sqrt{2}\) squared is \(81\times2=162\) (not 180), \(5\sqrt{6}\) squared is \(25\times6=150\), and \(3\sqrt{12}=3\sqrt{4\times3}=6\sqrt{3}\), which is different from \(6\sqrt{5}\). Exam tip: always factor out the largest perfect square from under the radical to simplify quickly.

Correct answer: A. Rational number

Explanation: With \(x=\sqrt{17}\), we have \(x^2=(\sqrt{17})^2=17\). Therefore \(x^2-5=17-5=12\). The number 12 is an integer and can be written as \(12/1\), so it is rational. Option B (irrational) is incorrect because irrational numbers cannot be expressed as a ratio of integers, whereas 12 can. Option C (non-real) is wrong because 12 is a real number. Option D (non-repeating decimal) is wrong because 12 is a terminating decimal (12.0), not a non-repeating non-terminating decimal. Exam tip: simplify powers of radicals first — squaring a square root removes the root, then evaluate the remaining arithmetic to classify the result.

Correct answer: A. Irrational number

Explanation: A rational number can be written as a fraction of integers, while an irrational number cannot be written in that form. The square root of a positive integer is irrational when the integer is not a perfect square. Since 13 is not a perfect square, \(\sqrt{13}\) is irrational. The number 4, however, is rational because it can be written as \(4/1\).

Adding a rational number to an irrational number always gives an irrational number. If the sum were rational, subtracting the rational number 4 would make \(\sqrt{13}\) rational, which is impossible. Therefore \(4+\sqrt{13}\) is irrational. It is consequently not an integer or a terminating decimal, so option A follows.