Irrational numbers

Correct answer: C. \(\sqrt{14}\)

Explanation: \(\sqrt{14}\) is irrational because 14 is not a perfect square. The square root of an integer is rational only when that integer is a perfect square. \(\frac{17}{19}\) is rational, \(0.090909\ldots\) is a recurring decimal and hence rational, and \(-8\) is an integer, so it is rational. Exam tip: Recurring decimals are always rational, whereas the square root of a non-perfect square is irrational.

Correct answer: B. Irrational

Explanation: Here \(x=\sqrt{7}\), so \(\frac{1}{x}=\frac{1}{\sqrt{7}}=\frac{\sqrt{7}}{7}\). Hence, \(x+\frac{1}{x}=\sqrt{7}+\frac{\sqrt{7}}{7}=\frac{8\sqrt{7}}{7}\). The product of a non-zero rational number, \(\frac{8}{7}\), and the irrational number \(\sqrt{7}\) is irrational. Therefore, the expression is not an integer or a natural number. Exam tip: the square root of a non-perfect square is irrational.

Correct answer: C. 0.101001000100001\ldots

Explanation: In option C, the number of zeros between successive 1s keeps increasing, so no fixed block of digits repeats. Its decimal expansion is non-terminating and non-repeating; therefore, it is irrational. Options A and B are non-terminating recurring decimals, so they are rational, while option D is a terminating decimal and is also rational. Exam tip: Every terminating or recurring decimal represents a rational number.

Correct answer: C. Irrational number

Explanation: 13 is not a perfect square. The square root of a natural number that is not a perfect square is irrational. Therefore, \(x=\sqrt{13}\) is an irrational number. In contrast, the square root of a perfect square, such as \(\sqrt{16}=4\), is an integer. Exam tip: first check whether the number under the square root is a perfect square.

Correct answer: B. Irrational number

Explanation: \(\sqrt{7}\) is irrational, while \(3\) is a non-zero rational number. The product of a non-zero rational number and an irrational number is irrational. Therefore, \(3\sqrt{7}\) is irrational. It cannot be an integer or a natural number. Exam tip: Multiplying an irrational number by zero is the exceptional case, as the result is the rational number 0.

Correct answer: B. \(\sqrt{2}+1\)

Explanation: \(\sqrt{2}\) is irrational, while \(1\) is rational. The sum of an irrational number and a rational number is always irrational, so \(\sqrt{2}+1\) is irrational. The other sums are \(5\), \(0\), and \(1\), all of which are rational. Exam tip: identify whether each term is rational or irrational first; irrational + rational is always irrational.

Correct answer: C. \(\sqrt{10}\)

Explanation: \(\sqrt{10}\) is irrational because 10 is not a perfect square. Therefore, its decimal expansion is non-terminating and non-recurring. \(\frac{3}{4}=0.75\) and \(\frac{7}{8}=0.875\) terminate, whereas \(\frac{2}{11}=0.1818\ldots\) is recurring. Exam tip: The square root of a natural number that is not a perfect square is irrational.

Correct answer: C. 2

Explanation: The intended irrational endpoints are \(\sqrt{2}\) and \(\sqrt{5}\). Since \(\sqrt{2}\approx1.414\) and \(\sqrt{5}\approx2.236\), we have \(1.414<2<2.236\). Therefore, 2 is the correct option. \(\sqrt{2}\) is an endpoint, not a number strictly between them, while 1 and 3 lie outside this interval. Exam tip: Use approximate decimal values to compare square roots quickly.

Correct answer: A. Irrational

Explanation: If p+q were rational, then subtracting the rational number p from it would make q rational. This contradicts the fact that q is irrational. Therefore, p+q is irrational. Integers and natural numbers are types of rational numbers, so they cannot be correct here. Exam tip: Adding or subtracting a rational number does not change an irrational number into a rational number.

Correct answer: B. 4

Explanation: Since \(16<17<25\), we get \(4<\sqrt{17}<5\). Also, \(\sqrt{17}\approx 4.12\), which is about 0.12 away from 4 but about 0.88 away from 5. Therefore, 4 is the closest number. Although 5 is the next integer, it is farther away. Exam tip: estimate a square root by locating the nearest perfect squares on either side.

Correct answer: B. \(\sqrt{11}-1\)

Explanation: \(\sqrt{11}\) is irrational, while \(1\) is rational. The difference between an irrational number and a rational number remains irrational, so \(\sqrt{11}-1\) is irrational. In contrast, \(7-2=5\), \(5-5=0\), and \(\frac{9}{2}-\frac{1}{2}=4\) are all rational. Exam tip: the square root of a number that is not a perfect square is generally irrational.

Correct answer: C. \(9\sqrt{3}\)

Explanation: Since \(48=16\times3\) and \(75=25\times3\), \(\sqrt{48}=4\sqrt{3}\) and \(\sqrt{75}=5\sqrt{3}\). Therefore, \(\sqrt{48}+\sqrt{75}=4\sqrt{3}+5\sqrt{3}=9\sqrt{3}\). A close option such as \(8\sqrt{3}\) may result from simplifying one radical incorrectly. Exam tip: simplify each surd first, then add like surds.

Correct answer: B. 7

Explanation: Since \(49<50<64\), we get \(\sqrt{49}<\sqrt{50}<\sqrt{64}\), so \(7<\sqrt{50}<8\). Also, \(\sqrt{50}\approx 7.07\), which is much closer to 7 than to 8. Therefore, 7 is correct. Exam tip: Compare with nearby perfect squares to estimate a square root quickly.

Correct answer: C. Rational or irrational

Explanation: If p=0, then pq=0, which is rational. However, if p is a non-zero rational number, for example p=2 and q=\(\sqrt{2}\), then pq=\(2\sqrt{2}\), which is irrational. Hence, the product can be either rational or irrational. “Always irrational” is true only when p is non-zero. Exam tip: Always check the special case p=0 in such questions.

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

Explanation: Since \(98=49\times2\), \(a=\sqrt{98}=7\sqrt{2}\); and since \(50=25\times2\), \(b=\sqrt{50}=5\sqrt{2}\). Therefore, \(a-b=7\sqrt{2}-5\sqrt{2}=2\sqrt{2}\). \(\sqrt{2}\) would result from incorrectly subtracting the coefficients. Exam tip: simplify surds to like radical terms before subtracting them.

Correct answer: A. (9+4\sqrt{5})

Explanation: The direct answer is A: \(9+4\sqrt5\). Multiply numerator and denominator by the denominator’s conjugate, \(\sqrt5+2\): \(\frac{\sqrt5+2}{\sqrt5-2}\times\frac{\sqrt5+2}{\sqrt5+2}=\frac{(\sqrt5+2)^2}{(\sqrt5)^2-2^2}=\frac{(\sqrt5+2)^2}{1}\). Now expand: \((\sqrt5+2)^2=5+4\sqrt5+4=9+4\sqrt5\). Thus A is correct. Option B, \(9-4\sqrt5\), has the wrong sign for the cross term. Option C, \(5+2\sqrt5\), is not the result of the square and misses terms. Option D, \(1+\sqrt5\), does not arise from rationalising or expanding. The key identity is \((a+b)(a-b)=a^2-b^2\), which makes the denominator 1 here. Check the sign carefully: both numerator and the conjugate used for multiplication contain plus 2.

Correct answer: A. \(0.\overline{27}\)

Explanation: \(0.\overline{27}=\frac{27}{99}=\frac{3}{11}\), so it is rational despite being non-terminating. \(\sqrt7\) and \(\pi\) are irrational. Exam tip: every repeating decimal is rational.

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

Explanation: As \(200=100\times2\), \(72=36\times2\), and \(18=9\times2\), we have \(\sqrt{200}=10\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\). Therefore, \(10\sqrt{2}-6\sqrt{2}+3\sqrt{2}=7\sqrt{2}\), so option C is correct. \(9\sqrt{2}\) results from incorrectly changing the subtraction sign into addition. Exam tip: extract perfect-square factors from each radical before combining like radicals.

Correct answer: C. \(qr\) is irrational.

Explanation: Option C is correct. If \(qr\) were rational, then since \(q\ne0\), \(r=\frac{qr}{q}\) would also be rational, a contradiction. Exam tip: test “always” claims using examples such as \(\sqrt2\) and \(-\sqrt2\).

Correct answer: C. 0.101001000100001…

Explanation: In 0.101001000100001…, the number of zeros between successive 1s keeps increasing, so the decimal neither terminates nor repeats. Hence it is irrational. 0.121212… repeats and is rational. Exam tip: every terminating or recurring decimal is rational.

Correct answer: B. \(\sqrt{5}+(-\sqrt{5})=0\)

Explanation: Both \(\sqrt{5}\) and \(-\sqrt{5}\) are irrational, but their sum is 0, which is rational. Therefore, the word “always” makes the claim false. Exam tip: one counterexample disproves a universal statement.

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

Explanation: To rationalise the denominator, multiply the fraction by the conjugate of the denominator. Since \(\sqrt{4}=2\), the expression is \(\frac{3}{\sqrt{7}+2}\). The conjugate of \(\sqrt{7}+2\) is \(\sqrt{7}-2\). Multiplying numerator and denominator by this conjugate gives \(\frac{3(\sqrt{7}-2)}{(\sqrt{7}+2)(\sqrt{7}-2)}\).

The denominator is a difference of squares: \((\sqrt{7})^2-2^2=7-4=3\). Thus the fraction becomes \(\frac{3(\sqrt{7}-2)}{3}=\sqrt{7}-2\). The denominator is now rational, so this is the rationalised form. Hence option B is correct. Option A is an unsimplified expression equal to the same value, but the listed rationalised answer is the simplified form in option B.

Correct answer: D. (6+\sqrt{7})

Explanation: The direct answer is D: \(6+\sqrt7\). Put \(x=\sqrt{6+\sqrt7}\). The given product is \(x\times x=x^2\). Since the expression inside the root is positive, the square of its principal square root is exactly the inside quantity. Therefore the answer is \(6+\sqrt7\), so option D is correct. Option A, \(36+7=43\), incorrectly squares the two inside terms separately; the original expression is not \((6+\sqrt7)^2\). Option B, \(\sqrt{13}\), has no valid square-root multiplication rule leading to it. Option C, \(6-\sqrt7\), changes the plus sign without justification and is a different number. The correct rule is \(\sqrt a\times\sqrt a=a\) for \(a\ge0\). Do not expand the expression inside the radical when the same radical is simply multiplied by itself.

Correct answer: A. It is irrational

Explanation: If \(rx\) were rational, then \(x=(rx)/r\) would be a quotient of two rational numbers and hence rational, a contradiction. Therefore, \(rx\) is irrational. Exam tip: the condition \(r\neq0\) is essential.

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

Explanation: The blocks \(1,01,001,0001,\ldots\) keep changing, so no fixed repeating block occurs. A non-terminating, non-repeating decimal is irrational. Exam tip: a repeating decimal always represents a rational number.