Irrational numbers

Correct answer: B. The statement is false because \(\sqrt{2}\times 2\sqrt{2}=4\), which is a rational number.

Explanation: Both \(\sqrt{2}\) and \(2\sqrt{2}\) are irrational, but \(\sqrt{2}\times2\sqrt{2}=4\), a rational number. Thus “always” in option A is false. Exam tip: one counterexample disproves a universal claim.

Correct answer: B. Its square is (6+\sqrt{11})

Explanation: Since (6+\sqrt{11}) is positive, its square root is real. Squaring it gives the inside number (6+\sqrt{11}).

Correct answer: D. (\frac{5(\sqrt{6}+1)}{5})

Explanation: Multiplying by the conjugate makes the denominator (6-1=5). So the form is (\frac{5(\sqrt{6}+1)}{5}), that is (\sqrt{6}+1).

Correct answer: A. (75)

Explanation: The direct answer is A: \(75\). First simplify the radicals: \(\sqrt{12}=\sqrt{4\times3}=2\sqrt3\), and \(\sqrt{27}=\sqrt{9\times3}=3\sqrt3\). Therefore, \(\sqrt{12}+\sqrt{27}=2\sqrt3+3\sqrt3=5\sqrt3\). Squaring gives \((5\sqrt3)^2=25\times3=75\). Option A is correct because it equals this result. Option B, 45, would come from an incorrect calculation and is not the square of \(5\sqrt3\). Option C, \(27+12\sqrt3\), is the expression obtained if the square is expanded incorrectly: the correct expansion is \(12+27+2\sqrt{12}\sqrt{27}=39+12\sqrt9=39+36=75\), not option C. Option D, 39, includes only the first two square terms and forgets the cross-product term. Remember: when squaring a sum, use \((a+b)^2=a^2+2ab+b^2\), not just \(a^2+b^2\).

Correct answer: D. (4.25025002500025\ldots)

Explanation: A decimal represents an irrational number only when it is non-terminating and non-repeating. A terminating decimal ends after a finite number of digits, while a repeating decimal has a fixed block of digits that continues again and again. Both types represent rational numbers because they can be written as fractions.

The first decimal repeats 25, the second terminates, and the third repeats 251. The fourth decimal continues forever but does not have one fixed repeating block: its zero groups and digits do not follow a constant cycle. Therefore it can represent an irrational number. Hence option D is correct. The important test is non-termination together with non-repetition.

Correct answer: B. The student is incorrect because \\(\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{9}=3\\), which is rational.

Explanation: The student’s rule is false: a quotient of two irrational numbers need not be irrational. Here \\(\sqrt{18}/\sqrt2=\sqrt9=3\\), so the result is rational. Exam tip: combine square roots first, then simplify.

Correct answer: A. (4)

Explanation: \(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\), so the bracket is \(2\sqrt{2}\). Multiplying by \(\sqrt{2}\) gives (4).

Correct answer: A. (\sqrt{7}-\sqrt{5})

Explanation: Multiplying by the conjugate makes the denominator (7-5=2). So (\frac{2(\sqrt{7}-\sqrt{5})}{2}=\sqrt{7}-\sqrt{5}).

Correct answer: A. (3+\sqrt{2})

Explanation: The expression contains the same square root multiplied by itself. For a nonnegative number, the principal square root satisfies \(\sqrt{x}\times\sqrt{x}=x\). Here the quantity inside the root is \(3+\sqrt{2}\), which is positive, so the rule applies directly. There is no need to expand the nested radical or approximate its value.

Applying the rule gives \(\sqrt{3+\sqrt{2}}\times\sqrt{3+\sqrt{2}}=3+\sqrt{2}\). Thus option A is correct. Option B incorrectly treats the expression as if the terms 3 and \(\sqrt{2}\) were separately squared and then added; that is not what the given product means. The result remains an exact expression, not a decimal approximation.

Correct answer: A. (11\sqrt{2})

Explanation: (\sqrt{98}=7\sqrt{2}), (\sqrt{72}=6\sqrt{2}), and (\sqrt{8}=2\sqrt{2}). Therefore the result is (11\sqrt{2}).

Correct answer: A. \(x+r\)

Explanation: If \(x+r\) were rational, subtracting the rational number \(r\) would make \(x\) rational, a contradiction. Hence A is correct. But \(x^2\) can be rational; for example, \((\sqrt{2})^2=2\). Exam tip: adding a rational number preserves irrationality.

Correct answer: A. (5)

Explanation: (\sqrt{27}=3\sqrt{3}) and (\sqrt{12}=2\sqrt{3}), so (s=5\sqrt{3}). Dividing by (\sqrt{3}) gives (5).

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

Explanation: Multiplying by the conjugate gives (\frac{(\sqrt{3}-1)^2}{2}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}). Rationalise the denominator.

Correct answer: A. \(\left(\sqrt{3},\sqrt{12}\right)\)

Explanation: \(\sqrt{3}\times\sqrt{12}=\sqrt{36}=6\), which is rational although both factors are irrational. Hence the claim is false. Exam tip: combine radicals first before deciding rationality.

Correct answer: D. (\sqrt{20}\div\sqrt{2})

Explanation: To simplify a quotient of square roots with positive radicands, combine them as \(\sqrt{a}\div\sqrt{b}=\sqrt{a/b}\). Then check whether the resulting number is a perfect square or has a remaining square-free factor. A result is rational when the radical simplifies completely to an integer or rational number. It is irrational when a non-square factor remains inside the square root.

The first three results are \(\sqrt{32/2}=\sqrt{16}=4\), \(\sqrt{18/2}=\sqrt{9}=3\), and \(\sqrt{45/5}=\sqrt{9}=3\). The fourth gives \(\sqrt{20/2}=\sqrt{10}\), and 10 is not a perfect square, so \(\sqrt{10}\) is irrational. Therefore option D is the result that is not rational.

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

Explanation: \(\sqrt{2}\) is irrational, but \((\sqrt{2})^2=2\), which is rational. Hence Riya’s statement is false. For instance, \((1+\sqrt{2})^2=3+2\sqrt{2}\) is still irrational, but one counterexample is enough to disprove a universal claim. In exams, test such claims with simple surds first.

Correct answer: A. (7)

Explanation: Area is ((3+\sqrt{2})(3-\sqrt{2})=9-2=7). Multiplying conjugate dimensions can give a rational area.

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

Explanation: (\sqrt{200}=10\sqrt{2}), (\sqrt{128}=8\sqrt{2}), and (\sqrt{18}=3\sqrt{2}). Therefore the result is (5\sqrt{2}).

Correct answer: A. 2√3

Explanation: The governing concept is rationalisation by using conjugate pairs. For the first fraction, multiply by √3 − √2. Its denominator becomes (√3 + √2)(√3 − √2) = 3 − 2 = 1, so the fraction equals √3 − √2. For the second fraction, multiply by √3 + √2; its denominator is again 3 − 2 = 1, so it equals √3 + √2. Adding gives (√3 − √2) + (√3 + √2) = 2√3 because the √2 terms cancel. Hence option A is correct. Option B retains the wrong radical, option C ignores the remaining terms, and option D incorrectly combines the radicands.

Correct answer: B. The statement is incorrect; \(\sqrt{45}+\sqrt{5}=4\sqrt{5}\), which is irrational.

Explanation: Since \(\sqrt{45}=3\sqrt{5}\), we get \(\sqrt{45}+\sqrt{5}=4\sqrt{5}\). As \(\sqrt{5}\) is irrational, its non-zero rational multiple is also irrational. Exam tip: never replace \(\sqrt{a}+\sqrt{b}\) with \(\sqrt{a+b}\).

Correct answer: A. \(x+1\)

Explanation: If \(x+1\) were rational, subtracting 1 would make \(x\) rational, which is a contradiction. Hence A is irrational. \(x-x=0\), \(x/x=1\), and \(x^2\) can be rational. Exam tip: check claims containing “always”.

Correct answer: B. The claim is incorrect; rationalising the denominator gives \(\frac{1}{\sqrt{5}+2}=\sqrt{5}-2\), which is irrational.

Explanation: Multiplying by \(\sqrt{5}-2\) makes the denominator \(5-4=1\), so the value is \(\sqrt{5}-2\). The difference of an irrational and a rational number is irrational. Exam tip: rationalise the denominator first.

Correct answer: A. (\sqrt{6.5})

Explanation: Since (6<6.5<7), (\sqrt{6.5}) lies between them. For positive square roots compare the numbers inside.

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

Explanation: If \(\frac{p+\sqrt{2}}{q}=r\) is rational, then \(\sqrt{2}=qr-p\). Since p, q and r are rational, the right side is rational, contradicting the irrationality of \(\sqrt{2}\). Exam tip: use closure of rational numbers under multiplication and subtraction.

Correct answer: A. (1)

Explanation: (\sqrt{45}=3\sqrt{5}) and (\sqrt{20}=2\sqrt{5}), so the numerator is (\sqrt{5}). Dividing gives (1).