Irrational numbers

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

Explanation: The square of a square root gives the number inside. So \(\left(\sqrt{2+\sqrt{3}}\right)^2=2+\sqrt{3}\).

Correct answer: D. \(0.101001000100001\ldots\)

Explanation: In \(0.101001000100001\ldots\), the number of zeros between successive 1s keeps increasing, so no repeating block is formed. Its decimal expansion is non-terminating and non-repeating; hence it is irrational. \(0.\overline{27}\) is rational because it repeats. Exam tip: non-terminating, non-repeating decimals are irrational.

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

Explanation: If \(x+r\) were rational, subtracting the rational number \(r\) would make \(x\) rational, a contradiction. Hence \(x+r\) is irrational. Exam tip: for \(r=0\), \(xr=0\), so a product is not always irrational.

Correct answer: A. \(\sqrt{2}\) and \(\sqrt{8}\)

Explanation: \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational, so A disproves the claim. In B, the product is \(\sqrt{6}\), still irrational. Exam tip: combine radicals before deciding the type of number.

Correct answer: A. The sum of an irrational number and a rational number is irrational.

Explanation: If the sum of an irrational and a rational number were rational, subtracting the same rational number would make the irrational number rational, which is impossible. Exam tip: test “always” claims with counterexamples.

Correct answer: A. ( \frac{11+6\sqrt{2}}{7})

Explanation: Multiplying by the conjugate gives denominator (9-2=7) and numerator ((3+\sqrt{2})^2=11+6\sqrt{2}). So the correct form is (\frac{11+6\sqrt{2}}{7}).

Correct answer: A. The claim is false because \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational.

Explanation: The product of irrational numbers need not be irrational. Here, \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), a rational number. Exam tip: combine square roots first before deciding the type of number.

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

Explanation: (\sqrt{108}=6\sqrt{3}), (\sqrt{48}=4\sqrt{3}), and (\sqrt{12}=2\sqrt{3}). Therefore the result is (4\sqrt{3}).

Correct answer: A. (\frac{\sqrt{7}-2}{3})

Explanation: The direct answer is A: \(\frac{\sqrt7-2}{3}\). To remove the radical from the denominator, multiply numerator and denominator by the conjugate of \(\sqrt7+2\), namely \(\sqrt7-2\). Thus \(p=\frac1{\sqrt7+2}\times\frac{\sqrt7-2}{\sqrt7-2}=\frac{\sqrt7-2}{(\sqrt7+2)(\sqrt7-2)}\). Using \((a+b)(a-b)=a^2-b^2\), the denominator is \(7-4=3\). Hence option A is correct. Option B, \(\sqrt7-2\), misses the denominator 3. Option C keeps the original sign and does not rationalise the denominator; it is not equal to the original fraction. Option D, \(2-\sqrt7\), is the negative of the numerator and therefore has the wrong sign. The conjugate changes the middle signs, and multiplying conjugates produces a difference of squares.

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

Explanation: The direct answer is A: \(5+\sqrt6\). Let \(x=\sqrt{5+\sqrt6}\). The expression is \(x\times x=x^2\). By the definition of a square root, the square of the principal square root of a non-negative number is the number inside it. Since \(5+\sqrt6>0\), the value is exactly \(5+\sqrt6\). Option A is correct. Option B, \(25+6=31\), incorrectly squares the two parts separately; the expression is not \((5+\sqrt6)^2\). Option C, \(\sqrt{11}\), has no valid rule behind it and does not represent the number inside the radical. Option D, \(5-\sqrt6\), changes the plus sign without reason. A common mistake is to think that multiplying two identical radicals requires expanding the inside. It does not: \(\sqrt a\times\sqrt a=a\) for \(a\ge0\).

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

Explanation: The number of zeros between successive 1s is 1, 2, 3, 4, ..., so no fixed repeating block occurs. A rational number has a terminating or recurring decimal expansion. Exam tip: check repetition, not merely the digits used.

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. \(x+r\)

Explanation: If \(x+r\) were rational, subtracting the rational number \(r\) would make \(x\) rational, a contradiction. But \(x^2\) need not be irrational; \((\sqrt{2})^2=2\). Exam tip: adding a rational number preserves irrationality.

Correct answer: B. Its square is (8+\sqrt{15})

Explanation: Since (8+\sqrt{15}) is positive, its square root is real. Squaring it gives the inside number (8+\sqrt{15}).

Correct answer: A. (\sqrt{10}+\sqrt{6})

Explanation: Multiplying by the conjugate makes the denominator (10-6=4). So (\frac{4(\sqrt{10}+\sqrt{6})}{4}=\sqrt{10}+\sqrt{6}).

Correct answer: A. (125)

Explanation: Option A is correct: the value is 125. First simplify the radicals: \(\sqrt{20}=\sqrt{4\cdot5}=2\sqrt5\) and \(\sqrt{45}=\sqrt{9\cdot5}=3\sqrt5\). Their sum is \(2\sqrt5+3\sqrt5=5\sqrt5\). Now square it: \((5\sqrt5)^2=25\times5=125\). Option A matches. Option B, 65, would result from an incorrect expansion or from mishandling the cross term. Option C, \(25\sqrt5\), stops after squaring the coefficient and forgets that \((\sqrt5)^2=5\); it is not the value of the full square. Option D, 100, may come from treating \(\sqrt{20}+\sqrt{45}\) incorrectly or ignoring the radical contribution. A safe method is to simplify each radical before adding like surds, then square. The memory cue is: extract perfect-square factors first; after combining, square both the number and the radical.

Correct answer: A. \(\frac{\sqrt{18}}{\sqrt{2}}=3\)

Explanation: Both \(\sqrt{18}\) and \(\sqrt{2}\) are irrational, but \(\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{9}=3\), which is rational. Thus the claim is false. Exam tip: one counterexample disproves an “always” statement.

Correct answer: A. (6)

Explanation: \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the bracket is \(2\sqrt{3}\). Multiplying by \(\sqrt{3}\) gives (6).

Correct answer: A. (\sqrt{8}-\sqrt{5})

Explanation: Multiplying by the conjugate makes the denominator (8-5=3). So (\frac{3(\sqrt{8}-\sqrt{5})}{3}=\sqrt{8}-\sqrt{5}).

Correct answer: D. Irrational number

Explanation: The sum of a rational number and an irrational number is always irrational. If their sum were rational, subtracting the rational number would make the irrational number rational, which is impossible. Exam tip: remember this rule for both addition and subtraction.

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

Explanation: (\sqrt{162}=9\sqrt{2}), (\sqrt{98}=7\sqrt{2}), and (\sqrt{50}=5\sqrt{2}). Therefore the result is (11\sqrt{2}).

Correct answer: A. (9)

Explanation: (\sqrt{48}=4\sqrt{3}) and (\sqrt{75}=5\sqrt{3}), so (s=9\sqrt{3}). Dividing by (\sqrt{3}) gives (9).

Correct answer: A. (\frac{7-2\sqrt{10}}{3})

Explanation: Multiplying by the conjugate gives (\frac{(\sqrt{5}-\sqrt{2})^2}{5-2}). So the answer is (\frac{7-2\sqrt{10}}{3}).

Correct answer: C. \(n\) is not a perfect square

Explanation: The square root of a positive integer is rational only when that integer is a perfect square. Hence, if \(n\) is not a perfect square, \(\sqrt{n}\) is irrational. For example, \(12\) is not a perfect square, so \(\sqrt{12}\) is irrational. Exam tip: being even or odd alone is not enough.

Correct answer: A. \(\sqrt{2}\times\sqrt{8}\)

Explanation: \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational. Since both factors are irrational, this is a counterexample to Reema’s “always” claim. Exam tip: combine square roots first before deciding the type of number.