Irrational numbers

Correct answer: B. Rational

Explanation: The square root of a perfect square is an integer, so it is rational. First check whether the number is a perfect square.

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

Explanation: \(\sqrt{3}\) is irrational, and adding an irrational number to a rational number gives an irrational result. If \(4+\sqrt{3}=r\) were rational, then \(\sqrt{3}=r-4\) would be rational, a contradiction. Exam tip: use this subtraction argument to test such claims.

Correct answer: C. It is irrational

Explanation: (\sqrt{5}) and (\sqrt{7}) are different irrational radicals and their sum is irrational. Different radicals are not added directly as (\sqrt{12}).

Correct answer: B. 4

Explanation: For dividing square roots, use \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\). Thus, \(\frac{\sqrt{112}}{\sqrt{7}}=\sqrt{\frac{112}{7}}=\sqrt{16}=4\). Therefore, option B is correct. Option A, 8, is incorrect because \(\sqrt{16}=4\), not 8. Exam tip: When dividing square roots, first simplify the quotient inside the radical.

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

Explanation: The direct answer is B: \(3\sqrt{2}\). These are like radicals because both terms contain the same radical \(\sqrt{2}\). Treat the radical as a common factor: \(8\sqrt{2} - 5\sqrt{2} = (8-5)\sqrt{2} = 3\sqrt{2}\). We subtract only the coefficients, 8 and 5; we do not subtract the numbers inside the square roots. Option A, \(13\sqrt{2}\), is wrong because 13 would result from addition, not from subtracting 5 from 8. Option B is correct because it follows the rule for subtracting like terms. Option C, 3, is wrong because the common factor \(\sqrt{2}\) cannot simply disappear. Option D, \(\sqrt{6}\), is wrong because subtraction of like radicals does not mean multiplying the radicands. The result is still irrational because \(3\sqrt{2}\) is a non-zero rational multiple of the irrational number \(\sqrt{2}\). Memory cue: same radical means subtract coefficients; never subtract or multiply the radicands in this situation.

Correct answer: A. (10)

Explanation: (\sqrt{8}=2\sqrt{2}) and (\sqrt{18}=3\sqrt{2}), so the bracket is (5\sqrt{2}) and the product is (10). Simplify the bracket first.

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

Explanation: Both \(\sqrt{5}\) and \(-\sqrt{5}\) are irrational, but their sum is \(0\), which is rational. Hence the word “always” makes the statement false. Exam tip: disprove such claims using one counterexample.

Correct answer: A. 6 + 5√3

Explanation: Use the distributive property to multiply √3 by each term inside the bracket: √3(2√3 + 5) = 2√3·√3 + 5√3. Since √3·√3 = 3, the first product becomes 2 × 3 = 6. The second product remains 5√3, so the simplified expression is 6 + 5√3, making option A correct. Option B incorrectly treats √3·√3 as 1 or fails to multiply the first coefficient correctly. Option C incorrectly removes the radical from the second term, and option D combines unlike terms or applies multiplication incorrectly. The rational term 6 and the irrational term 5√3 cannot be added further because they are unlike terms.

Correct answer: B. (\sqrt{3}-1)

Explanation: The direct answer is B: \(\sqrt{3}-1\). To rationalise \(\frac{2}{\sqrt{3}+1}\), multiply numerator and denominator by the conjugate of the denominator, \(\sqrt{3}-1\). Thus \(\frac{2}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} = \frac{2(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\). The denominator is a difference of squares: \(3-1=2\). Therefore the expression becomes \(\frac{2(\sqrt{3}-1)}{2}=\sqrt{3}-1\). Option A is wrong because it is the original denominator expression and does not rationalise the denominator. Option B is correct because its denominator is 1, a rational number, and it is the simplified result. Option C, \(2\sqrt{3}-2\), is wrong because it is twice the correct result. Option D is wrong because it leaves a radical in the denominator and is not the rationalised form. Memory cue: for \(a+\sqrt b\), multiply by the conjugate \(a-\sqrt b\).

Correct answer: A. 16

Explanation: Adding the expressions gives \(8+\sqrt{17}+8-\sqrt{17}=16\). The terms \(\sqrt{17}\) and \(-\sqrt{17}\) cancel each other, so only the rational parts remain. Option B represents the sum of the two square-root terms, not the sum of the complete expressions. Exam tip: the sum of conjugate expressions \((a+b)\) and \((a-b)\) is directly \(2a\).

Correct answer: A. It is irrational because the number of zeros between successive 1s keeps increasing, so there is no fixed repeating block.

Explanation: The blocks of zeros after 1 have lengths 1, 2, 3, 4, ...; hence no fixed digit block repeats. The decimal is non-terminating and non-repeating, so it is irrational. Exam tip: check repetition, not merely whether a decimal continues forever.

Correct answer: C. Irrational

Explanation: Subtracting irrational (\sqrt{2}) from rational (5) gives an irrational number. The irrational part remains.

Correct answer: B. Legs (1) and (2)

Explanation: The direct answer is B, a right triangle with legs 1 and 2. The Pythagorean theorem says that for a right triangle, the square of the hypotenuse equals the sum of the squares of the legs: \(c^2=a^2+b^2\). With legs 1 and 2, \(c=\sqrt{1^2+2^2}=\sqrt{1+4}=\sqrt5\). This is exactly the length required for constructing \(\sqrt5\) on the number line. Option A gives \(\sqrt{1^2+1^2}=\sqrt2\), not \(\sqrt5\). Option B gives \(\sqrt5\), so it is correct. Option C gives \(\sqrt{2^2+2^2}=\sqrt8=2\sqrt2\), not \(\sqrt5\). Option D gives \(\sqrt{3^2+1^2}=\sqrt{10}\), also not \(\sqrt5\). After drawing this triangle, the hypotenuse can be transferred to the number line with a compass. Memory cue: for \(\sqrt5\), think of the Pythagorean pair 1 and 2.

Correct answer: B. 3

Explanation: Using the division rule for square roots, \(\frac{\sqrt{54}}{\sqrt{6}}=\sqrt{\frac{54}{6}}=\sqrt{9}=3\). Therefore, the correct answer is 3. Option A gives only 1, while options C and D are other numbers rather than the evaluated quotient. Exam tip: For square roots with the same index, divide the radicands first and then simplify the root.

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

Explanation: The area of a square is \((\text{side})^2\), so its side is \(\sqrt{50}\). Since \(50=25\times2\), we get \(\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}\). Therefore, option C is correct. Exam tip: take the greatest perfect-square factor outside the square root; \(2\sqrt{5}\) is incorrect because its square is 20, not 50.

Correct answer: A. (\frac{4+\sqrt{7}}{9})

Explanation: To rationalise the denominator, multiply the fraction by the conjugate of 4-\sqrt{7}, which is 4+\sqrt{7}. This operation does not change the value because the same nonzero expression is used in the numerator and denominator. The denominator becomes (4-\sqrt{7})(4+\sqrt{7}).

Using the difference of squares, this product is 4^2-(\sqrt{7})^2=16-7=9. The numerator becomes 4+\sqrt{7}, so the fraction is \frac{4+\sqrt{7}}{9}. The denominator is now rational, which is the required rationalised form. Hence option A is correct; option D is only the conjugate expression and does not rationalise the original denominator.

Correct answer: A. यह अपरिमेय है, क्योंकि इसका दशमलव प्रसार न समाप्त होता है और न आवर्ती होता है।

Explanation: The number of zeros between successive 1s is 1, 2, 3, 4, ..., so no fixed repeating block can occur. A non-terminating, non-repeating decimal is irrational. Exam tip: a repeating decimal must have a fixed period.

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

Explanation: (\sqrt{28}=2\sqrt{7}), so (3\sqrt{7}+4\sqrt{7}=7\sqrt{7}). Watch both coefficients and radicals carefully.

Correct answer: C. (\sqrt{37})

Explanation: (\sqrt{37}) is irrational because (37) is not a perfect square. An irrational number has a non-terminating non-repeating decimal.

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

Explanation: The zeros between 1s occur in groups of 1, 2, 3, 4, ... so no fixed repeating block can occur. It is a non-terminating, non-repeating decimal and hence irrational. Exam tip: an infinite decimal is rational only if it eventually repeats.

Correct answer: A. \(0.333\ldots\)

Explanation: \(0.333\ldots\) is non-terminating, but the digit 3 repeats and the number equals \(\frac{1}{3}\); hence it is rational. In exams, check whether a non-terminating decimal has a repeating pattern.

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

Explanation: Both \(\sqrt{5}\) and \(-\sqrt{5}\) are irrational, but their sum is \(0\), which is rational. Hence, “always” is false. Exam tip: test universal claims by finding one counterexample.

Correct answer: A. Non-terminating and non-repeating

Explanation: An irrational number has a decimal expansion that neither ends nor repeats a fixed block of digits, so A is correct. A non-terminating recurring decimal is rational. Exam tip: look for the word “non-repeating”.

Correct answer: B. (8)

Explanation: (\sqrt{150}=5\sqrt{6}) and (\sqrt{54}=3\sqrt{6}), so division gives (8). First convert the numerator into like radicals.

Correct answer: B. It is irrational because the number of zeros between successive 1s keeps increasing, so no repeating block is formed.

Explanation: The groups of zeros between 1s have lengths 1, 2, 3, 4, \(\ldots\), so there is no fixed repeating block. Its decimal expansion is non-terminating and non-repeating; hence it is irrational. Exam tip: identify such decimals as irrational.