Irrational numbers

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

\(\sqrt{3}\) 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.

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

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.

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.

(\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.

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.

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.

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 \(a-\sqrt b\).

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 \(2a\).

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.

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

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.

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.

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.

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.

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.

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

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

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.

\(0.333\ldots\) 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.

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.

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”.

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

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.