Irrational numbers

If \(5-\sqrt{7}\) were rational, then \(5-(5-\sqrt{7})=\sqrt{7}\) would also be rational, which is impossible. Hence the difference is irrational. Exam tip: rational ± irrational is irrational.

(\sqrt{588}=14\sqrt{3}), (\sqrt{300}=10\sqrt{3}), and (\sqrt{192}=8\sqrt{3}). Therefore the result is (12\sqrt{3}).

Adding the two terms gives numerator (2\sqrt{10}) and denominator (10-6=4). So the value is (\frac{\sqrt{10}}{2}).

The correct expression is \(a+b\). If \(a+b\) were rational, then \(b=(a+b)-a\) would also be rational, a contradiction. However, \(ab=0\) when \(a=0\). Exam tip: test zero when checking an “always” claim.

(\sqrt{392}=14\sqrt{2}) and (\sqrt{200}=10\sqrt{2}), so the numerator is (24\sqrt{2}). Dividing gives (24).

(r^2-s^2=(r-s)(r+s)), where (r-s=2\sqrt{11}) and (r+s=2\sqrt{17}). So the value is (4\sqrt{187}).

In option A, \(\sqrt{11}\) is irrational, but \((\sqrt{11})^2=11\) is rational, so the claim is false. In option B, \(\sqrt[3]{4}\) remains irrational. Exam tip: one counterexample is enough to disprove an “every” statement.

(\sqrt{500}=10\sqrt{5}), (\sqrt{180}=6\sqrt{5}), and (\sqrt{320}=8\sqrt{5}). Therefore (P=8\sqrt{5}).

(\sqrt{125}=5\sqrt{5}) and (\sqrt{320}=8\sqrt{5}), so the sum is (13\sqrt{5}). Its square is (845).

Option A is correct. The blocks of zeros between 1s keep increasing, so no fixed block repeats periodically. A rational number has a terminating or recurring decimal expansion. Exam tip: check periodic repetition, not merely the digits used.

Both \(\sqrt{2}\) and \(\sqrt{8}\) are irrational, but \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational. Hence the word “always” makes the statement false. Exam tip: disprove universal claims using one counterexample.

Multiplying by the conjugate gives numerator (50+2\sqrt{589}) and denominator (12). The simplified form is (\frac{25+\sqrt{589}}{6}).

Both \(\sqrt{2}\) and \(\sqrt{8}\) are irrational, but \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational. In the other options, the product under the radical is not a perfect square. Exam tip: combine radicals before classifying the result.

(\sqrt{726}=11\sqrt{6}), (\sqrt{486}=9\sqrt{6}), and (\sqrt{150}=5\sqrt{6}). Therefore the result is (7\sqrt{6}).

To rationalise a denominator containing a difference of square roots, multiply the fraction by the conjugate of that denominator. The conjugate of \(\sqrt{43}-\sqrt{31}\) is \(\sqrt{43}+\sqrt{31}\). Thus the denominator becomes \((\sqrt{43})^2-(\sqrt{31})^2=43-31=12\), which is rational.

The numerator then becomes \(12(\sqrt{43}+\sqrt{31})\). Cancelling the common factor 12 with the denominator leaves \(\sqrt{43}+\sqrt{31}\). Hence option C is correct. Option A reverses the factor instead of multiplying by it, while option D uses an incorrect product of the radicands. The supplied answer and explanation are correct.

If rx were rational, dividing it by the non-zero rational number r would make x rational, a contradiction. Hence rx is irrational. Option A fails because √2 + (−√2) = 0. Exam tip: test “always” statements using a counterexample.

Area is ((\sqrt{41}+\sqrt{17})(\sqrt{41}-\sqrt{17})=41-17=24). Conjugate dimensions give rational area.

(\frac{\sqrt{75}}{\sqrt{3}}=\sqrt{25}=5), which is rational. Check whether division forms a perfect square.

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but they cancel to give \(0\), which is rational. Hence “always” is false. Exam tip: one counterexample is enough to disprove a universal statement.

Since \(3\) is not a perfect square, \(\sqrt{3}\) is irrational. The decimal expansion of an irrational number is non-terminating and non-recurring, so option C is correct. Option B is wrong because a non-terminating recurring decimal represents a rational number, such as \(0.333\ldots\). Exam tip: The square root of a positive integer is rational only if the integer is a perfect square.