Irrational numbers

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.

(\sqrt{40}=2\sqrt{10}), so the bracket is (-\sqrt{10}) and the product is (-10). Simplify the bracket first.

Multiplying numerator and denominator by (\sqrt{19}) gives (\frac{6\sqrt{19}}{19}). Rationalisation removes the radical from the denominator.

(\sqrt{12}=2\sqrt{3}) and (\sqrt{108}=6\sqrt{3}), so the sum is (8\sqrt{3}). Find simplified forms before comparing.

It has no fixed repeating block, so it is irrational. Identifying non-terminating non-repeating decimals is important.

Multiplying by the conjugate makes the denominator (25-6=19). So the rationalised form is (\frac{5-\sqrt{6}}{19}).

The number of zeros between successive 1s increases as 1, 2, 3, 4... Hence, no fixed block repeats and the number is irrational. Exam tip: a non-terminating, non-repeating decimal is irrational.

(\sqrt{48}=4\sqrt{3}) and (\sqrt{27}=3\sqrt{3}), so the sum is (7\sqrt{3}). Simplify before adding like radicals.

The numbers of zeros between successive 1s are 1, 2, 3, 4, …, so no fixed digit block can repeat forever. A non-terminating, non-repeating decimal is irrational. In exams, check repetition, not merely whether the decimal is infinite.

The governing concept is rationalisation of a denominator containing a surd. To remove √6 + 1 from the denominator, multiply the numerator and denominator by its conjugate, √6 − 1. Thus 1/(√6 + 1) × (√6 − 1)/(√6 − 1) = (√6 − 1)/[(√6)^2 − 1^2]. Applying the difference-of-squares identity gives the denominator 6 − 1 = 5, so the rationalised form is (√6 − 1)/5. Therefore option C is correct. Option A uses the original expression instead of the conjugate, so it does not produce the required rational denominator. Option B leaves the denominator irrational, and option D is not the equivalent result of multiplying by the appropriate conjugate. The value is unchanged; only its representation has been altered.

Since \(\sqrt{36}=6=\frac{6}{1}\), it is rational, so Riya’s statement is false. \(2\), \(5\), and \(11\) are not perfect squares, so their square roots are irrational. Exam tip: check for perfect squares first.

Both \(\sqrt{3}\) and \(-\sqrt{3}\) are irrational, but their sum is \(0\), a rational number. Hence the word “always” makes the claim false. Exam tip: one counterexample is enough to disprove a universal statement.

(\sqrt{98}=7\sqrt{2}) and (\sqrt{50}=5\sqrt{2}), so the difference is (2\sqrt{2}). First take out perfect-square factors.

(\sqrt{28}=2\sqrt{7}) and (\sqrt{63}=3\sqrt{7}), so the total is (6\sqrt{7}). Convert all terms to like radicals.

This decimal has no fixed repeating block, so it is irrational. Identify non-terminating non-repeating decimals.

Using the quotient property of square roots, \(\frac{\sqrt{108}}{\sqrt{3}}=\sqrt{\frac{108}{3}}=\sqrt{36}=6\). Therefore, option A is correct. Option D represents only \(\sqrt{108}=6\sqrt{3}\) and does not complete the division by \(\sqrt{3}\). Exam tip: For positive radicands, use \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\) to simplify such expressions.

(\sqrt{162}=9\sqrt{2}) and (\sqrt{72}=6\sqrt{2}), so the sum is (15\sqrt{2}). Add coefficients of like radicals.