Irrational numbers

C is correct. Zero blocks between 1s have lengths 1, 2, 3, 4, ..., so no fixed block repeats. A non-terminating, non-repeating decimal is irrational. Exam tip: check repetition, not digits.

Using the distributive law, \(\sqrt{2}(3\sqrt{2}+1)=3\sqrt{2}\times\sqrt{2}+\sqrt{2}\). Since \(\sqrt{2}\times\sqrt{2}=2\), the simplified form is \(3\times2+\sqrt{2}=6+\sqrt{2}\). Option C incorrectly doubles the term containing \(\sqrt{2}\). Exam tip: use \(\sqrt{a}\times\sqrt{a}=a\) when multiplying identical square roots.

Multiplying by the conjugate (\sqrt{2}-1) gives (3(\sqrt{2}-1)). So the form is (3\sqrt{2}-3).

Adding the two expressions gives \((7+\sqrt{13})+(7-\sqrt{13})=7+7+\sqrt{13}-\sqrt{13}=14\). The irrational terms \(\sqrt{13}\) and \(-\sqrt{13}\) cancel each other, so the result is the rational number 14. Exam tip: when adding conjugate forms \((a+b)\) and \((a-b)\), combine the matching terms first.

Option A is correct. Zeros between successive 1s are 1, 2, 3, 4, ...; no fixed block repeats. Thus this non-terminating decimal is irrational. Exam tip: check repetition, not digit types.

Adding irrational (\sqrt{5}) to rational (4) gives an irrational number. The irrational part remains.

(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4), which is rational. The product of two irrationals can be rational.

The side of a square is the square root of its area. Thus, side \(=\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}\) units, so option C is correct. The distractor \(2\sqrt{3}\) is incorrect because its square is \(12\), not \(18\). Exam tip: factor the number under the square root into a perfect square and the remaining factor before simplifying.

A rational number has a terminating or recurring decimal expansion. Here, the zeros between successive 1s are 1, 2, 3, 4, …, so no repeating block occurs. In exams, check recurrence, not merely the digits used.

(\sqrt{3}+\sqrt{27}=4\sqrt{3}) and (\sqrt{48}=4\sqrt{3}). Simplified form makes comparison easy.

(\sqrt{20}=2\sqrt{5}), so (2\sqrt{5}+6\sqrt{5}=8\sqrt{5}). Watch both coefficients and radicals carefully.

Here, a 1 appears after 1, 2, 3, 4, … zeros successively, so no fixed block of digits repeats. A non-terminating, non-repeating decimal is irrational. Exam tip: digits 0 and 1 alone do not make a number rational.

\(\sqrt{32}=\sqrt{16\times2}=4\sqrt{2}\). Therefore, \(\sqrt{2}+\sqrt{32}=\sqrt{2}+4\sqrt{2}=5\sqrt{2}\). Option D is incorrect because the sum of separate square roots cannot generally be written as \(\sqrt{34}\). Exam tip: simplify each radical first, then combine terms with the same radical part.

In \(0.333\ldots\), the digit 3 repeats, so \(0.333\ldots=\frac{1}{3}\), which is rational. Hence, an infinite decimal need not be irrational. Exam tip: check whether the decimal digits repeat.

Multiplying numerator and denominator by (\sqrt{7}) gives (\frac{\sqrt{7}}{7}). Rationalisation should not change the value.

In conjugate multiplication (rs=(\sqrt{15})^2-(\sqrt{6})^2=9). Use (a^2-b^2) in such questions.

(\sqrt{98}=7\sqrt{2}) and (\sqrt{50}=5\sqrt{2}), so division gives (12). Simplify the numerator first.

Digits 0 and 1 alone do not make a number rational. The gaps of zeros are 1, 2, 3, 4,..., so no block repeats. Thus the decimal is irrational. Exam tip: look for a fixed repeating block.

(\sqrt{20}=2\sqrt{5}), so inside becomes (2\sqrt{5}-6\sqrt{5}=-4\sqrt{5}) and the product is (-20). Simplify the bracket first.

Multiplying numerator and denominator by (\sqrt{11}) gives (\frac{4\sqrt{11}}{11}). Rationalisation must keep the value equal.