Irrational numbers

Since (21<25<30), (\sqrt{25}) lies between them. Compare square roots using the numbers inside.

The zero blocks between successive 1s have lengths 1, 2, 3, 4, …, so no fixed block repeats. It is a non-terminating, non-recurring decimal and hence irrational. Exam tip: check repetition, not just the digits used.

\(0.\overline{27}=27/99=3/11\), so it is rational despite being non-terminating. Only non-terminating, non-repeating decimals are irrational. In exams, convert a repeating decimal into a fraction.

Multiplying numerator and denominator by (\sqrt{3}) gives (\frac{7\sqrt{3}}{3}). Rationalisation removes the radical from the denominator.

The governing concept is simplification of surds before applying an exponent. First, √12 = √(4 × 3) = 2√3. Therefore √12 + √3 = 2√3 + √3 = 3√3. Squaring gives (3√3)^2 = 3^2(√3)^2 = 9 × 3 = 27, so option B is correct. The result can also be checked by using (a + b)^2 = a^2 + 2ab + b^2, but simplifying the radicals first is more efficient and avoids unnecessary expansion. Option A may result from adding the radicands, option D may result from forgetting that the radical is also squared, and option C is only an unsimplified or incorrect partial expression, not the value of the complete square. Thus 27 is the unique correct value.

The zeros between 1s increase as 1, 2, 3, 4, …, so no fixed block repeats. A non-terminating, non-repeating decimal is irrational. Exam tip: check repetition, not merely which digits occur.

(\sqrt{147}=7\sqrt{3}), (\sqrt{75}=5\sqrt{3}) and (\sqrt{27}=3\sqrt{3}), so the answer is (9\sqrt{3}). Add and subtract coefficients of like radicals.

\(\frac{1}{3}=0.333\ldots\) has an infinite decimal expansion, but the digit 3 repeats, so it is rational. \(\sqrt{2}\) is non-terminating and non-repeating. Exam tip: recurring decimals are rational.

The square root of a positive integer is irrational when it is not a perfect square. Check for a perfect square first.

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but their sum is \(0\), which is rational. Hence, the word “always” makes the statement false. Exam tip: a single counterexample disproves an “always” claim.

(\sqrt{13}) and (\sqrt{17}) are different irrational radicals and their sum is irrational. Different radicals are not added directly.

The governing method is to extract the largest perfect-square factor from under the radical. Factor 392 as 196 × 2, and observe that 196 = 14². Therefore √392 = √(196 × 2) = √196 × √2 = 14√2. Hence option A is the fully simplified form. Option C, 7√8, is numerically equivalent to 14√2, but it is not in simplest form because √8 = √(4 × 2) = 2√2, giving 7√8 = 14√2. Option B incorrectly doubles the required coefficient; its square is 28² × 2 = 1568, not 392. Option D mistakes the perfect-square factor 196 for its square root and is also far too large. Squaring 14√2 gives 196 × 2 = 392, confirming option A.

\(0.\overline{36}\) repeats, so it is rational: \(0.\overline{36}=\frac{36}{99}=\frac{4}{11}\). Unlike it, \(\sqrt{7}\) is non-repeating. Exam tip: repeating decimals are rational.

Subtracting coefficients of like radicals gives (7\sqrt{3}). It is irrational because (\sqrt{3}) remains.

(\sqrt{45}=3\sqrt{5}) and (\sqrt{80}=4\sqrt{5}), so the bracket is (7\sqrt{5}) and the product is (35). Simplify the bracket first.

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but their sum is \(0\), which is rational. Hence the word “always” makes Reena’s claim false. Exam tip: disprove universal statements using one counterexample.

The statement is true in some cases and false in others. For example, \(\sqrt{2}+\sqrt{3}\) is irrational, but \(\sqrt{2}+(-\sqrt{2})=0\), which is rational. In exams, test “always” statements using a counterexample.

Multiplying by the conjugate gives (\frac{4(\sqrt{5}-1)}{5-1}=\sqrt{5}-1). Make the denominator rational.

\(0.272727\ldots\) is non-terminating but recurring. It can be written as \(27/99=3/11\), so it is rational. \(\sqrt{7}\), \(\pi\), and \(\sqrt{11}\) are irrational. Exam tip: every recurring decimal is rational.

The two expressions are conjugates. Using \((a+b)(a-b)=a^2-b^2\), we get \((10+\sqrt{29})(10-\sqrt{29})=10^2-(\sqrt{29})^2=100-29=71\). Therefore, option B is correct. Exam tip: when conjugate expressions are multiplied, apply the difference-of-squares identity directly instead of expanding every term.