Irrational numbers

The governing concept is rationalisation of a denominator containing a surd. The conjugate of √17 + 4 is √17 − 4. Multiply numerator and denominator by that conjugate: r = [1/(√17 + 4)]×[(√17 − 4)/(√17 − 4)]. The denominator becomes (√17 + 4)(√17 − 4) = (√17)² − 4² = 17 − 16 = 1. Hence r = (√17 − 4)/1 = √17 − 4, so option A is correct. Option B is the original denominator expression, option C reverses the sign and is negative, and option D leaves a surd in the denominator rather than rationalising it. Since √17 is slightly greater than 4, the final value is positive, confirming the sign.

(\sqrt{27}=3\sqrt{3}) and (\sqrt{75}=5\sqrt{3}), so the sum is (8\sqrt{3}). Its square is (192).

Multiplying by the conjugate makes the denominator (14-5=9). So (\frac{3(\sqrt{14}-\sqrt{5})}{9}=\frac{\sqrt{14}-\sqrt{5}}{3}).

(\sqrt{18}=3\sqrt{2}) and (\sqrt{72}=6\sqrt{2}) so the numerator is (9\sqrt{2}). Dividing by (\sqrt{2}) gives (9).

(\sqrt{12}=2\sqrt{3}), (\sqrt{75}=5\sqrt{3}), and (\sqrt{48}=4\sqrt{3}). Therefore (a-b=3\sqrt{3}).

Since \(12=4\times3\), \(\sqrt{12}=2\sqrt{3}\). Thus the difference is \(2\sqrt{3}-\sqrt{3}=\sqrt{3}\), which is irrational. Exam tip: never use \(\sqrt a-\sqrt b=\sqrt{a-b}\).

Multiplying by the conjugate makes the denominator (13-5=8). So (\frac{4(\sqrt{13}+\sqrt{5})}{8}=\frac{\sqrt{13}+\sqrt{5}}{2}).

Area is ((\sqrt{11}+\sqrt{6})(\sqrt{11}-\sqrt{6})=11-6=5). Conjugate dimensions can give a rational area.

(\sqrt{12}\times\sqrt{3}=\sqrt{36}=6), which is rational. Check whether multiplication creates a perfect square.

The denominator conjugate is (\sqrt{5}+2) and the denominator becomes (1). So the value is ((\sqrt{5}+2)^2=9+4\sqrt{5}).

\(0.\overline{27}=\frac{27}{99}=\frac{3}{11}\), so it is rational despite being non-terminating. \(\sqrt7\) and \(\pi\) are irrational. Exam tip: every repeating decimal is rational.

As \(200=100\times2\), \(72=36\times2\), and \(18=9\times2\), we have \(\sqrt{200}=10\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\). Therefore, \(10\sqrt{2}-6\sqrt{2}+3\sqrt{2}=7\sqrt{2}\), so option C is correct. \(9\sqrt{2}\) results from incorrectly changing the subtraction sign into addition. Exam tip: extract perfect-square factors from each radical before combining like radicals.

Option C is correct. If \(qr\) were rational, then since \(q\ne0\), \(r=\frac{qr}{q}\) would also be rational, a contradiction. Exam tip: test “always” claims using examples such as \(\sqrt2\) and \(-\sqrt2\).

In 0.101001000100001…, the number of zeros between successive 1s keeps increasing, so the decimal neither terminates nor repeats. Hence it is irrational. 0.121212… repeats and is rational. Exam tip: every terminating or recurring decimal is rational.

Both \(\sqrt{5}\) and \(-\sqrt{5}\) are irrational, but their sum is 0, which is rational. Therefore, the word “always” makes the claim false. Exam tip: one counterexample disproves a universal statement.

To rationalise the denominator, multiply the fraction by the conjugate of the denominator. Since \(\sqrt{4}=2\), the expression is \(\frac{3}{\sqrt{7}+2}\). The conjugate of \(\sqrt{7}+2\) is \(\sqrt{7}-2\). Multiplying numerator and denominator by this conjugate gives \(\frac{3(\sqrt{7}-2)}{(\sqrt{7}+2)(\sqrt{7}-2)}\).

The denominator is a difference of squares: \((\sqrt{7})^2-2^2=7-4=3\). Thus the fraction becomes \(\frac{3(\sqrt{7}-2)}{3}=\sqrt{7}-2\). The denominator is now rational, so this is the rationalised form. Hence option B is correct. Option A is an unsimplified expression equal to the same value, but the listed rationalised answer is the simplified form in option B.

Multiplying the same square root by itself gives the number inside. Therefore the value is (6+\sqrt{7}).

If \(rx\) were rational, then \(x=(rx)/r\) would be a quotient of two rational numbers and hence rational, a contradiction. Therefore, \(rx\) is irrational. Exam tip: the condition \(r\neq0\) is essential.

The blocks \(1,01,001,0001,\ldots\) keep changing, so no fixed repeating block occurs. A non-terminating, non-repeating decimal is irrational. Exam tip: a repeating decimal always represents a rational number.

If \(rx\) were rational, then \(x=(rx)/r\) would be rational, a contradiction. Hence \(rx\) is irrational. But \(x^2\) need not be irrational: for \(x=\sqrt{2}\), \(x^2=2\). Exam tip: use contradiction by dividing by a non-zero rational number.