Irrational numbers

If \(p+x\) were rational, then \(x=(p+x)-p\) would also be rational, which is a contradiction. Hence A is correct; B, C and D wrongly call the result rational. Exam tip: use contradiction for such properties.

(\frac{6}{\sqrt{31}+5}=\sqrt{31}-5) because the denominator becomes (31-25=6). So the sum is (2\sqrt{31}).

(\sqrt{300}=10\sqrt{3}) and (\sqrt{108}=6\sqrt{3}), so the numerator is (4\sqrt{3}). Dividing gives (4).

Assume \(rx\) is rational. Since \(r\ne0\) is rational, \(x=(rx)/r\) would also be rational, giving a contradiction. Hence \(rx\) is irrational. In exams, always check that the rational multiplier is non-zero.

(\sqrt{847}=11\sqrt{7}), (\sqrt{363}=7\sqrt{7}), and (\sqrt{147}=3\sqrt{7}). So the result is (11\sqrt{7}-7\sqrt{7}+3\sqrt{7}=7\sqrt{7}).

Multiplying by the conjugate makes the denominator (27-11=16). So the form is (\frac{8(\sqrt{27}+\sqrt{11})}{16}).

In A, \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational, so the claim fails. In B, \(\sqrt{6}\) remains irrational. Exam tip: test “always” using one counterexample.

The square of a square root gives the number inside. So \(\left(\sqrt{16+\sqrt{45}}\right)^2=16+\sqrt{45}\).

Area is ((\sqrt{30}+\sqrt{11})(\sqrt{30}-\sqrt{11})=30-11=19). Conjugate dimensions give rational area.

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but \(\sqrt{2}+(-\sqrt{2})=0\), which is rational. Hence one counterexample disproves the claim. Exam tip: test words such as “always” using a counterexample.

The governing concept is simplifying radicals by expressing each radicand as a perfect square multiplied by the same number. Since 75 = 25 × 3, √75 = 5√3. Since 147 = 49 × 3, √147 = 7√3. Thus y = 5√3 + 7√3 = 12√3. Dividing by √3 gives y/√3 = 12√3/√3 = 12. Therefore option B is correct. Option A or C could result from an incorrect addition of the coefficients, while option D may come from using an incorrect square factor in one of the radicals. The important point is that only like surds can be combined directly, so both terms must first be written as multiples of √3.

Multiplying by the conjugate gives (\frac{(\sqrt{10}-\sqrt{6})^2}{4}=4-\sqrt{15}). Make the denominator rational.

\(x=\sqrt{2}\) is irrational, but \(x^2=(\sqrt{2})^2=2\) is rational. Thus, this counterexample disproves the claim. Exam tip: test “every” statements by finding one counterexample.

\(\sqrt{539}=7\sqrt{11}\) and \(\sqrt{275}=5\sqrt{11}\), so the bracket is \(2\sqrt{11}\). Multiplying by \(\sqrt{11}\) gives (22).

The number of zeros between successive 1s keeps increasing, so no fixed block repeats. A non-terminating, non-repeating decimal is irrational. Exam tip: check for repetition, not just the digits used.

Multiplying by the conjugate makes the denominator (26-17=9). So the form is (\frac{5(\sqrt{26}-\sqrt{17})}{9}).

Suppose \(p-q\) were rational. Then \((p+q)+(p-q)=2p\) would be rational, forcing \(p\) to be rational, which is a contradiction. Thus A must be irrational. Exam tip: take \(p=\sqrt2,q=-\sqrt2\); B, C and D then become rational.

Since (26<28<30), (\sqrt{28}) lies between them. Compare square roots using the numbers inside.

\(0.272727\ldots\) is non-terminating but repeats 27, so it is rational. If \(x=0.272727\ldots\), then \(100x-x=27\), giving \(x=3/11\). Exam tip: only non-terminating, non-repeating decimals are irrational.

Here \(\sqrt{12}+\sqrt{27}=2\sqrt3+3\sqrt3=5\sqrt3\), so the conclusion is true. The reason is false: \(\sqrt3-\sqrt3=0\). Exam tip: test the claim and its justification separately.