Irrational numbers

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.

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.

If \(5-\sqrt{7}\) were rational, then \(5-(5-\sqrt{7})=\sqrt{7}\) would also be rational, which is impossible. Hence the difference is irrational. Exam tip: rational ± irrational is irrational.

(\sqrt{588}=14\sqrt{3}), (\sqrt{300}=10\sqrt{3}), and (\sqrt{192}=8\sqrt{3}). Therefore the result is (12\sqrt{3}).

Adding the two terms gives numerator (2\sqrt{10}) and denominator (10-6=4). So the value is (\frac{\sqrt{10}}{2}).

The correct expression is \(a+b\). If \(a+b\) were rational, then \(b=(a+b)-a\) would also be rational, a contradiction. However, \(ab=0\) when \(a=0\). Exam tip: test zero when checking an “always” claim.

(\sqrt{392}=14\sqrt{2}) and (\sqrt{200}=10\sqrt{2}), so the numerator is (24\sqrt{2}). Dividing gives (24).

(r^2-s^2=(r-s)(r+s)), where (r-s=2\sqrt{11}) and (r+s=2\sqrt{17}). So the value is (4\sqrt{187}).

In option A, \(\sqrt{11}\) is irrational, but \((\sqrt{11})^2=11\) is rational, so the claim is false. In option B, \(\sqrt[3]{4}\) remains irrational. Exam tip: one counterexample is enough to disprove an “every” statement.

(\sqrt{500}=10\sqrt{5}), (\sqrt{180}=6\sqrt{5}), and (\sqrt{320}=8\sqrt{5}). Therefore (P=8\sqrt{5}).

(\sqrt{125}=5\sqrt{5}) and (\sqrt{320}=8\sqrt{5}), so the sum is (13\sqrt{5}). Its square is (845).

Option A is correct. The blocks of zeros between 1s keep increasing, so no fixed block repeats periodically. A rational number has a terminating or recurring decimal expansion. Exam tip: check periodic repetition, not merely the digits used.

Both \(\sqrt{2}\) and \(\sqrt{8}\) are irrational, but \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational. Hence the word “always” makes the statement false. Exam tip: disprove universal claims using one counterexample.

Multiplying by the conjugate gives numerator (50+2\sqrt{589}) and denominator (12). The simplified form is (\frac{25+\sqrt{589}}{6}).

Both \(\sqrt{2}\) and \(\sqrt{8}\) are irrational, but \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational. In the other options, the product under the radical is not a perfect square. Exam tip: combine radicals before classifying the result.