Irrational numbers

Adding any rational number \(q\) to \(\sqrt{2}\) keeps the result irrational. If \(\sqrt{2}+q\) were rational, subtracting \(q\) would make \(\sqrt{2}\) rational, a contradiction. In \(q\sqrt{2}\), taking \(q=0\) gives 0. Exam tip: rational ± irrational is always irrational.

If \(x+r\) were rational, then \(x=(x+r)-r\) would be rational, which is a contradiction. However, \(x^2\) need not be irrational: \((\sqrt{2})^2=2\). Exam tip: rational ± irrational is always irrational.

A is correct: if \(r+q\) were rational, subtracting \(q\) would make \(r\) rational, a contradiction. For \(r=\sqrt2\), \(r^2=2\), so C fails. Tip: test “always” claims with a counterexample.

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

The governing concept is simplification of surds by extracting perfect-square factors. Since 108=36×3, √108=6√3. Since 300=100×3, √300=10√3. Thus s=6√3+10√3=16√3. Dividing by √3 gives s/√3=(16√3)/√3=16, because √3 is non-zero. Hence option C is correct. The irrational factor cancels only after both radicals have been expressed as like surds. Option A could result from losing one coefficient, while options B and D do not agree with the coefficient sum 6+10. Directly adding 108 and 300 would also be invalid, because square roots cannot generally be combined by adding their radicands.

Multiplying the same square root by itself gives the number inside. Therefore the value is (8+\sqrt{11}).

Option C is correct. A rational number has a terminating or eventually repeating decimal expansion. Here, the zeros between successive 1s keep increasing, so no fixed repeating block exists. Exam tip: check repetition, not merely the digits used.

Multiplying by the conjugate gives numerator (16+2\sqrt{39}) and denominator (10). So the answer is (\frac{8+\sqrt{39}}{5}).

An irrational number has a non-terminating, non-repeating decimal expansion; no fixed block of digits repeats forever. Terminating and repeating decimals are rational. Exam tip: a recurring digit pattern always indicates a rational number.

\(\frac{\sqrt{45}}{\sqrt{5}}=\sqrt{9}=3\), so it is rational. The other expressions reduce to \(3\sqrt3\), \(\sqrt3\), and \(\sqrt5\), which are irrational. Exam tip: rewrite radicands using perfect-square factors first.

(\frac{7}{\sqrt{23}+4}=\sqrt{23}-4) because the denominator becomes (23-16=7). So the sum is (2\sqrt{23}).

(\sqrt{98}=7\sqrt{2}) and (\sqrt{50}=5\sqrt{2}), so the numerator is (2\sqrt{2}). Dividing gives (2).

If \(x+q\) were rational, then \(x=(x+q)-q\) would be rational, a contradiction. But \(x^2\) need not be irrational: for \(x=\sqrt{2}\), it is 2. Exam tip: use contradiction.

(\sqrt{539}=7\sqrt{11}), (\sqrt{275}=5\sqrt{11}), and (\sqrt{99}=3\sqrt{11}). So the result is (7\sqrt{11}-5\sqrt{11}+3\sqrt{11}=5\sqrt{11}).

Multiplying by the conjugate makes the denominator (19-10=9). So the form is (\frac{6(\sqrt{19}+\sqrt{10})}{9}).

0.272727... is a recurring decimal. If \(y=0.272727...\), then \(100y-y=27\), so \(y=27/99=3/11\). Therefore, it is rational. Exam tip: convert recurring decimals into fractions to check rationality.

The square of a square root gives the number inside. So \(\left(\sqrt{13+\sqrt{42}}\right)^2=13+\sqrt{42}\).

Area is ((\sqrt{22}+\sqrt{7})(\sqrt{22}-\sqrt{7})=22-7=15). Conjugate dimensions can give a rational area.

If \(2x\) were rational, then \(x=(2x)/2\) would also be rational, a contradiction. Hence \(2x\) is irrational. But \(x^2\) need not be irrational: for \(x=\sqrt{2}\), \(x^2=2\). In exams, test claims using a counterexample.

The governing concept is simplification of surds by extracting perfect-square factors. Since 27 = 9 × 3, √27 = √9 × √3 = 3√3. Similarly, 75 = 25 × 3, so √75 = 5√3. Therefore y = 3√3 + 5√3 = 8√3, because like surds are added by adding their rational coefficients. Dividing by √3 gives y/√3 = 8√3/√3 = 8, since √3 is non-zero. Thus option B is correct. Option A or C may result from adding the radicands or simplifying one radical incorrectly, while option D has no valid basis. The essential method is to express both radicals with the same √3 factor before combining them.