Irrational numbers

The claim is correct. If \(r+\sqrt{7}\) were rational, then \(\sqrt{7}=(r+\sqrt{7})-r\) would also be rational, a contradiction. Exam tip: rational ± irrational is always irrational.

(\sqrt{242}=11\sqrt{2}), (\sqrt{128}=8\sqrt{2}), and (\sqrt{72}=6\sqrt{2}). Therefore the result is (9\sqrt{2}).

In the first option, both terms are irrational but they are additive inverses. Their sum is 0, which is rational. Hence, the sum of two irrational numbers need not be irrational. Exam tip: test an “always” statement by looking for one counterexample.

Option B is correct. If \(r+\sqrt{2}\) were rational, subtracting the rational number \(r\) would make \(\sqrt{2}\) rational, which is impossible. Exam tip: rational ± irrational is always irrational.

Since 7 is not a perfect square, \(\sqrt{7}\) is irrational. If the sum were rational, subtracting rational 5 would make \(\sqrt{7}\) rational, a contradiction. Exam tip: use closure under subtraction.

Suppose \(rx\) were rational. Then \(x=(rx)/r\) would also be rational because \(r\ne0\), which is a contradiction. Hence \(rx\) is irrational. Since \(\sqrt2\cdot\sqrt2=2\), option D is not always true. Exam tip: test every “always” statement using a counterexample.

(\sqrt{363}=11\sqrt{3}), (\sqrt{147}=7\sqrt{3}), and (\sqrt{75}=5\sqrt{3}). Therefore (P=13\sqrt{3}).

The denominator \(\sqrt{26}+\sqrt{17}\) contains two square roots, so its conjugate \(\sqrt{26}-\sqrt{17}\) is used. This changes the denominator into a difference of squares and makes it rational. Multiplying by the conjugate in both numerator and denominator keeps the fraction equal to its original value.

The denominator becomes \((\sqrt{26}+\sqrt{17})(\sqrt{26}-\sqrt{17})=26-17=9\). Hence \(\frac{9}{\sqrt{26}+\sqrt{17}}=\frac{9(\sqrt{26}-\sqrt{17})}{9}=\sqrt{26}-\sqrt{17}\). Therefore option A is correct. The cancellation of the factor 9 is valid because the denominator difference is exactly 9.

\(p+q\) must be irrational. If it were rational, then \(q=(p+q)-p\) would be the difference of two rational numbers and hence rational, a contradiction. But \(q^2\) need not be irrational: \((\sqrt{2})^2=2\). Exam tip: adding a rational number to an irrational number gives an irrational result.

Multiplying by the conjugate gives numerator (17+2\sqrt{66}) and denominator (5). Rationalise the denominator and simplify.

\(2-\sqrt{2}\) is irrational, yet \(\sqrt{2}+(2-\sqrt{2})=2\), which is rational. Hence the statement is false. Exam tip: test “always” claims using a counterexample.

(\sqrt{605}=11\sqrt{5}), (\sqrt{320}=8\sqrt{5}), and (\sqrt{125}=5\sqrt{5}). Therefore the result is (8\sqrt{5}).

\(\sqrt{5}\) is irrational, but \((\sqrt{5})^2=5\), which is rational. Thus, a rational square does not guarantee a rational number. In exams, check the nature of both the number and its square.

(\sqrt{20}=2\sqrt{5}), (\sqrt{45}=3\sqrt{5}), and (\sqrt{80}=4\sqrt{5}). So (y=9\sqrt{5}) and (y^2=405).

To remove the irrational denominator \(\sqrt{30}-\sqrt{23}\), multiply by its conjugate \(\sqrt{30}+\sqrt{23}\). A conjugate changes only the sign between the two terms. Their product is a difference of squares, so the radicals disappear from the denominator.

The denominator becomes \((\sqrt{30}-\sqrt{23})(\sqrt{30}+\sqrt{23})=30-23=7\). Therefore \(\frac{7}{\sqrt{30}-\sqrt{23}}=\frac{7(\sqrt{30}+\sqrt{23})}{7}=\sqrt{30}+\sqrt{23}\). This is the expression represented by option C. The other choices do not simplify to the required result or retain unnecessary factors.

The governing concept is extraction of perfect-square factors from radicals followed by addition of like surds. Since 242 = 121 × 2, √242 = 11√2. Also, 128 = 64 × 2, so √128 = 8√2. Therefore s = 11√2 + 8√2 = 19√2. Dividing by √2 gives s/√2 = 19√2/√2 = 19, because √2 is non-zero. Hence option B is correct. The values 17, 21, and 23 can arise from errors such as subtracting or misadding the coefficients, or extracting an incorrect square factor. It is not valid to add 242 and 128 inside one square root; each radical must first be simplified separately.

Multiplying the same square root by itself gives the number inside. Therefore the value is (15+\sqrt{26}).

In option C, the decimal continues forever without repeating a fixed block; the number of zeros keeps increasing. Hence it is irrational. A terminates and B repeats, so both are rational. Exam tip: a non-terminating, non-repeating decimal represents an irrational number.

Multiplying by the conjugate gives numerator (22+2\sqrt{85}) and denominator (12). The simplified form is (\frac{11+\sqrt{85}}{6}).

\(x-x=0\), and \(0=0/1\), so it is rational. The expressions \(x+1\), \(3x\), and \(x/2\) remain irrational. Exam tip: identify exact cancellation of like terms quickly.