Irrational numbers

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.

(\frac{1}{\sqrt{17}+4}=\sqrt{17}-4) because the denominator becomes (17-16=1). So the sum is (2\sqrt{17}).

An irrational number has a non-terminating, non-repeating decimal expansion, so A is correct. B fails since \((\sqrt{2})^2=2\) is rational. Exam tip: repeating decimals are rational.

The square of a square root gives the number inside. So \(\left(\sqrt{10+\sqrt{21}}\right)^2=10+\sqrt{21}\).

Area is ((\sqrt{14}+\sqrt{5})(\sqrt{14}-\sqrt{5})=14-5=9). Conjugate dimensions can give a rational area.

An irrational number has an endless decimal expansion with no repeating block. In contrast, \(1/8=0.125\) terminates, so it is rational. Exam tip: choose non-terminating, non-repeating decimals.

\(\sqrt{12}\) and \(2-\sqrt{12}\) are irrational; otherwise subtracting from 2 would make \(\sqrt{12}\) rational. Their sum is \(2\), which is rational. Option A gives \(5\sqrt{3}\). Check both conditions in such questions.

\(x+r\) is irrational: if it were rational, subtracting the rational number \(r\) would make \(x\) rational, a contradiction. Also, \(x/x=1\) is rational. Exam tip: adding a rational number does not change irrationality.

Since (7<7.5<8), (\sqrt{7.5}) lies between them. Compare square roots using the numbers inside.

\(\sqrt{3}\) is irrational because 3 is not a perfect square, yet \(\sqrt{3}\times\sqrt{3}=3\), a rational number. Option B still gives irrational \(\sqrt6\). Exam tip: one valid counterexample disproves an “always” statement.

Since \(m\) is not a perfect square, \(\sqrt{m}\) is irrational. A rational \(a+b\sqrt{m}\) would give \((a+b\sqrt{m}-a)/b=\sqrt{m}\) rational, a contradiction. Tip: check \(b\ne0\).

(\sqrt{300}=10\sqrt{3}), (\sqrt{108}=6\sqrt{3}), and (\sqrt{75}=5\sqrt{3}). Therefore the result is (9\sqrt{3}).

If \(rx\) were rational, then \(x=\frac{rx}{r}\) would also be rational because \(r\neq0\) is rational. This is a contradiction. However, \(x^2\) can be rational, for example when \(x=\sqrt{2}\). Exam tip: multiplying by a non-zero rational preserves irrationality.

(\sqrt{72}=6\sqrt{2}) and (\sqrt{128}=8\sqrt{2}), so the numerator is (14\sqrt{2}). Dividing gives (14).

The direct answer is A, 4√14. Use the identity r²−s²=(r−s)(r+s). With r=√7+√2 and s=√7−√2, r−s = (√7+√2)−(√7−√2)=2√2. Also, r+s=(√7+√2)+(√7−√2)=2√7. Multiplying gives (2√2)(2√7)=4√14, so option A is correct. Option B, 9, is not obtained and is too small for the exact expression. Option C, 2√14, misses a factor of 2. Option D, 14, incorrectly removes the radical and treats the result as a whole number. The difference-of-squares identity is much safer than expanding both squares separately. Memory cue: in r²−s², first find r−s and r+s; the middle terms then disappear automatically.

(\sqrt{242}=11\sqrt{2}), (\sqrt{50}=5\sqrt{2}), and (\sqrt{98}=7\sqrt{2}). Therefore (P=9\sqrt{2}).

The direct answer is A, 175. Simplify first: √28 = √(4×7)=2√7 and √63=√(9×7)=3√7. Their sum is 5√7. Therefore (5√7)^2=25×7=175, so option A is correct. Option B, 91, is not the result of squaring the simplified sum. Option C, 25√7, stops after squaring the coefficient and forgets that (√7)^2=7; the radical must disappear. Option D, 700, is four times the correct value and usually comes from an incorrect treatment of the cross term. A direct check gives 28+63+2√(28×63)=91+2√1764=91+84=175. The key idea is to take perfect-square factors out of radicals before adding like radical terms.

If \(x+r\) were rational, subtracting the rational number \(r\) would make \(x\) rational, which is a contradiction. \(x^2\) is not always irrational; for \(x=\sqrt{2}\), it equals 2. Exam tip: rational ± irrational is irrational.

If \(x+3\) were rational, subtracting the rational number 3 would make \(x\) rational, a contradiction. However, \(x^2\) can be rational, for example when \(x=\sqrt{2}\). Exam tip: rational ± irrational is always irrational.

Both \(\sqrt{2}\) and \(\sqrt{8}\) are irrational, but their product is \(\sqrt{16}=4\), a rational number. Thus the word “always” makes the claim false. Exam tip: one valid counterexample is enough to disprove a universal statement.

Multiplying by the conjugate makes the denominator (17-8=9). So the form is (\frac{5(\sqrt{17}-\sqrt{8})}{9}).

((\sqrt{14}-\sqrt{7})^2=14+7-2\sqrt{98}). Since (\sqrt{98}=7\sqrt{2}), the answer is (21-14\sqrt{2}).

(\sqrt{45}=3\sqrt{5}) and (\sqrt{80}=4\sqrt{5}), so the numerator is (7\sqrt{5}). Dividing by (\sqrt{5}) gives (7).

\(x^2\) need not be irrational. For example, \(x=\sqrt{2}\) is irrational, but \(x^2=2\) is rational. Adding a rational number to, or multiplying/dividing an irrational number by a non-zero rational, keeps it irrational. Exam tip: test such claims using \(\sqrt{2}\).

Option A is correct. If \(qx\) were rational, then \(x=(qx)/q\) would also be rational because \(q\ne0\) is rational. This contradicts that \(x\) is irrational. Exam tip: always check the non-zero condition.