Irrational numbers

Multiplying by the conjugate gives (\frac{(\sqrt{6}-\sqrt{2})^2}{4}=2-\sqrt{3}). Make the denominator rational.

Both \(\sqrt{2}\) and \(\sqrt{8}\) are irrational, but \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational. Hence the claim is false. Exam tip: multiply radicands and check for a perfect square.

\(\sqrt{175}=5\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so the bracket is \(2\sqrt{7}\). Multiplying by \(\sqrt{7}\) gives (14).

Let \(x\) be irrational and \(r\) be rational. If \(x+r\) were rational, then \((x+r)-r=x\) would be rational, a contradiction. Option B fails since \(\sqrt{2}+(-\sqrt{2})=0\). Exam tip: test “always” claims with a counterexample.

The governing concept is rationalisation of a denominator containing a binomial surd. The conjugate of √17 + √8 is √17 − √8, so multiply both numerator and denominator by that conjugate. This gives 3(√17 − √8)/[(√17 + √8)(√17 − √8)]. Using the difference-of-squares identity, the denominator becomes (√17)² − (√8)² = 17 − 8 = 9. Hence the expression is 3(√17 − √8)/9 = (√17 − √8)/3. Therefore option C is correct. Option A misses the factor 1/3, option B leaves the denominator irrational and does not rationalise it, and option D is not algebraically equivalent. The conjugate is essential because it changes the denominator into a rational number.

If \(a+b\) were rational, subtracting the rational number \(a\) would make \(b\) rational, a contradiction. Thus \(a+b\) is irrational. Exam tip: test an “always” statement using contradiction.

Since (18<19<20), (\sqrt{19}) lies between them. Compare square roots using the numbers inside.

If \(p+x\) were rational, subtracting the rational number \(p\) would make \(x\) rational, a contradiction. B fails because \(\sqrt2+(-\sqrt2)=0\). Exam tip: use contradiction for such properties.

If \(r+x\) were rational, subtracting rational \(r\) would make \(x\) rational, a contradiction. But \((\sqrt{2})^2=2\). Exam tip: use closure of rational numbers under subtraction.

(\sqrt{432}=12\sqrt{3}), (\sqrt{192}=8\sqrt{3}), and (\sqrt{48}=4\sqrt{3}). Therefore the result is (8\sqrt{3}).

After rationalising the terms become (\frac{\sqrt{7}-\sqrt{5}}{2}) and (\frac{\sqrt{7}+\sqrt{5}}{2}). The sum is (\sqrt{7}).

\(r+q\) is always irrational; otherwise, subtracting the rational number \(q\) would make \(r\) rational, a contradiction. But \(r\times q\) can be 0 when \(q=0\). Exam tip: adding or subtracting a rational number from an irrational number remains irrational.

(\sqrt{200}=10\sqrt{2}) and (\sqrt{72}=6\sqrt{2}), so the numerator is (16\sqrt{2}). Dividing gives (16).

(r^2-s^2=(r-s)(r+s)) where (r-s=2\sqrt{7}) and (r+s=2\sqrt{13}). So the value is (4\sqrt{91}).

To rationalise a denominator containing \(\sqrt{18}-\sqrt{10}\), multiply the fraction by the conjugate \(\sqrt{18}+\sqrt{10}\) in both numerator and denominator. This preserves the value of the fraction while changing the denominator into a difference of squares.

The denominator becomes \((\sqrt{18}-\sqrt{10})(\sqrt{18}+\sqrt{10})=18-10=8\). Thus the expression becomes \(\frac{8(\sqrt{18}+\sqrt{10})}{8}=\sqrt{18}+\sqrt{10}\). The rationalised result is therefore the sum of the two square roots, corresponding to option A as supplied. The essential step is using the conjugate and the identity \((a-b)(a+b)=a^2-b^2\).

(\sqrt{363}=11\sqrt{3}), (\sqrt{108}=6\sqrt{3}), and (\sqrt{192}=8\sqrt{3}). Therefore (P=9\sqrt{3}).

(\sqrt{44}=2\sqrt{11}) and (\sqrt{99}=3\sqrt{11}), so the sum is (5\sqrt{11}). Its square is (275).

In D, \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so it becomes \(5\sqrt{3}\), which is irrational. A becomes \(\sqrt{36}=6\). Exam tip: extract perfect-square factors first.

Option A is correct. The 1s occur at positions 1, 3, 6, 10, …, so the zero-gaps 1, 2, 3, … keep growing and no fixed block repeats. B is false because using two digits does not ensure repetition. Exam tip: rational decimals terminate or repeat.

For a denominator of the form \(\sqrt{29}+2\), use its conjugate \(\sqrt{29}-2\). Multiplying by the conjugate is useful because the middle terms cancel. The denominator then becomes a rational number, so no square root remains below the fraction line.

Multiply numerator and denominator by \(\sqrt{29}-2\). The denominator is \((\sqrt{29}+2)(\sqrt{29}-2)=29-4=25\). The result is \(\frac{5(\sqrt{29}-2)}{25}=\frac{\sqrt{29}-2}{5}\). This matches option B. Options that leave an irrational denominator or fail to include the correct factor do not give the simplified rationalised form.