Irrational numbers

Multiplying by \(\sqrt{5}-2\) makes the denominator \(5-4=1\), so the value is \(\sqrt{5}-2\). The difference of an irrational and a rational number is irrational. Exam tip: rationalise the denominator first.

Since (6<6.5<7), (\sqrt{6.5}) lies between them. For positive square roots compare the numbers inside.

If \(\frac{p+\sqrt{2}}{q}=r\) is rational, then \(\sqrt{2}=qr-p\). Since p, q and r are rational, the right side is rational, contradicting the irrationality of \(\sqrt{2}\). Exam tip: use closure of rational numbers under multiplication and subtraction.

(\sqrt{45}=3\sqrt{5}) and (\sqrt{20}=2\sqrt{5}), so the numerator is (\sqrt{5}). Dividing gives (1).

The square of a square root gives the number inside. So \(\left(\sqrt{2+\sqrt{3}}\right)^2=2+\sqrt{3}\).

In \(0.101001000100001\ldots\), the number of zeros between successive 1s keeps increasing, so no repeating block is formed. Its decimal expansion is non-terminating and non-repeating; hence it is irrational. \(0.\overline{27}\) is rational because it repeats. Exam tip: non-terminating, non-repeating decimals are irrational.

If \(x+r\) were rational, subtracting the rational number \(r\) would make \(x\) rational, a contradiction. Hence \(x+r\) is irrational. Exam tip: for \(r=0\), \(xr=0\), so a product is not always irrational.

The governing concept is the difference-of-squares identity, (a+b)(a−b)=a²−b². Let a = √8 and b = √18. Then the product is (√8)² − (√18)² = 8 − 18 = −10. Therefore option A is correct. A second check is possible by simplifying first: √8 = 2√2 and √18 = 3√2. The product becomes (5√2)(−√2) = −5×2 = −10. Option B reverses the subtraction, option C adds 8 and 18 instead of subtracting them, and option D incorrectly multiplies radicands while ignoring the conjugate structure. Recognising the identity makes the calculation shorter and safer.

\(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational, so A disproves the claim. In B, the product is \(\sqrt{6}\), still irrational. Exam tip: combine radicals before deciding the type of number.

If the sum of an irrational and a rational number were rational, subtracting the same rational number would make the irrational number rational, which is impossible. Exam tip: test “always” claims with counterexamples.

Multiplying by the conjugate gives denominator (9-2=7) and numerator ((3+\sqrt{2})^2=11+6\sqrt{2}). So the correct form is (\frac{11+6\sqrt{2}}{7}).

The product of irrational numbers need not be irrational. Here, \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), a rational number. Exam tip: combine square roots first before deciding the type of number.

(\sqrt{108}=6\sqrt{3}), (\sqrt{48}=4\sqrt{3}), and (\sqrt{12}=2\sqrt{3}). Therefore the result is (4\sqrt{3}).

Multiplying by the conjugate (\sqrt{7}-2) makes the denominator (7-4=3). So (p=\frac{\sqrt{7}-2}{3}).

Multiplying the same square root by itself gives the number inside. Therefore the value is (5+\sqrt{6}).

The number of zeros between successive 1s is 1, 2, 3, 4, ..., so no fixed repeating block occurs. A rational number has a terminating or recurring decimal expansion. Exam tip: check repetition, not merely the digits used.

Multiplying by the conjugate gives (\frac{(\sqrt{3}+1)^2}{2}=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}). Rationalise the denominator.

Answer: option A, (7 + 2√10)/3. Start with x/y = (√5 + √2)/(√5 − √2). The denominator contains a difference of square roots, so multiply the numerator and denominator by its conjugate, √5 + √2. The denominator becomes (√5 − √2)(√5 + √2) = (√5)² − (√2)² = 5 − 2 = 3. The numerator becomes (√5 + √2)² = 5 + 2 + 2√10 = 7 + 2√10. Therefore x/y = (7 + 2√10)/3. Option B is missing the denominator 3. Option C has the wrong sign for the middle term; squaring a sum gives a positive cross term. Option D is only the rationalised denominator, not the complete quotient. Since √5 − √2 is positive and nonzero, the division is valid. Memory cue: use the conjugate, and remember (a+b)² has +2ab.

If \(x+r\) were rational, subtracting the rational number \(r\) would make \(x\) rational, a contradiction. But \(x^2\) need not be irrational; \((\sqrt{2})^2=2\). Exam tip: adding a rational number preserves irrationality.

Since (8+\sqrt{15}) is positive, its square root is real. Squaring it gives the inside number (8+\sqrt{15}).