Irrational numbers

Multiplying by the conjugate (2+\sqrt{3}) makes the denominator (1) and numerator ((2+\sqrt{3})^2). So the answer is (7+4\sqrt{3}).

\(x+\frac{1}{3}\) must be irrational: if it were rational, subtracting the rational number \(\frac{1}{3}\) would make \(x\) rational. But \(x^2\) need not be irrational; for \(x=\sqrt{2}\), \(x^2=2\). Exam tip: adding a rational number to an irrational number keeps it irrational.

(\sqrt{45}=3\sqrt{5}), (\sqrt{20}=2\sqrt{5}), and (\sqrt{80}=4\sqrt{5}). Therefore the result is (\sqrt{5}).

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but their sum is \(0\), which is rational. Hence, the sum of two irrational numbers need not be irrational. Exam tip: test “always” statements using a counterexample.

The square root of a positive number is real and its square gives the same number. Here (3+\sqrt{5}) is positive.

The governing concept is reducing radicals to like surds before adding or subtracting them. Since 8 = 4×2, √8 = 2√2; since 18 = 9×2, √18 = 3√2. Thus m = 2√2 + 3√2 = 5√2. Also, 50 = 25×2, so n = √50 = 5√2. Consequently, m − n = 5√2 − 5√2 = 0, and option A is correct. Option B may result from subtracting an unrelated leftover radical. Option C reflects incomplete simplification or an arithmetic error. Option D is the common value of m and n, not their difference. The cancellation is exact because equal surds with equal coefficients have opposite signs in the subtraction.

Answer: 2√2 - 1, so option B is correct. To simplify x=1/(√2+1), rationalise the denominator by multiplying the numerator and denominator by the conjugate of √2+1. Its conjugate is √2-1. Thus x=[1(√2-1)]/[(√2+1)(√2-1)]. The denominator is a difference of squares: (√2)^2-1^2=2-1=1. Therefore x=√2-1. Now add √2: x+√2=(√2-1)+√2=2√2-1. Option A, 1, incorrectly treats the radical terms as if they cancel. Option C, 2, loses both the radical and the constant term. Option D is just the original denominator and is not the simplified value of the expression. The conjugate method works because (a+b)(a-b)=a²-b², which removes the surd from the denominator. Always change the sign between the two terms when choosing a conjugate.

This decimal is non-terminating and non-repeating: the number of zeros between successive 1s is 1, 2, 3, 4, ... . Hence no fixed period is possible. Exam tip: non-terminating recurring decimals are rational.

Take \(x=\sqrt{2}\). It is irrational, but \(x^2=2\) is rational. If \(x+1\), \(2x\), or \(1/x\) were rational, then \(x\) would also be rational, a contradiction. Exam tip: test a claim using a counterexample.

Since \(x-x=0\), and 0 is rational, option D is always rational. In contrast, \(x^2\) may be rational, as \((\sqrt{2})^2=2\), or irrational. Exam tip: first cancel identical terms before classifying a number.

(\sqrt{48}=4\sqrt{3}), (\sqrt{75}=5\sqrt{3}), and (\sqrt{27}=3\sqrt{3}). Therefore the result is (6\sqrt{3}).

(\sqrt{72}=6\sqrt{2}) and (\sqrt{50}=5\sqrt{2}), so the numerator is (11\sqrt{2}). Dividing by (\sqrt{2}) gives (11).

Option C is correct. For \(x=\sqrt{2}\), \(x\) is irrational but \(x^2=2\) is rational. Adding 2 or multiplying or dividing by non-zero rational 3 preserves irrationality. Exam tip: test “always” claims with a counterexample.

The first square adds (2\sqrt{10}) and the second subtracts (2\sqrt{10}). The difference is (4\sqrt{10}).

Multiplying by the conjugate gives denominator (9-5=4) and numerator ((3+\sqrt{5})^2=14+6\sqrt{5}). So the answer is (\frac{7+3\sqrt{5}}{2}).

The conjugate of the denominator is (\sqrt{3}+\sqrt{2}) and the denominator becomes (1). So the value is ((\sqrt{3}+\sqrt{2})^2=5+2\sqrt{6}).

(\sqrt{7}+\sqrt{28}=3\sqrt{7}), which is irrational. In the other options the radicals simplify to rational values.

Both \(\sqrt{2}\) and \(2\sqrt{2}\) are irrational, but \(\sqrt{2}\times2\sqrt{2}=4\), a rational number. Thus “always” in option A is false. Exam tip: one counterexample disproves a universal claim.

Since (6+\sqrt{11}) is positive, its square root is real. Squaring it gives the inside number (6+\sqrt{11}).

The governing concept is extraction of perfect-square factors from radicals followed by combining like surds. Since 80 = 16×5, √80 = 4√5. Since 45 = 9×5, √45 = 3√5. Thus u = 4√5 − 3√5 = √5. The question gives v = √5, so u + v = √5 + √5 = 2√5. Therefore option A is correct. Option B results from adding coefficients without first evaluating u, option C is only the value of u and not u + v, and option D would require opposite signs. The important rule is that coefficients of identical surds may be added, while unlike radicals cannot be combined directly.