Real Numbers

First find ( (2+\sqrt{3})^2=7+4\sqrt{3} ). Multiplying by (2+\sqrt{3}) gives (26+15\sqrt{3}).

First find ( (3-\sqrt{2})^2=11-6\sqrt{2} ). Multiplying by (3-\sqrt{2}) gives (45-29\sqrt{2}).

An irrational number has a non-terminating, non-repeating decimal expansion: its digits continue indefinitely without a fixed repeating pattern. Therefore, option C is correct. In option B, the decimal expansion is non-terminating but recurring, which represents a rational number. Exam tip: Decimal expansions of rational numbers are either terminating or non-terminating recurring.

Since \(\sqrt{m}\) is positive, squaring every part of the inequality preserves the order. Squaring gives \(15^2<m<16^2\), i.e. \(225<m<256\). Option C (\(196<m<225\)) would correspond to squaring \(14<\sqrt{m}<15\) and so is incorrect; option B simply restates the original range for \(\sqrt{m}\) rather than for \(m\); option D corresponds to squaring \(16<\sqrt{m}<17\). Exam tip: when all terms are nonnegative you may square an inequality termwise — then compute the numerical squares to get the correct interval for the variable inside the square.

Generally ( \sqrt{a}+\sqrt{b} \neq \sqrt{a+b} ). For example, ( \sqrt{4}+\sqrt{9}=5 ), not ( \sqrt{13} ).

Multiplying the square root of the same non-negative number by itself gives the number. Therefore the value is (a).

Both \(\sqrt{5}\) and \(-\sqrt{5}\) are irrational, but \(\sqrt{5}+(-\sqrt{5})=0\), and 0 is rational. Hence, the sum of two irrational numbers need not always be irrational. The sums in A, C and D are irrational; in particular, \(\sqrt{2}+\sqrt{8}=3\sqrt{2}\), which is still irrational. Exam tip: To test an ‘always’ statement about irrational numbers, try a number and its additive inverse.

( \sqrt{2}\times\sqrt{8}=\sqrt{16}=4 ), which is rational. The other products do not become square roots of perfect squares.

Combine the two fractions using a common denominator. The denominator is (2 + √3)(2 − √3) = 4 − 3 = 1. The numerator is (2 − √3) + (2 + √3) = 4 because the surd terms cancel. Therefore the entire expression equals 4, making option A correct. Options B and D incorrectly retain a radical, while C ignores the numerator after rationalisation.

Since (1125=225\times5), ( \sqrt{1125}=15\sqrt{5} ). First take out the largest perfect-square factor.

\(r+x\) is always irrational. If \(r+x\) were rational, then \(x=(r+x)-r\) would be rational as the difference of two rational numbers. This contradicts the fact that \(x\) is irrational. \(rx\) is not always irrational because it becomes 0 when \(r=0\). Also, for \(x=\sqrt{2}\), \(x^2=2\) is rational, and \(x/x=1\). Exam tip: Adding or subtracting a rational number and an irrational number always gives an irrational number.

(5\sqrt{20}=10\sqrt{5}), (2\sqrt{125}=10\sqrt{5}), and ( \sqrt{45}=3\sqrt{5} ). Therefore the result is (3\sqrt{5}).

(3\sqrt{75}=15\sqrt{3}), (4\sqrt{48}=16\sqrt{3}), and (2\sqrt{27}=6\sqrt{3}). Therefore (25\sqrt{3}) is correct.

The student's statement is incorrect. In \(0.101001000100001\ldots\), the number of zeros between successive 1s keeps increasing, so no fixed block of digits repeats. Its decimal expansion is non-terminating and non-repeating; therefore, it is irrational. Having only the digits 0 and 1 does not make a number rational. Exam tip: A rational number has either a terminating or a non-terminating recurring decimal expansion.

The first square is (48) and the second square is (18). The middle term is (2\times4\sqrt{3}\times3\sqrt{2}=24\sqrt{6}).

\(x^2=(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}\). Since \(\sqrt{6}\) is irrational, \(5+2\sqrt{6}\) is also irrational. Hence, \(x^2\) is irrational. If \(x\) were rational, then \(x^2\) would have to be rational, which is a contradiction. Therefore, \(x\) is also irrational. Option B is the closest distractor, but \(x^2\) is not rational. Exam tip: To test whether a number can be rational, use the fact that the square of every rational number is rational.

Option B is correct. The decimal expansion of a rational number is either terminating or non-terminating recurring, meaning that a fixed block of digits repeats. Here, the number of zeros between successive 1s keeps increasing, so no fixed repeating block exists. Option D reaches the right classification for this number but gives a wrong reason: a non-terminating recurring decimal such as \(0.333\ldots\) is rational. Exam tip: For a non-terminating decimal to be rational, its repeating block must have a fixed length.

Multiplying by the conjugate gives denominator (48-45=3). So the form is ( \frac{4\sqrt{3}-3\sqrt{5}}{3} ).

Multiplying by the conjugate gives denominator (13-8=5). The numerator also has (5), so the answer is ( \sqrt{13}+\sqrt{8} ).

Apply the conjugate-denominator method. Let a=4+√7 and b=4−√7. The sum a/b+b/a has common denominator ab. Calculate ab=(4+√7)(4−√7)=4²−(√7)²=16−7=9. The numerator is a²+b²=(4+√7)²+(4−√7)². Expanding gives (16+8√7+7)+(16−8√7+7)=46, because the opposite cross-terms cancel. Therefore the value is 46/9, so option A is correct. Option B is exactly half the correct numerator and suggests that one squared expression was omitted. Option C comes from an incorrect expansion of the radical terms. Option D incorrectly treats the two fractions as if their sum were simply 2, ignoring their non-unit denominators. Direct substitution also confirms that both original denominators are nonzero.