Real Numbers

The fraction \(\frac{13}{120}\) is already in lowest terms because 13 and 120 have no common factor. A rational number has a terminating decimal expansion only when, in lowest form, its denominator has no prime factors other than 2 and/or 5. Here, \(120=2^3\times3\times5\); the factor 3 makes the decimal expansion non-terminating recurring. Option A is incorrect because the mere presence of 2 and 5 is not enough; no other prime factor may occur. Exam tip: first reduce the fraction, then prime-factorise its denominator.

\(a^2=(\sqrt{3}+\sqrt{2})^2=5+2\sqrt{6}\). Also, \((\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=1\), so \(\frac{1}{a}=\sqrt{3}-\sqrt{2}\) and \(\frac{1}{a^2}=5-2\sqrt{6}\). Therefore, \(a^2+\frac{1}{a^2}=(5+2\sqrt{6})+(5-2\sqrt{6})=10\). The \(2\sqrt{6}\) terms cancel out. Exam tip: use the conjugate surd first to find the reciprocal in such questions.

Substituting \(x=-7\), we get \(x^2=(-7)^2=49\). Hence, \(\sqrt{x^2}=\sqrt{49}=7\), since the principal square root is non-negative. Also, \(-2x=-2(-7)=14\). Therefore, \(\sqrt{x^2}-2x=7+14=21\). Taking \(-7\) for the square root would be incorrect because \(\sqrt{x^2}=|x|\), not always \(x\). Exam tip: Rewrite \(\sqrt{x^2}\) as \(|x|\) to avoid sign errors.

The direct answer is option A: \(\sqrt{20}-3\). Since \(\sqrt{20}=2\sqrt{5}\) and \(20>9\), we have \(\sqrt{20}>3\). Therefore \(3-\sqrt{20}\) is negative. The absolute-value rule says \(|y|=-y\) when \(y<0\), so \(|3-\sqrt{20}|=-(3-\sqrt{20})=\sqrt{20}-3\). Option A has exactly this value. Option B forgets to change the sign. Option C changes a minus into a plus, which is not what absolute value does. Option D has the same positive terms but in the wrong order, so it represents \(\sqrt{20}+3\), a different number. Always check whether the expression inside absolute value is positive or negative first.

Statement A is not always true. For example, \(\sqrt{2}\) and \(-\sqrt{2}\) are both irrational, but their sum is \(0\), which is rational. Statement B is always true: if the sum of an irrational and a rational number were rational, subtracting the rational number would make the irrational number rational, which is impossible. Similarly, multiplying or dividing an irrational number by a non-zero rational number keeps it irrational. Exam tip: Never assume that the sum or difference of two irrational numbers must be irrational.

( \sqrt{27}+\sqrt{12}=3\sqrt{3}+2\sqrt{3}=5\sqrt{3} ). Also ( \sqrt{75}=5\sqrt{3} ), so both are equal.

In this decimal, the number of zeros between successive 1s keeps increasing, so no fixed block of digits repeats. Therefore, its decimal expansion is non-terminating and non-repeating, which identifies an irrational number. Using only 0 and 1 does not make a number rational. Exam tip: A rational number has a decimal expansion that is either terminating or non-terminating recurring.

The governing concept is combining fractions with conjugate denominators and applying the difference-of-squares identity. Use the common denominator (√3−√2)(√3+√2). The product of these conjugates is (√3)² − (√2)² = 3 − 2 = 1. The numerator becomes (√3+√2) − (√3−√2) = √3 + √2 − √3 + √2 = 2√2. Hence the expression equals (2√2)/1 = 2√2, so option A is correct. Option B comes from failing to cancel the two √3 terms. Option C confuses multiplication of radicals with the subtraction required here, and option D omits the factor √2. The original denominators are valid because √3 and √2 are unequal, so neither difference nor sum creates a zero denominator.

Taking a common denominator gives \((\sqrt{11}+3)(\sqrt{11}-3)=11-9=2\). The numerator is \((\sqrt{11}-3)+(\sqrt{11}+3)=2\sqrt{11}\). Hence, the value is \(\frac{2\sqrt{11}}{2}=\sqrt{11}\). The option \(2\sqrt{11}\) results from forgetting to divide by the denominator 2. Exam tip: use \((a+b)(a-b)=a^2-b^2\) to simplify such expressions quickly.

The direct answer is option A: \(26+15\sqrt{3}\). Use \((a+b)^2=a^2+2ab+b^2\): \((2+\sqrt{3})^2=4+4\sqrt{3}+3=7+4\sqrt{3}\). Multiply this by \(2+\sqrt{3}\): \((7+4\sqrt{3})(2+\sqrt{3})=14+7\sqrt{3}+8\sqrt{3}+12=26+15\sqrt{3}\). Option A matches. Option B is only the first-term cube expansion and misses cross terms. Option C has incorrect coefficients. Option D has the correct constant term but the radical coefficient is too small. The safe method is to square first and then multiply, collecting rational and radical terms separately.

The direct answer is A: \(45-29\sqrt2\). Use \((a-b)^3=a^3-3a^2b+3ab^2-b^3\), with \(a=3,b=\sqrt2\). Then \(3^3=27\), \(-3(3^2)(\sqrt2)=-27\sqrt2\), \(3(3)(\sqrt2)^2=9\cdot2=18\), and \(-(sqrt2)^3=-2\sqrt2\). Combining rational terms gives \(27+18=45\); combining surd terms gives \(-27\sqrt2-2\sqrt2=-29\sqrt2\). Thus A is correct. Option B, \(27-6\sqrt2\), is only a partial or incorrectly expanded result. Option C, \(33-11\sqrt2\), does not follow from the cube expansion. Option D, \(45-27\sqrt2\), includes the rational part correctly but misses the extra \(-2\sqrt2\) from the final term. Another check is to first square: \((3-sqrt2)^2=11-6\sqrt2\), then multiply by \(3-sqrt2\). Memory cue: in a cube, do not forget the fourth term \(-b^3\).

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.

The direct answer is C: \(15\sqrt5\). To simplify a square root, separate the largest perfect-square factor. Factor 1125 as \(225\times5\), and since \(225=15^2\), \(\sqrt{1125}=\sqrt{225\times5}=\sqrt{225}\sqrt5=15\sqrt5\). Option C is correct. Option A, \(25\sqrt5\), would square to \(625\times5=3125\), not 1125. Option B, \(15\sqrt3\), would represent \(\sqrt{675}\), not \(\sqrt{1125}\); the remaining factor is 5, not 3. Option D, \(75\sqrt5\), is five times too large and also squares to the wrong number. A numerical check gives \(15\sqrt5\approx33.54\), while \(\sqrt{1125}\) is also about 33.54. The key rule is \(\sqrt{a^2b}=a\sqrt b\) for positive a. Memory cue: find the largest square factor before simplifying.

\(r+x\) 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 direct answer is option A: \(66+24\sqrt{6}\). Use the identity \((a+b)^2=a^2+2ab+b^2\). Here \(a=4\sqrt{3}\) and \(b=3\sqrt{2}\). First, \(a^2=(4\sqrt{3})^2=16\times3=48\). Next, \(b^2=(3\sqrt{2})^2=9\times2=18\). The middle term is \(2ab=2(4\sqrt{3})(3\sqrt{2})=24\sqrt{6}\). Adding gives \(48+24\sqrt{6}+18=66+24\sqrt{6}\). Option A is correct because it contains both square terms and the middle term. Option B, \(30+24\sqrt{6}\), has the wrong constant term; 48 and 18 add to 66, not 30. Option C, \(66+12\sqrt{6}\), halves the middle term incorrectly. Option D, \(48+18\sqrt{6}\), does not correctly combine either the constant and radical parts. Remember: square both terms, calculate twice their product, and then add all three parts.

\(x^2=(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}\). 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.