Real Numbers

A number with a non-terminating, non-recurring decimal expansion is irrational. The decimal expansion of every rational number either terminates or repeats, so option A is not correct. Integers and natural numbers are special types of rational numbers. Exam tip: Link “non-terminating, non-recurring” directly with irrational numbers.

( \sqrt{30}\approx5.477 ) and ( \frac{11}{2}=5.5 ). So ( -\sqrt{30} ) is closer to zero and is greater.

( \frac{39}{312}=\frac{1}{8} ) and (8=2^3). A decimal terminates when the denominator has only (2) and (5) as prime factors.

First simplify the fraction by dividing its numerator and denominator by their common factor 11: \\(\frac{22}{77}=\frac{2}{7}\\). A rational number has a terminating decimal expansion only when, after simplification, its denominator has no prime factors other than 2 and 5. The denominator 7 does not satisfy this condition. Therefore, the decimal expansion of \\(\frac{2}{7}\\) cannot end; instead, its digits repeat in a recurring pattern.

Indeed, \\(\frac{2}{7}=0.285714285714\ldots\\), where the block 285714 repeats indefinitely. Hence option B, non-terminating repeating decimal, is correct. It is not terminating because the denominator is not of the required form, and it is not non-repeating irrational because \\(\frac{2}{7}\\) is still a rational number. It is also not an integer because the numerator is not a multiple of the denominator.

For \(\frac{11}{18}\), the prime factorisation of the denominator is \(18=2\times 3^2\). In lowest form, the denominator contains the prime factor 3 in addition to 2 and 5, so its decimal expansion is non-terminating recurring. In contrast, the denominators of \(\frac{7}{40}\), \(\frac{13}{125}\), and \(\frac{9}{64}\) contain only 2 and/or 5, so they have terminating decimal expansions. Exam tip: In lowest form, a rational number has a non-terminating recurring decimal when its denominator has a prime factor other than 2 or 5.

The block (23) repeats, so the decimal is repeating. Every repeating decimal is a rational real number.

\(\sqrt{12}=\sqrt{4\times3}=2\sqrt{3}\), \(\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}\), and \(\sqrt{48}=\sqrt{16\times3}=4\sqrt{3}\). Therefore, the sum is \(\sqrt{3}+2\sqrt{3}+3\sqrt{3}+4\sqrt{3}=(1+2+3+4)\sqrt{3}=10\sqrt{3}\). \(8\sqrt{3}\) would result from adding the coefficients incorrectly. Exam tip: factor out the greatest perfect square from each radicand before combining like surds.

Use the difference-of-reciprocals structure and take a common denominator. The denominator is \\( (\sqrt{5}-2)(\sqrt{5}+2)=(\sqrt{5})^2-2^2=5-4=1\\). The numerator is the second denominator minus the first: \\( (\sqrt{5}+2)-(\sqrt{5}-2)=4\\). Therefore the whole expression equals \\(\frac{4}{1}=4\\). The radical terms cancel in the numerator, so no irrational term remains.

Option A is correct. A careful sign is important: because the original expression subtracts the second reciprocal from the first, the numerator becomes \\(\sqrt{5}+2-(\sqrt{5}-2)\\), not the reverse. The denominator is positive and equal to 1, so the result is exactly 4. The alternatives involving \\(\sqrt{5}\\), \\(2\sqrt{5}\\), or 8 do not follow from this simplification.

The governing concept is addition of algebraic fractions using a common denominator and the conjugate identity (a − b)(a + b) = a² − b². The common denominator is (3 − √7)(3 + √7) = 3² − (√7)² = 9 − 7 = 2. The numerator is the sum of the opposite denominators: (3 + √7) + (3 − √7) = 6, because the irrational terms cancel. Therefore the complete expression equals 6/2 = 3, so option A is correct. Option B is only the numerator before division by the common denominator. Option C comes from an incorrect division, while option D fails to combine the two fractions and ignores the cancellation.

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.

The governing concept is simplifying each radical by extracting its largest perfect-square factor and then combining like surds with their signs. We have √12 = √(4×3) = 2√3, √75 = √(25×3) = 5√3, √48 = √(16×3) = 4√3, and √27 = √(9×3) = 3√3. Substitution gives 2√3 + 5√3 − 4√3 − 3√3 = (2 + 5 − 4 − 3)√3 = 0√3 = 0. Hence option A is correct. The negative signs before √48 and √27 must be retained. Options B, C, and D result from ignoring one or both negative signs, adding all coefficients, or making an arithmetic error. The cancellation is valid only after all radicals have been expressed as multiples of √3.

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

The governing concept is using a conjugate to rationalise the reciprocal of a surd expression. Since x = √5 + 2, multiply the numerator and denominator of 1/x by √5 − 2: 1/x = (√5 − 2)/[(√5 + 2)(√5 − 2)] = (√5 − 2)/(5 − 4) = √5 − 2. Now add this to x: x + 1/x = (√5 + 2) + (√5 − 2) = 2√5. Therefore option A is correct. Option B incorrectly combines constants and irrational terms, option C leaves out one √5 term, and option D is merely the given value of x rather than the requested sum. The cancellation of +2 and −2 is the key step.

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.

Since ( \sqrt{20}>3 ), (3-\sqrt{20}) is negative. Absolute value changes it to ( \sqrt{20}-3 ).

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.