Real Numbers

The governing concept is rationalising a denominator by multiplying the numerator and denominator by the conjugate of the denominator. The conjugate of 2 − √3 is 2 + √3. Thus, (2 + √3)/(2 − √3) = [(2 + √3)(2 + √3)]/[(2 − √3)(2 + √3)]. The numerator is (2 + √3)² = 4 + 4√3 + 3 = 7 + 4√3. The denominator is a difference of squares: 2² − (√3)² = 4 − 3 = 1. Consequently, the value is (7 + 4√3)/1 = 7 + 4√3, so option A is correct. Option B has the wrong sign in the irrational term. Option C ignores the numerator expansion and cannot result from cancellation because the numerator and denominator are not identical. Option D does not follow from either the square identity or the conjugate product.

The governing concept is rationalisation using the conjugate of a binomial denominator. The conjugate of 3 − √2 is 3 + √2, so multiply both numerator and denominator by 3 + √2. The numerator becomes (3 + √2)² = 3² + 2(3)(√2) + (√2)² = 9 + 6√2 + 2 = 11 + 6√2. The denominator becomes (3 − √2)(3 + √2) = 3² − (√2)² = 9 − 2 = 7. Therefore the rationalised value is (11 + 6√2)/7, which is option A. Option B reverses the numerator and denominator results. Option C incorrectly assumes cancellation even though the two binomials are not identical. Option D has an incorrect expansion because it omits the factor 2 in the middle term and uses the wrong coefficient of √2.

(2\sqrt{147}=14\sqrt{3}), (3\sqrt{75}=15\sqrt{3}), and ( \sqrt{300}=10\sqrt{3} ). The coefficient is (14-15+10=9), so the value is (9\sqrt{3}).

(4\sqrt{72}=24\sqrt{2}), (2\sqrt{200}=20\sqrt{2}), and (5\sqrt{50}=25\sqrt{2}). Therefore the result should be (19\sqrt{2}).

(2\sqrt{8}=4\sqrt{2}) and ( \sqrt{50}=5\sqrt{2} ), so the bracket is ( -\sqrt{2} ). ( \sqrt{18}\times(-\sqrt{2})=-\sqrt{36}=-6 ).

The fraction \(\frac{77}{480}\) is already in lowest terms, and \(480=2^5\times3\times5\). A rational number has a terminating decimal expansion only when the denominator in lowest form has no prime factors other than 2 and 5. Since 3 is also present, its decimal expansion is non-terminating recurring. Option C is incorrect because non-terminating non-recurring decimals represent irrational numbers. Exam tip: First reduce the fraction, then check the prime factors of its denominator.

( (5\sqrt{3})^2=75 ) and ( (2\sqrt{5})^2=20 ). The middle term (2\times5\sqrt{3}\times2\sqrt{5}=20\sqrt{15}) is subtracted.

\(\sqrt{2}+\sqrt{3}\) is irrational. If it were equal to a rational number \(r\), then on squaring we would get \(r^2=5+2\sqrt{6}\). This would make \(\sqrt{6}\) rational, which is impossible; hence the sum is irrational. In option B, \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), so it is rational. Exam tip: To test a sum of square roots, assume it is rational and square both sides.

Multiplying by the conjugate gives denominator (18-20=-2). So the form can be written as ( \frac{2\sqrt{5}-3\sqrt{2}}{2} ).

Multiplying by the conjugate ( \sqrt{7}+\sqrt{5} ) gives denominator (7-5=2). The numerator also has (2), so the answer is ( \sqrt{7}+\sqrt{5} ).

Use the conjugate-pair identity with a common denominator. Let a=3+√2 and b=3−√2. Then a/b+b/a=(a²+b²)/(ab). The denominator is ab=(3+√2)(3−√2)=3²−(√2)²=9−2=7. For the numerator, expand a²+b²: (3+√2)²+(3−√2)²=(9+6√2+2)+(9−6√2+2)=22. The irrational cross-terms cancel because they have opposite signs. Hence the original value is 22/7, so option A is correct. Option B is half of the correct numerator, usually caused by dropping one squared term. Option C results from an incomplete or incorrect expansion. Option D ignores the denominator and the actual values of the conjugate expressions, so it cannot be correct. The denominator is nonzero, so the operations are valid.

Use the identity \((a+b)^2-(a-b)^2=4ab\). Here, \(a=\sqrt{5}\) and \(b=\sqrt{3}\). Therefore, the value is \(4\times\sqrt{5}\times\sqrt{3}=4\sqrt{15}\). The option 16 may result from ignoring the effect of the middle terms in the two squares. Exam tip: For expressions of this form, applying the identity is quicker and safer than expanding both squares separately.

( \sqrt{12}=2\sqrt{3} ) and ( \sqrt{27}=3\sqrt{3} ). Dividing (5\sqrt{3}) by ( \sqrt{3} ) gives (5).

Adding a rational number to an irrational number always gives an irrational number. Hence \(x+5\) is irrational because 5 is rational. Option B is not always true: if \(x=\sqrt{2}\), then \(x^2=2\), which is rational. In options C and D, \(x\neq0\), so the value is 1, a rational number. Exam tip: irrational ± rational is always irrational.

The correct answer is \(\frac{21}{84}\), because the fraction must first be reduced to lowest terms: \(\frac{21}{84}=\frac{1}{4}=0.25\). Its denominator in lowest form is \(4=2^2\), so its decimal expansion terminates. The student's error is checking prime factors before cancelling common factors. In contrast, \(\frac{7}{12}\) has a factor of \(3\) in its denominator in lowest terms, so its decimal expansion is non-terminating recurring. Exam tip: Always reduce a fraction to lowest terms before applying the terminating-decimal rule.

In \(0.101001000100001\ldots\), the number of zeros between successive 1s keeps increasing, so no fixed block repeats. Its decimal expansion is non-terminating and non-repeating; therefore, it is irrational. Although \(0.272727\ldots\) is non-terminating, the block 27 repeats, so it is rational. Also, \(0.125\) terminates and \(\sqrt{121}=11\) is an integer. Exam tip: A non-terminating decimal is irrational only when it has no repeating pattern.

( \frac{11}{5}=2.2 ), ( \sqrt{5}\approx2.236 ), and then (2.3). So the ascending order is ( \frac{11}{5}<\sqrt{5}<2.3 ).

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.

Direct answer: option B, rational real number. A real number is rational if it can be written as a fraction of integers. A decimal that repeats forever in a fixed block is always rational. Here 23 repeats: \(5.232323\ldots=5.\overline{23}\). Therefore it can be converted into a fraction, so it is rational. Option A is wrong because an irrational decimal neither terminates nor repeats regularly. Option B is correct because this decimal repeats regularly. Option C is wrong because an integer has no nonzero decimal part; this number lies between 5 and 6. Option D is wrong because the expression is defined and has a clear decimal value. Exam cue: terminating or recurring decimal means rational; non-terminating non-recurring means irrational.

\(\sqrt{12}=\sqrt{4\times3}=2\sqrt{3}\), \(\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}\), 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.