Real Numbers

Since \(6.25=2.5^2\), we have \(\sqrt{6.25}=2.5\). Similarly, \(0.64=0.8^2\), so \(\sqrt{0.64}=0.8\). Therefore, the value is \(2.5-0.8=1.7\), making option A correct. Exam tip: When finding the square root of a decimal, check whether it is the square of a simple decimal number before calculating.

The first term is ( \frac{5}{7} ) and the second is ( \frac{4}{9} ). The sum is ( \frac{45}{63}+\frac{28}{63}=\frac{73}{63} ).

( \sqrt{\frac{100}{169}}=\frac{10}{13} ) and ( \sqrt{\frac{9}{64}}=\frac{3}{8} ). The difference is ( \frac{80}{104}-\frac{39}{104}=\frac{41}{104} ).

Since (6^2<43<7^2), ( \sqrt{43} ) lies between (6) and (7). Since (43) is not a perfect square, it is irrational.

Since \(63<64\), we have \(\sqrt{63}<\sqrt{64}=8\). Therefore, \(8\) is greater. Option A is incorrect because \(\sqrt{63}\) is slightly less than 8, and option C is incorrect because the two numbers are not equal. Exam tip: For non-negative numbers, comparing their squares is a quick way to compare their square roots.

(10^2=100) and (11^2=121), so ( \sqrt{101} ) lies between (10) and (11). Compare nearby perfect squares.

(11^2=121) and (12^2=144), so ( \sqrt{130} ) lies between (11) and (12). Check the range by squaring.

The greater number is -8.1, so option B is correct. First estimate the square root: since 8^2=64 and 8.1^2=65.61,
sqrt(65)
is slightly greater than 8 and approximately 8.062. Therefore -
sqrt(65)
is approximately -8.062. On a number line, among negative numbers the number closer to zero is greater. Because -8.062 is closer to zero than -8.1, we have -
sqrt(65)
>-8.1. Option A is wrong because it chooses the less negative number incorrectly. Option B is correct. Option C is wrong because the values are not equal; 65 is not exactly 8.1 squared. Option D is wrong because both numbers are negative, not positive. Memory trick: when comparing negative numbers, compare their positive sizes and reverse the sign order; the smaller absolute value is greater.

In 4.616161..., the block 61 repeats indefinitely, so it is a recurring decimal. Every recurring decimal is rational; in fact, 4.616161... = 457/99. Therefore, it is a rational real number. It is not an integer because its decimal part is not zero. Exam tip: terminating and recurring decimals are rational, whereas non-terminating, non-recurring decimals are irrational.

A non-terminating decimal without regular repetition is irrational. Only fixed repeating decimals are rational.

\(\frac{37}{54}\) is already in lowest terms because the greatest common divisor of 37 and 54 is 1. The denominator is \(54=2\times3^3\), so it contains the factor 3 along with 2. A reduced fraction has a terminating decimal only when its denominator contains no prime factors other than 2 and 5. Therefore, \(\frac{37}{54}=0.685185185\ldots\) is non-terminating and repeating. Thus, the ‘Terminating’ option is incorrect. Exam tip: If a reduced denominator has any prime factor other than 2 or 5, the decimal expansion is non-terminating repeating.

\(\sqrt{2}\) is an irrational number, so its decimal expansion is non-terminating and non-repeating. \(\frac{7}{16}\) has a terminating decimal expansion, while \(0.363636\ldots\) is non-terminating but repeating; both are rational numbers. Exam tip: A non-terminating, non-repeating decimal expansion identifies an irrational number.

The fraction \(\frac{13}{375}\) is already in lowest terms, and \(375=3\times5^3\). A rational number \(\frac{p}{q}\) has a terminating decimal expansion only when, in lowest terms, the prime factors of \(q\) are only 2 and 5. Since 3 is also a factor here, the decimal expansion is non-terminating recurring, not non-terminating non-recurring. Exam tip: First reduce the fraction, then check whether the denominator has only 2 and 5 as prime factors.

The number of zeros between successive 1s keeps increasing: 1 zero, then 2, then 3, then 4, and so on. Hence, no fixed block of digits repeats periodically. Its decimal expansion is non-terminating and non-repeating, so the number is irrational. The repeated occurrence of a digit alone does not make a number rational; a fixed repeating block is required. Exam tip: A rational number has either a terminating decimal expansion or a non-terminating recurring decimal expansion.

Both \(\sqrt{5}\) and \(-\sqrt{5}\) are irrational numbers, but their sum is \(0\), which is rational. Hence, the statement that the sum of two irrational numbers is always irrational is false. In option D, the sum is \(3\sqrt{7}\), which is still irrational, so it does not disprove the statement. Exam tip: To disprove a statement containing words such as “always,” one valid counterexample is enough.

Direct answer: option A,
\(\frac{19+8\sqrt{3}}{13}\). To rationalise a denominator containing a surd, multiply numerator and denominator by its conjugate. The conjugate of \(4-\sqrt{3}\) is \(4+\sqrt{3}\). Thus \(\frac{4+\sqrt3}{4-
\sqrt3}\times\frac{4+\sqrt3}{4+
\sqrt3}=\frac{(4+
\sqrt3)^2}{4^2-(\sqrt3)^2}=\frac{16+8\sqrt3+3}{16-3}=\frac{19+8\sqrt3}{13}\). Option A has this exact result. Option B reverses the 13 and 19 positions, so it is incorrect. Option C says 1, but the numerator and denominator are not equal. Option D has an incorrect expansion of the square. Memory cue: conjugates make the denominator a difference of squares.

Multiplying by the conjugate ( \sqrt{11}+3 ) makes the denominator (11-9=2). Use difference of squares in the conjugate method.

\(\frac{7}{11}=0.636363\ldots\) has an infinite decimal expansion, but the block 63 repeats. Hence, it is a non-terminating recurring decimal and is rational. In contrast, \(\sqrt{2}\) and \(\pi\) have non-terminating, non-recurring decimal expansions, so they are irrational. Exam tip: Do not classify an infinite decimal as irrational until you check whether it repeats.

\(\sqrt{36}=6\) and \(\sqrt{121}=11\). Therefore, \(\sqrt{36}-\sqrt{121}=6-11=-5\). The absolute value of a number is never negative, so \(\left|-5\right|=5\). Option C is a close distractor because it is the value before applying the absolute value. Exam tip: evaluate the square roots first, then take the absolute value of the difference.

In real numbers, \(\sqrt{a^2}=|a|\), because the principal square root is always non-negative. Therefore, \(\sqrt{(-12)^2}=|-12|=12\). The option \(-12\) is incorrect: although its square is 144, the principal square root of 144 is 12. Exam tip: Do not simplify \(\sqrt{x^2}\) directly as \(x\); write it as \(|x|\).

Use \(\sqrt{x^2}=|x|\), not simply \(x\) in every case. Since \(x=-15\), \(\sqrt{x^2}=|-15|=15\). Therefore, \(\sqrt{x^2}-x=15-(-15)=30\). The option \(0\) results from the incorrect assumption that \(\sqrt{x^2}=x\) for a negative value of \(x\). Exam tip: the principal square root is always non-negative, so write \(\sqrt{x^2}=|x|\).

Multiplying by the conjugate (6-\sqrt{35}) makes the denominator (36-35=1). So the rationalised form is (6-\sqrt{35}).

( \sqrt{500}=10\sqrt{5} ), ( \sqrt{180}=6\sqrt{5} ), and ( \sqrt{45}=3\sqrt{5} ). Therefore (10\sqrt{5}-6\sqrt{5}+3\sqrt{5}=7\sqrt{5}).

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 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.