Real Numbers

For \(\frac{11}{12}\), the denominator is \(12=2^2\times3\). Since it contains the prime factor 3, its decimal expansion is non-terminating recurring: \(0.91\overline{6}\). In contrast, the denominators 8, 20 and 125 have only 2 and/or 5 as prime factors, so their decimal expansions terminate. Exam tip: First reduce a fraction to its simplest form, then check the prime factors of its denominator.

Both \(\sqrt{5}\) and \(-\sqrt{5}\) are irrational, but \(\sqrt{5}+(-\sqrt{5})=0\). Since 0 is rational, this is a counterexample to the claim that the sum of two irrational numbers is always irrational. Option A does give an irrational sum, but it does not disprove an “always” statement. Exam tip: one counterexample is enough to disprove a universal statement.

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.

( \sqrt{65}\approx8.06 ), so ( -\sqrt{65}\approx-8.06 ). ( -8.06 ) is greater than ( -8.1 ).

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.

Multiply by the conjugate (4+\sqrt{3}). The numerator is ((4+\sqrt{3})^2=19+8\sqrt{3}) and the denominator is (13).

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.