Real Numbers

( \sqrt{\frac{81}{121}}=\frac{9}{11} ) and ( \sqrt{\frac{4}{49}}=\frac{2}{7} ). The difference is ( \frac{63}{77}-\frac{22}{77}=\frac{41}{77} ).

Since (5^2<31<6^2), ( \sqrt{31} ) lies between (5) and (6). Since (31) is not a perfect square, it is irrational.

Compare the squares of the two positive numbers: \((\sqrt{45})^2=45\), whereas \(\left(\frac{13}{2}\right)^2=\frac{169}{4}=42.25\). Since \(45>42.25\), we have \(\sqrt{45}>\frac{13}{2}\). Therefore, option A is correct. Exam tip: For positive square roots or positive numbers, comparing their squares is often quicker than using decimal approximations.

(8^2=64) and (9^2=81), so ( \sqrt{68} ) lies between (8) and (9). Compare nearby perfect squares.

The governing concept is locating a principal square root by comparing the number with consecutive perfect squares. The nearby perfect squares are 9² = 81 and 10² = 100. Because 81 < 86 < 100, taking the positive square root preserves the order and gives 9 < √86 < 10. Therefore √86 lies between 9 and 10, making option C correct. Its approximate value is about 9.27, which provides an additional check. The interval 7 to 8 is too low because 8² is only 64. The interval 8 to 9 is also too low because 9² is 81. The interval 10 to 11 is too high because √86 is less than 10.

( \sqrt{50}\approx7.07 ), so ( -\sqrt{50}\approx-7.07 ). It is greater than ( -7.2 ) because it is closer to zero.

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

A rational number has a terminating decimal when, after reducing the fraction, the denominator has no prime factors other than 2 and 5. This happens because powers of 2 and 5 can be multiplied to make a power of 10. Therefore the decimal form in this question ends after a finite number of digits, so option A is correct. It is not irrational or undefined.

Here, the denominator is 80, and the factorisation is \(80=2^4\times5\). The numerator 21 has no common factor with 80, so the fraction is already in simplest form. Since its denominator contains only 2 and 5, \(21/80=0.2625\), which terminates. Thus the correct statement is that it is a terminating decimal.

The diagonal of the square is \(\sqrt{1^2+1^2}=\sqrt{2}\) cm. Since 2 is not a perfect square, \(\sqrt{2}\) cannot be expressed exactly as a ratio of two integers; therefore, it is irrational. It is neither an integer nor a natural number, so options C and D are incorrect. Exam tip: the square root of a non-perfect square is generally irrational.

The diagonal of a square is given by side × \(\sqrt{2}\), so its length is \(5\sqrt{2}\) cm. Since \(\sqrt{2}\) is irrational and 5 is a non-zero rational number, \(5\sqrt{2}\) is also irrational. Therefore, option B is correct. Exam tip: multiplying an irrational number by a non-zero rational number keeps the result irrational.

In the decimal expansion \(0.101001000100001\ldots\), the number of zeros between successive 1s keeps increasing, so no fixed block repeats regularly. Hence it is a non-terminating, non-recurring decimal and therefore irrational. The use of only the digits 0 and 1 does not make a number rational. Exam tip: a decimal is rational only if it terminates or eventually repeats a fixed pattern.

The decimal expansion \(0.101001000100001\ldots\) is non-terminating, and the number of zeros between successive 1s keeps increasing. Hence, no fixed block of digits repeats regularly. A rational number has either a terminating or an infinitely repeating decimal expansion, so this number is irrational. Exam tip: an infinite decimal is not automatically irrational; an infinite recurring decimal, such as \(0.333\ldots\), is rational.

The two expressions are conjugates, so use \((a+b)(a-b)=a^2-b^2\): \((3+\sqrt{14})(3-\sqrt{14})=3^2-(\sqrt{14})^2=9-14=-5\). Option A results from missing the negative sign. Exam tip: whenever conjugate pairs appear, apply the difference-of-squares identity directly.

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.

Multiplying by the conjugate ( \sqrt{7}+2 ) makes the denominator (7-4=3). Use difference of squares in the conjugate method.

Since \(\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}\), we get \(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\), not \(\sqrt{10}\). Because \(\sqrt{2}\) is irrational, its product with the non-zero rational number 3 is also irrational. Exam tip: simplify radicals first and combine only like radical terms; in general, \(\sqrt{a}+\sqrt{b}\neq\sqrt{a+b}\).

\(\sqrt{25}=5\) and \(\sqrt{100}=10\). Therefore, \(\left|\sqrt{25}-\sqrt{100}\right|=|5-10|=|-5|=5\). Option B results from adding the two values instead of subtracting them, while option C ignores that an absolute value cannot be negative. Exam tip: the absolute value of any real number is always non-negative.

For real numbers,
\(\sqrt{x^2}=|x|\). Therefore,
\(\sqrt{(-9)^2}=|-9|=9\). The principal square root is always non-negative, so
\(-9\) is not the answer. Exam tip: simplify
\(\sqrt{x^2}\) as
\(|x|\), not automatically as
\(x\).

For every real number \(x\), \(\sqrt{x^2}=|x|\), not always \(x\). Thus, for \(x=-8\), \(\sqrt{x^2}+x=|-8|+(-8)=8-8=0\). Option B results from the common mistake of taking \(\sqrt{x^2}=x\) directly. Exam tip: In such questions, first rewrite \(\sqrt{x^2}\) as the absolute value \(|x|\).

Multiplying by the conjugate (5-2\sqrt{6}) makes the denominator (25-24=1). So the rationalised form is (5-2\sqrt{6}).