Rational and Irrational Numbers

Q1: Which of the following is a rational number?

Step 1: Examine option (a) \(\pi\).
\(\pi\) is a well-known irrational number because its decimal expansion is non-terminating and non-repeating.
Step 2: Examine option (b) \(\sqrt{2}\).
The square root of any prime number is irrational. \(\sqrt{2}\) cannot be expressed as a simple fraction.
Step 3: Examine option (c) 3.4.
3.4 is a terminating decimal. It can be written as \(\frac{34}{10}\) or \(\frac{17}{5}\). Since it fits the \(\frac{p}{q}\) form, it is rational.
Step 4: Examine option (d) 1.010010001…
The pattern shows that the number of zeros increases each time. This decimal is non-terminating and non-repeating, making it irrational.
Answer: c. 3.4

Q2: Which of the following is an irrational number?

Step 1: Examine option (a) 2.7.
2.7 is a terminating decimal. It can be written as \(\frac{27}{10}\). Therefore, it is a rational number.
Step 2: Examine option (b) \(2.7\overline{2}\).
This is a non-terminating repeating decimal. All repeating decimals can be expressed as fractions. Therefore, it is a rational number.
Step 3: Examine option (c) \(\sqrt{11}\).
The square root of any prime number (like 11) is always irrational because its decimal expansion is non-terminating and non-repeating.
Step 4: Examine option (d) \(\frac{2}{7}\).
This is already in the form of \(\frac{p}{q}\), where p and q are integers and q ≠ 0. Therefore, it is a rational number.
Answer: c. \(\sqrt{11}\)

Q3: Which of the following is a prime number?

Step 1: Examine option (a) 51.
The sum of digits is 5 + 1 = 6. Since 6 is divisible by 3, 51 is divisible by 3 (3 × 17 = 51). It is not a prime number.
Step 2: Examine option (b) 57.
The sum of digits is 5 + 7 = 12. Since 12 is divisible by 3, 57 is divisible by 3 (3 × 19 = 57). It is not a prime number.
Step 3: Examine option (c) 71.
71 is not divisible by 2 (it’s odd), 3 (7+1=8), or 5 (doesn’t end in 0 or 5). Checking further, it is not divisible by 7. It has no factors other than 1 and 71.
Step 4: Examine option (d) 81.
81 is a perfect square (9 × 9 = 81) and is also divisible by 3. It is not a prime number.
Answer: c. 71

Q4: When \(8.\overline{32}\) is expressed as a vulgar fraction, then it becomes:

Step 1: Let \(x = 8.3232…\) (Eq. 1)
Step 2: Since two digits are repeating, multiply both sides by 100.
\(100x = 832.3232…\) (Eq. 2)
Step 3: Subtract Eq. 1 from Eq. 2.
\(100x – x = 832.3232… – 8.3232…\)
\(99x = 824\)
Step 4: Solve for \(x\).
\(x = \frac{824}{99}\)
Answer: b. \(\frac{824}{99}\)

Q5: 0.3 when expressed as a ratio of two integers, becomes:

Step 1: Convert 0.3 to a fraction.
0.3 can be expressed as \( \frac{3}{10}\).
Step 2: To express \(\frac{3}{10}\) as a ratio of two integers, we can multiply both the numerator and denominator by 33 to find a common denominator with the options given.
\(\frac{3}{10} \times \frac{33}{33} = \frac{99}{330}\)
Step 3: Now we can compare \(\frac{99}{330}\) with the options provided.
– a. \(\frac{103}{330}\)
– b. \(\frac{52}{165}\)
– c. \(\frac{103}{111}\)
– d. \(\frac{104}{333}\)
Step 4: To find a common ratio, we can simplify the options:
– a. \(\frac{103}{330}\) is already in simplest form.
– b. \(\frac{52}{165}\) can be simplified to \(\frac{52 \times 13}{165 \times 13} = \frac{4}{15}\).
– c. \(\frac{103}{111}\) is already in simplest form.
– d. \(\frac{104}{333}\) is already in simplest form.
Step 5: None of the options match \(\frac{3}{10}\) or \(\frac{99}{330}\). However, we can check if any of the options can be converted back to a decimal.
– a. \(\frac{103}{330} \approx 0.3121\)
– b. \(\frac{52}{165} \approx 0.3152\)
– c. \(\frac{103}{111} \approx 0.9270\)
– d. \(\frac{104}{333} \approx 0.3123\)
Step 6: The closest option to 0.3 is option a. \(\frac{103}{330}\).
Answer: a. \(\frac{103}{330}\)

Q6: Only by inspecting the prime factors of the denominator, state which of the following fractions will be a terminating decimal?

a. \(\frac{7}{12}\)

Step 1: Identify the denominator, which is 12.
Step 2: Prime factorise the denominator: \(12 = 2^2 \times 3\).
Step 3: Since the denominator contains the prime factor 3 (other than 2 or 5), it is a non-terminating recurring decimal.
Answer: Non-terminating

b. \(\frac{2}{15}\)

Step 1: Identify the denominator, which is 15.
Step 2: Prime factorise the denominator: \(15 = 3 \times 5\).
Step 3: The presence of the prime factor 3 makes this decimal non-terminating.
Answer: Non-terminating

c. \(\frac{3}{16}\)

Step 1: Identify the denominator, which is 16.
Step 2: Prime factorise the denominator: \(16 = 2^4\).
Step 3: Since the denominator consists only of powers of 2, it results in a terminating decimal.
Answer: c. Terminating

d. \(\frac{4}{21}\)

Step 1: Identify the denominator, which is 21.
Step 2: Prime factorise the denominator: \(21 = 3 \times 7\).
Step 3: The prime factors 3 and 7 (other than 2 or 5) make this non-terminating.
Answer: Non-terminating

Q7: Only by inspecting the prime factors of the denominators, state which of the following fractions will be a recurring decimal?

a. \(\frac{7}{16}\)

Step 1: Factorise the denominator: \(16 = 2^4\).
Step 2: Since the denominator has only 2 as a prime factor, it is a terminating decimal.
Answer: Terminating

b. \(\frac{8}{51}\)

Step 1: Factorise the denominator: \(51 = 3 \times 17\).
Step 2: The prime factors are 3 and 17 (other than 2 or 5).
Step 3: Therefore, it results in a non-terminating recurring decimal.
Answer: b. Recurring Decimal

c. \(\frac{3}{25}\)

Step 1: Factorise the denominator: \(25 = 5^2\).
Step 2: Since the denominator has only 5 as a prime factor, it is a terminating decimal.
Answer: Terminating

d. \(\frac{11}{20}\)

Step 1: Factorise the denominator: \(20 = 2^2 \times 5\).
Step 2: The prime factors are only 2 and 5.
Step 3: Therefore, it is a terminating decimal.
Answer: Terminating

Q8: The number which is to be subtracted from \(\sqrt{72}\) to get \(\sqrt{32}\) is:

Step 1: Simplify \(\sqrt{72}\) by finding its factors.
\(\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\)
Step 2: Simplify \(\sqrt{32}\) by finding its factors.
\(\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}\)
Step 3: Let the required number be \(x\). According to the question:
\(6\sqrt{2} – x = 4\sqrt{2}\)
Step 4: Rearrange the equation to solve for \(x\):
\(x = 6\sqrt{2} – 4\sqrt{2}\)
\(x = 2\sqrt{2}\)
Answer: d. 2\(\sqrt{2}\)

Q9: If \(x=3+2\sqrt{2}\), then find the value of \(x+\frac{1}{x}\)

Step 1: Write the reciprocal of \(x\) and multiply the numerator and denominator by its conjugate \(3 – 2\sqrt{2}\).
\(\frac{1}{3+2\sqrt{2}} \times \frac{3-2\sqrt{2}}{3-2\sqrt{2}}\)
Step 2: Use the identity \((a+b)(a-b) = a^2 – b^2\) for the denominator.
Denominator: \(3^2 – (2\sqrt{2})^2 = 9 – (4 \times 2) = 9 – 8 = 1\).
So, \(\frac{1}{x} = 3 – 2\sqrt{2}\).
Step 3: Add the value of \(x\) and \(\frac{1}{x}\).
\(x + \frac{1}{x} = (3 + 2\sqrt{2}) + (3 – 2\sqrt{2})\)
Step 4: Cancel out the irrational terms (\(2\sqrt{2}\) and \(-2\sqrt{2}\)).
\(3 + 3 = 6\).
Answer: c. 6

Q10: If \(x=2-\sqrt{2}\) , then find the value of \(x-\frac{1}{x}\)

Step 1: Write the reciprocal of \(x\) and rationalize the denominator by multiplying with the conjugate \(2 + \sqrt{2}\).
\(\frac{1}{2-\sqrt{2}} \times \frac{2+\sqrt{2}}{2+\sqrt{2}}\)
Step 2: Use the identity \((a-b)(a+b) = a^2 – b^2\) for the denominator.
Denominator: \(2^2 – (\sqrt{2})^2 = 4 – 2 = 2\).
So, \(\frac{1}{x} = \frac{2+\sqrt{2}}{2} = 1 + \frac{\sqrt{2}}{2}\).
Step 3: Subtract the values of \(x\) and \(\frac{1}{x}\).
\(x – \frac{1}{x} = (2 – \sqrt{2}) + (1 – \frac{\sqrt{2}}{2})\)
Step 4: Combine the rational and irrational terms.
\((2 – 1) + (-\sqrt{2} – \frac{\sqrt{2}}{2})\)
\(1 – \frac{3\sqrt{2}}{2}\).
Answer: 1 – \(\frac{3\sqrt{2}}{2}\)

Q11: If \(x=5+2\sqrt{6}\), find the value of \(x^2 + \frac{1}{x^2}\)

Step 1: Write \(\frac{1}{x}\) and rationalise the denominator by multiplying with the conjugate \(5 – 2\sqrt{6}\).
\(\frac{1}{5 + 2\sqrt{6}} \times \frac{5 – 2\sqrt{6}}{5 – 2\sqrt{6}}\)
Step 2: Use the identity \((a + b)(a – b) = a^2 – b^2\) for the denominator.
Denominator: \(5^2 – (2\sqrt{6})^2 = 25 – (4 \times 6) = 25 – 24 = 1\).
So, \(\frac{1}{x} = 5 – 2\sqrt{6}\).
Step 3: Add the values of \(x\) and \(\frac{1}{x}\).
\(x + \frac{1}{x} = (5 + 2\sqrt{6}) + (5 – 2\sqrt{6})\)
\(x + \frac{1}{x} = 10\).
Step 4: Use the algebraic identity: \(x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 – 2\).
Substitute the value \(x + \frac{1}{x} = 10\) into the identity:
\(x^2 + \frac{1}{x^2} = (10)^2 – 2\)
Step 5: Perform the final calculation.
\(100 – 2 = 98\).
Answer: a. 98

Q12: Two rational numbers between \(-\frac{3}{7}\) and \(-\frac{1}{7}\) are:

Step 1: Multiply the numerator and denominator of both fractions by 2 to get the denominator 14.
\(-\frac{3}{7} = \frac{-3 \times 2}{7 \times 2} = -\frac{6}{14}\)
\(-\frac{1}{7} = \frac{-1 \times 2}{7 \times 2} = -\frac{2}{14}\)
Step 2: Look for integers between -6 and -2 for the numerator.
The integers are -5, -4, and -3.
Step 3: Write the corresponding rational numbers with denominator 14:
\(-\frac{5}{14}, -\frac{4}{14}, \text{ and } -\frac{3}{14}\).
Step 4: Compare these results with the given options.
Option (d) contains \(-\frac{4}{14}\) and \(-\frac{3}{14}\), which lie between the two values.
Answer: d. \(-\frac{4}{14}, -\frac{3}{14}\)

Q13: The correct ascending order of \(\sqrt{3}, \sqrt[3]{6}, \sqrt[4]{7}\) is:

Step 1: Identify the indices of the roots: 2, 3, and 4. The L.C.M. of 2, 3, and 4 is 12.
Step 2: Convert each surd to have an index of 12:
\(\sqrt{3} = 3^{1/2} = 3^{6/12} = \sqrt[12]{3^6} = \sqrt[12]{729}\)
\(\sqrt[3]{6} = 6^{1/3} = 6^{4/12} = \sqrt[12]{6^4} = \sqrt[12]{1296}\)
\(\sqrt[4]{7} = 7^{1/4} = 7^{3/12} = \sqrt[12]{7^3} = \sqrt[12]{343}\)
Step 3: Compare the values inside the roots: \(343 < 729 < 1296\).
Step 4: Arrange the original surds in the same order:
\(\sqrt[12]{343} < \sqrt[12]{729} < \sqrt[12]{1296}\)
\(\Rightarrow \sqrt[4]{7} < \sqrt{3} < \sqrt[3]{6}\)
Answer: b. \(\sqrt[4]{7}, \sqrt{3}, \sqrt[3]{6}\)

Q14: The mixed surd for \(\sqrt[3]{432}\) is:

Step 1: Find the prime factors of 432.
432 = 2 × 216
432 = 2 × 2 × 108
432 = 2 × 2 × 2 × 54
432 = 2 × 2 × 2 × 2 × 27
432 = 2^4 × 3^3
432 = (2^3 × 3^3) × 2
Step 2: Apply the cube root to the factors.
\(\sqrt[3]{432} = \sqrt[3]{2^3 \times 3^3 \times 2}\)
Step 3: Take out the terms that form perfect cubes (powers of 3).
\(\sqrt[3]{2^3} \times \sqrt[3]{3^3} \times \sqrt[3]{2}\)
2 × 3 × \(\sqrt[3]{2}\)
Step 4: Multiply the coefficients outside the radical.
6\(\sqrt[3]{2}\)
Answer: b. 6\(\sqrt[3]{2}\)

Q15: What is the pure surd for \(5\sqrt[3]{2}\)?

Step 1: Identify the index of the root, which is 3 (cube root).
Step 2: To move the coefficient 5 inside the cube root, raise it to the power of 3.
\(5^3 = 5 \times 5 \times 5 = 125\).
Step 3: Multiply the value inside the radical by this result.
\(\sqrt[3]{125 \times 2}\)
Step 4: Simplify the multiplication under the radical.
\(\sqrt[3]{250}\)
Answer: c. \(\sqrt[3]{250}\)