Rational and Irrational Numbers

Q1: Multiple Choice Questions (Number Systems)

a. The negative of an irrational number is:

Step 1: An irrational number is a number that cannot be written in the form \( \frac{p}{q} \).
Step 2: If we take an irrational number like \( \sqrt{2} \), its negative is \( -\sqrt{2} \).
Step 3: Since \( -\sqrt{2} \) also cannot be expressed as a ratio of integers, it remains irrational.
Answer: ii. an irrational number

b. \( \sqrt{8}(\sqrt{8}-1) \) is always:

Step 1: Apply the distributive law: \( \sqrt{8} \times \sqrt{8} – \sqrt{8} \times 1 \).
Step 2: We know that \( \sqrt{a} \times \sqrt{a} = a \), so \( \sqrt{8} \times \sqrt{8} = 8 \).
Step 3: The expression becomes \( 8 – \sqrt{8} \).
Step 4: Since 8 is rational and \( \sqrt{8} \) is irrational, their difference is always irrational.
Answer: ii. Irrational

c. For the given figure length of OA is:

Step 1: Identify the dimensions from the image. The base (along the x-axis) is from 0 to 1, so \( \text{Base} = 1 \).
Step 2: The height (perpendicular line) is given as 2 units, so \( \text{Height} = 2 \).
Step 3: Use Pythagoras Theorem: \( OA^2 = \text{Base}^2 + \text{Height}^2 \).
Step 4: \( OA^2 = 1^2 + 2^2 \)
\( OA^2 = 1 + 4 = 5 \).
Step 5: Taking square root, \( OA = \sqrt{5} \).
Answer: i. \( \sqrt{5} \)

d. \( 2\sqrt{3} \times 3\sqrt{8} \) is:

Step 1: Multiply the rational parts: \( 2 \times 3 = 6 \).
Step 2: Multiply the irrational parts: \( \sqrt{3} \times \sqrt{8} = \sqrt{24} \).
Step 3: Combine them: \( 6\sqrt{24} \).
Step 4: Simplify \( \sqrt{24} \) as \( \sqrt{4 \times 6} = 2\sqrt{6} \).
Step 5: Total value = \( 6 \times 2\sqrt{6} = 12\sqrt{6} \). Since \( \sqrt{6} \) is a non-perfect square, the result is irrational.
Answer: ii. Irrational

e. Two irrational numbers between 8 and 11 are:

Step 1: Convert the boundaries into square root form. \( 8 = \sqrt{64} \) and \( 11 = \sqrt{121} \).
Step 2: We need to find numbers between \( \sqrt{64} \) and \( \sqrt{121} \) that are not perfect squares.
Step 3: Look at option (i): \( \sqrt{65} \) and \( \sqrt{120} \).
Step 4: Check the range: \( \sqrt{64} < \sqrt{65} < \sqrt{121} \) and \( \sqrt{64} < \sqrt{120} < \sqrt{121} \). Both values fall within the range and are irrational.
Answer: i. \( \sqrt{65} \) and \( \sqrt{120} \)

Q2: State whether the following numbers are rational or not:

i. \( \left(2+\sqrt{2}\right)^2 \)

Step 1: Use identity \[ (a+b)^2 = a^2 + 2ab + b^2 \]Step 2: Apply formula \[ (2+\sqrt{2})^2 = 2^2 + 2 \cdot 2 \cdot \sqrt{2} + (\sqrt{2})^2 \\ = 4 + 4\sqrt{2} + 2 = 6 + 4\sqrt{2} \]Step 3: Conclusion
Since \( \sqrt{2} \) is irrational, \( 6 + 4\sqrt{2} \) is irrational
Answer: Irrational

ii. \( \left(3-\sqrt{3}\right)^2 \)

Step 1: Use identity \[ (a-b)^2 = a^2 – 2ab + b^2 \]Step 2: Apply formula \[ (3-\sqrt{3})^2 = 3^2 – 2 \cdot 3 \cdot \sqrt{3} + (\sqrt{3})^2 \\ = 9 – 6\sqrt{3} + 3 = 12 – 6\sqrt{3} \]Step 3: Conclusion
Contains \( \sqrt{3} \) ⇒ irrational
Answer: Irrational

iii. \( \left(5+\sqrt{5}\right)\left(5-\sqrt{5}\right) \)

Step 1: Use identity \[ (a+b)(a-b) = a^2 – b^2 \]Step 2: Apply formula \[ (5+\sqrt{5})(5-\sqrt{5}) = 5^2 – (\sqrt{5})^2 \\ = 25 – 5 = 20 \]Step 3: Conclusion
20 is a rational number
Answer: Rational

iv. \( \left(\sqrt{3}-\sqrt{2}\right)^2 \)

Step 1: Use identity \[ (a-b)^2 = a^2 – 2ab + b^2 \]Step 2: Apply formula \[ (\sqrt{3}-\sqrt{2})^2 = (\sqrt{3})^2 – 2\sqrt{3}\sqrt{2} + (\sqrt{2})^2 \\ = 3 – 2\sqrt{6} + 2 = 5 – 2\sqrt{6} \]Step 3: Conclusion
Contains \( \sqrt{6} \) ⇒ irrational
Answer: Irrational

Q3: Find the square of the following:

i. \( \frac{3\sqrt{5}}{5} \)

Step 1: Square the fraction \[ \left(\frac{3\sqrt{5}}{5}\right)^2 = \frac{(3\sqrt{5})^2}{5^2} \]Step 2: Simplify numerator \[ (3\sqrt{5})^2 = 9 \times 5 = 45 \\ \frac{45}{25} = \frac{9}{5} \]Answer: \( \frac{9}{5} \)

ii. \( \sqrt{3} + \sqrt{2} \)

Step 1: Use identity \[ (a+b)^2 = a^2 + 2ab + b^2 \]Step 2: Apply formula \[ (\sqrt{3}+\sqrt{2})^2 = (\sqrt{3})^2 + 2\sqrt{3}\sqrt{2} + (\sqrt{2})^2 \\ = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6} \]Answer: \( 5 + 2\sqrt{6} \)

iii. \( \sqrt{5} – 2 \)

Step 1: Use identity \[ (a-b)^2 = a^2 – 2ab + b^2 \]Step 2: Apply formula \[ (\sqrt{5} – 2)^2 = (\sqrt{5})^2 – 2 \cdot \sqrt{5} \cdot 2 + 2^2 \\ = 5 – 4\sqrt{5} + 4 = 9 – 4\sqrt{5} \]Answer: \( 9 – 4\sqrt{5} \)

iv. \( 3 + 2\sqrt{5} \)

Step 1: Use identity \[ (a+b)^2 = a^2 + 2ab + b^2 \]Step 2: Apply formula \[ (3 + 2\sqrt{5})^2 = 3^2 + 2 \cdot 3 \cdot 2\sqrt{5} + (2\sqrt{5})^2 \\ = 9 + 12\sqrt{5} + 4 \times 5 = 9 + 12\sqrt{5} + 20 \\ = 29 + 12\sqrt{5} \]Answer: \( 29 + 12\sqrt{5} \)

Q4: State, in each case, whether true or false:

i. \( \sqrt{2} + \sqrt{3} = \sqrt{5} \)

Step 1: Square both sides to check \[ (\sqrt{2}+\sqrt{3})^2 = 2 + 3 + 2\sqrt{6} = 5 + 2\sqrt{6} \\ (\sqrt{5})^2 = 5 \]Step 2: Compare \[ 5 + 2\sqrt{6} \ne 5 \]Answer: False

ii. \( 2\sqrt{4} + 2 = 6 \)

Step 1: Simplify \[ \sqrt{4} = 2 \\ 2\sqrt{4} + 2 = 2 \times 2 + 2 = 4 + 2 = 6 \]Answer: True

iii. \( 3\sqrt{7} – 2\sqrt{7} = \sqrt{7} \)

Step 1: Combine like terms \[ 3\sqrt{7} – 2\sqrt{7} = (3-2)\sqrt{7} = \sqrt{7} \]Answer: True

iv. \( \frac{2}{7} \) is an irrational number.

Step 1: Definition
A rational number is of the form \( \frac{p}{q} \), where \( q \ne 0 \)
Step 2: Check \[ \frac{2}{7} \text{ is of the form } \frac{p}{q} \]Answer: False

v. \( \frac{5}{11} \) is a rational number.

Step 1: Check definition \[ \frac{5}{11} \text{ is of the form } \frac{p}{q},\ q \ne 0 \]Answer: True

vi. All rational numbers are real numbers.

Step 1: Recall sets
Rational numbers are a subset of real numbers
Answer: True

vii. All real numbers are rational numbers.

Step 1: Counter example \[ \sqrt{2} \text{ is a real number but irrational} \]Answer: False

viii. Some real numbers are rational numbers.

Step 1: Explanation
Numbers like \(1, \frac{1}{2}, -3\) are real as well as rational
Answer: True

Q5: Given universal set \( = \{-6,\ -5\frac{3}{4},\ -\sqrt{4},\ -\frac{3}{5},\ -\frac{3}{8},\ 0,\ \frac{4}{5},\ 1,\ 1\frac{2}{3},\ \sqrt{8},\ 3.01,\ \pi,\ 8.47\} \). From the given sets, find:

i. Set of rational numbers

Step 1: Recall definition
Rational numbers are numbers that can be written in the form \( \frac{p}{q},\ q \neq 0 \)
Step 2: Identify rational numbers \[ -6,\ -5\frac{3}{4},\ -\sqrt{4},\ -\frac{3}{5},\ -\frac{3}{8},\ 0,\ \frac{4}{5},\ 1,\ 1\frac{2}{3},\ 3.01,\ 8.47 \]Note: \( -\sqrt{4} = -2 \) which is rational
Answer: \( \{-6,\ -5\frac{3}{4},\ -\sqrt{4},\ -\frac{3}{5},\ -\frac{3}{8},\ 0,\ \frac{4}{5},\ 1,\ 1\frac{2}{3},\ 3.01,\ 8.47\} \)

ii. Set of irrational numbers

Step 1: Recall definition
Irrational numbers cannot be written in the form \( \frac{p}{q} \)
Step 2: Identify irrational numbers \[ \sqrt{8},\ \pi \]Answer: \( \{\sqrt{8},\ \pi\} \)

iii. Set of integers

Step 1: Recall integers
Integers are …, -2, -1, 0, 1, 2, …
Step 2: Identify integers \[ -6,\ -\sqrt{4},\ 0,\ 1 \]Note: \( -\sqrt{4} = -2 \)
Answer: \( \{-6,\ -\sqrt{4},\ 0,\ 1\} \)

iv. Set of non-negative integers

Step 1: Definition
Non-negative integers are \( 0, 1, 2, 3, … \)
Step 2: Identify from set \[ 0,\ 1 \]Answer: \( \{0,\ 1\} \)

Q6: Prove each of the following numbers is irrational:

i. \( \sqrt{3} + \sqrt{2} \)

Step 1: Assume contrary
Assume \( \sqrt{3} + \sqrt{2} \) is rational
Step 2: Let \[ \sqrt{3} + \sqrt{2} = r \quad (\text{where } r \text{ is rational}) \]Step 3: Rearranging \[ \sqrt{3} = r – \sqrt{2} \]Step 4: Square both sides \[ (\sqrt{3})^2 = (r – \sqrt{2})^2 \\ 3 = r^2 – 2r\sqrt{2} + 2 \\ 3 – 2 = r^2 – 2r\sqrt{2} \\ 1 = r^2 – 2r\sqrt{2} \]Step 5: Rearranging \[ 2r\sqrt{2} = r^2 – 1 \\ \sqrt{2} = \frac{r^2 – 1}{2r} \]Step 6: Conclusion
Right side is rational ⇒ \( \sqrt{2} \) becomes rational (contradiction)
Hence, assumption is wrong
Answer: \( \sqrt{3} + \sqrt{2} \) is irrational

ii. \( 3 – \sqrt{2} \)

Step 1: Assume contrary
Assume \( 3 – \sqrt{2} \) is rational
Step 2: Let \[ 3 – \sqrt{2} = r \quad (\text{where } r \text{ is rational}) \]Step 3: Rearranging \[ \sqrt{2} = 3 – r \]Step 4: Conclusion
Right side is rational ⇒ \( \sqrt{2} \) becomes rational (contradiction)
Hence, assumption is wrong
Answer: \( 3 – \sqrt{2} \) is irrational

Q7: Write a pair of irrational numbers whose sum is irrational.

Step 1: Choose two irrational numbers \[ \sqrt{2} \text{ and } \sqrt{3} \]Step 2: Find their sum \[ \sqrt{2} + \sqrt{3} \]Step 3: Check nature
Since both \( \sqrt{2} \) and \( \sqrt{3} \) are irrational and their sum cannot be simplified into a rational number, \[ \sqrt{2} + \sqrt{3} \text{ is irrational} \]Answer: A required pair is \( \sqrt{2} \) and \( \sqrt{3} \)

Q8: Write a pair of irrational numbers whose sum is rational.

Step 1: Choose two irrational numbers \[ \sqrt{2} \text{ and } -\sqrt{2} \]Step 2: Find their sum \[ \sqrt{2} + (-\sqrt{2}) \\ = 0 \]Step 3: Check nature
\( 0 \) is a rational number
Answer: A required pair is \( \sqrt{2} \) and \( -\sqrt{2} \)

Q9: Write a pair of irrational numbers whose difference is irrational.

Step 1: Choose two irrational numbers \[ \sqrt{3} \text{ and } \sqrt{2} \]Step 2: Find their difference \[ \sqrt{3} – \sqrt{2} \]Step 3: Check nature
Since \( \sqrt{3} \) and \( \sqrt{2} \) are irrational and their difference cannot be simplified into a rational number, \[ \sqrt{3} – \sqrt{2} \text{ is irrational} \]Answer: A required pair is \( \sqrt{3} \) and \( \sqrt{2} \)

Q10: Write a pair of irrational numbers whose difference is rational.

Step 1: Choose two irrational numbers \[ 2 + \sqrt{3} \text{ and } \sqrt{3} \]Step 2: Find their difference \[ (2 + \sqrt{3}) – \sqrt{3} \\ = 2 \]Step 3: Check nature
\( 2 \) is a rational number
Answer: A required pair is \( 2 + \sqrt{3} \) and \( \sqrt{3} \)

Q11: Write a pair of irrational numbers whose product is irrational.

Step 1: Choose two irrational numbers \[ \sqrt{2} \text{ and } \sqrt{3} \]Step 2: Find their product \[ \sqrt{2} \times \sqrt{3} \\ = \sqrt{6} \]Step 3: Check nature \[ \sqrt{6} \text{ is irrational} \]Answer: A required pair is \( \sqrt{2} \) and \( \sqrt{3} \)

Q12: Write a pair of irrational numbers whose product is rational.

Step 1: Choose two irrational numbers \[ \sqrt{2} \text{ and } \sqrt{2} \]Step 2: Find their product \[ \sqrt{2} \times \sqrt{2} \\ = (\sqrt{2})^2 = 2 \]Step 3: Check nature \[ 2 \text{ is a rational number} \]Answer: A required pair is \( \sqrt{2} \) and \( \sqrt{2} \)