Table of Contents
Exercise 1.4 Solutions – Class X Mathematics
These solutions are based on the Telangana State Class X Mathematics textbook, focusing on proving irrational numbers. Mathematical expressions are rendered using MathJax.
1. Prove that the following are irrational.
(i) \( \frac{1}{\sqrt{2}} \)
Assume \( \frac{1}{\sqrt{2}} \) is rational, i.e., \( \frac{1}{\sqrt{2}} = \frac{a}{b} \), where \( a \) and \( b \) are co-prime integers, \( b \neq 0 \).
Then, \( \sqrt{2} = \frac{b}{a} \).
Square both sides: \( 2 = \frac{b^2}{a^2} \), so \( b^2 = 2a^2 \).
\( b^2 \) is even, so \( b \) is even. Let \( b = 2c \).
Substitute: \( (2c)^2 = 2a^2 \), \( 4c^2 = 2a^2 \), \( a^2 = 2c^2 \).
\( a^2 \) is even, so \( a \) is even, contradicting co-primality.
Thus, \( \frac{1}{\sqrt{2}} \) is irrational.
Conclusion: \( \frac{1}{\sqrt{2}} \) is irrational.
(ii) \( \sqrt{3} + \sqrt{5} \)
Assume \( \sqrt{3} + \sqrt{5} \) is rational, i.e., \( \sqrt{3} + \sqrt{5} = \frac{a}{b} \), where \( a \) and \( b \) are co-prime integers, \( b \neq 0 \).
Square both sides: \( (\sqrt{3} + \sqrt{5})^2 = \frac{a^2}{b^2} \).
\( 3 + 5 + 2\sqrt{15} = \frac{a^2}{b^2} \), so \( 8 + 2\sqrt{15} = \frac{a^2}{b^2} \).
\( 2\sqrt{15} = \frac{a^2}{b^2} – 8 \), \( \sqrt{15} = \frac{a^2 – 8b^2}{2b^2} \).
Left side irrational, right side rational, contradiction (since \( \sqrt{15} \) is irrational).
Thus, \( \sqrt{3} + \sqrt{5} \) is irrational.
Conclusion: \( \sqrt{3} + \sqrt{5} \) is irrational.
(iii) \( 6 + \sqrt{2} \)
Assume \( 6 + \sqrt{2} \) is rational, i.e., \( 6 + \sqrt{2} = \frac{a}{b} \), where \( a \) and \( b \) are co-prime integers, \( b \neq 0 \).
Then, \( \sqrt{2} = \frac{a}{b} – 6 = \frac{a – 6b}{b} \).
Square both sides: \( 2 = \frac{(a – 6b)^2}{b^2} \), so \( (a – 6b)^2 = 2b^2 \).
Left side integer, right side \( 2b^2 \) (even if \( b \) is odd), but \( a – 6b \) must be even, leading to \( a \) and \( b \) both even, contradicting co-primality.
Thus, \( 6 + \sqrt{2} \) is irrational.
Conclusion: \( 6 + \sqrt{2} \) is irrational.
(iv) \( \sqrt{5} \)
Assume \( \sqrt{5} \) is rational, i.e., \( \sqrt{5} = \frac{a}{b} \), where \( a \) and \( b \) are co-prime integers, \( b \neq 0 \).
Square both sides: \( 5 = \frac{a^2}{b^2} \), so \( a^2 = 5b^2 \).
\( a^2 \) is multiple of 5, so \( a \) is multiple of 5. Let \( a = 5c \).
Substitute: \( (5c)^2 = 5b^2 \), \( 25c^2 = 5b^2 \), \( b^2 = 5c^2 \).
\( b^2 \) is multiple of 5, so \( b \) is multiple of 5, contradicting co-primality.
Thus, \( \sqrt{5} \) is irrational.
Conclusion: \( \sqrt{5} \) is irrational.
(v) \( 3 + 2\sqrt{5} \)
Assume \( 3 + 2\sqrt{5} \) is rational, i.e., \( 3 + 2\sqrt{5} = \frac{a}{b} \), where \( a \) and \( b \) are co-prime integers, \( b \neq 0 \).
Then, \( 2\sqrt{5} = \frac{a}{b} – 3 = \frac{a – 3b}{b} \), \( \sqrt{5} = \frac{a – 3b}{2b} \).
Square both sides: \( 5 = \frac{(a – 3b)^2}{4b^2} \), so \( 20b^2 = (a – 3b)^2 \).
Left side even, right side must be even, implying \( a – 3b \) even, leading to \( a \) and \( b \) both even, contradicting co-primality.
Thus, \( 3 + 2\sqrt{5} \) is irrational.
Conclusion: \( 3 + 2\sqrt{5} \) is irrational.
2. Prove that \( \sqrt{p} + \sqrt{q} \) is irrational, where \( p, q \) are primes.
Assume \( \sqrt{p} + \sqrt{q} \) is rational, i.e., \( \sqrt{p} + \sqrt{q} = \frac{a}{b} \), where \( a \) and \( b \) are co-prime integers, \( b \neq 0 \), and \( p, q \) are distinct primes.
Square both sides: \( (\sqrt{p} + \sqrt{q})^2 = \frac{a^2}{b^2} \).
\( p + q + 2\sqrt{pq} = \frac{a^2}{b^2} \), so \( 2\sqrt{pq} = \frac{a^2}{b^2} – p – q \), \( \sqrt{pq} = \frac{a^2 – b^2(p + q)}{2b^2} \).
Left side \( \sqrt{pq} \) (where \( pq \) is not a perfect square since \( p \neq q \)) is irrational, but right side is rational, a contradiction.
Thus, \( \sqrt{p} + \sqrt{q} \) is irrational.
Conclusion: \( \sqrt{p} + \sqrt{q} \) is irrational when \( p, q \) are primes.
