GSEB Class 10 Maths Solutions Chapter 1 Real Numbers Exercise 1.3

Question 1. Prove that \( \sqrt{5} \) is irrational.
Answer: Let us assume \( \sqrt{5} \) is a rational number. So, we can find two integers a, b (where \( b \neq 0 \)) such that \( \frac{a}{b} = \sqrt{5} \). Suppose a and b possess a common factor other than 1. If we divide both by this common factor, Then, we get \( \frac{p}{q} = \sqrt{5} \) ...(1) [Here, p and q are co-prime integers] Squaring both sides gives: \( \frac{p^{2}}{q^{2}} = 5 \) \( p^2 = 5q^2 \) ...(2) This implies \( p^2 \) is divisible by 5, which means p will also be divisible by 5. Let \( p = 5r \). Substituting this value of p into equation (2), we get: \( (5r)^2 = 5q^2 \) \( 25r^2 = 5q^2 \) Dividing both sides by 5: \( 5r^2 = q^2 \) This means \( q^2 \) is divisible by 5, so q will also be divisible by 5. This shows that p and q share a common factor of 5. This contradicts our initial assumption that p and q are co-prime. Therefore, our initial assumption was incorrect, and \( \sqrt{5} \) is irrational.
In simple words: To prove \( \sqrt{5} \) is irrational, we first pretend it's rational. This lets us write it as a fraction \( \frac{a}{b} \). By squaring both sides and simplifying, we find that both 'a' and 'b' must have 5 as a common factor. But this goes against our first assumption that 'a' and 'b' had no common factors, meaning our initial pretend was wrong. So, \( \sqrt{5} \) must be irrational.

Exam Tip: For proofs by contradiction, clearly state the assumption, derive a contradiction by logical steps, and then explicitly state that the initial assumption must be false. Use proper mathematical notation and ensure each step is clear.

Question 2. Prove that \( 3 + 2\sqrt{5} \) is irrational. (CBSE)
Answer: Let us assume \( 3 + 2\sqrt{5} \) is a rational number. Then, we can write \( \frac{a}{b} = 3 + 2\sqrt{5} \). [Here, a and b are integers, with \( b \neq 0 \)]
\( \implies \) \( \frac{a}{b} - 3 = 2\sqrt{5} \)
\( \implies \) \( \frac{1}{2} \left( \frac{a}{b} - 3 \right) = \sqrt{5} \) Since a and b are integers, the expression \( \frac{1}{2} \left( \frac{a}{b} - 3 \right) \) is rational. This means that \( \sqrt{5} \) should also be rational. However, this goes against the known fact that \( \sqrt{5} \) is an irrational number. Therefore, our assumption that \( 3 + 2\sqrt{5} \) is rational must be false. Hence, \( 3 + 2\sqrt{5} \) is an irrational number.
In simple words: If we think \( 3 + 2\sqrt{5} \) is rational, we can write it as a fraction \( \frac{a}{b} \). By moving numbers around, we end up showing that \( \sqrt{5} \) would also have to be a rational fraction. But we already know \( \sqrt{5} \) is irrational, so this is a clash. This means our first guess was wrong, and \( 3 + 2\sqrt{5} \) is irrational.

Exam Tip: When proving irrationality for sums or products involving known irrationals, always try to isolate the irrational term on one side of the equation. This makes it easier to show how the rational operations lead to a contradiction.

Question 3. Prove that the following are irrationals:
(1) \( \frac { 1 }{\sqrt { 2 } } \)
(2) \( 7\sqrt {5} \)
(3) \( 6 + \sqrt {2} \)
Answer:
(1) Let us assume that \( \frac{1}{\sqrt{2}} \) is a rational number. So, we can find two integers a, b (where \( b \neq 0 \)) such that \( \frac{1}{\sqrt{2}} = \frac{a}{b} \). Rearranging this equation, we get \( \sqrt{2} = \frac{b}{a} \). Since a and b are integers, \( \frac{b}{a} \) is a rational number. This would mean that \( \sqrt{2} \) is also rational, which contradicts the established fact that \( \sqrt{2} \) is an irrational number. Therefore, our initial assumption is incorrect, and \( \frac{1}{\sqrt{2}} \) is irrational.
(2) Let us suppose that \( 7\sqrt{5} \) is a rational number. Therefore, we can write \( 7\sqrt{5} = \frac{a}{b} \). [Here, a and b are integers, with \( b \neq 0 \)] Dividing both sides by 7, we get \( \sqrt{5} = \frac{a}{7b} \). Since a and b are integers, and \( b \neq 0 \), then \( \frac{a}{7b} \) must be a rational number. This suggests that \( \sqrt{5} \) should be rational. However, this contradicts the known fact that \( \sqrt{5} \) is an irrational number. Therefore, our initial supposition is incorrect, and \( 7\sqrt{5} \) is irrational.
(3) Let us assume that \( 6 + \sqrt{2} \) is a rational number. Therefore, we can write \( 6 + \sqrt{2} = \frac{a}{b} \). [Here, a and b are integers, with \( b \neq 0 \)] Subtracting 6 from both sides, we get \( \sqrt{2} = \frac{a}{b} - 6 \). Since a and b are integers, the expression \( \frac{a}{b} - 6 \) is a rational number. This implies that \( \sqrt{2} \) should also be rational. However, this contradicts the established fact that \( \sqrt{2} \) is an irrational number. Therefore, our initial assumption is incorrect, and \( 6 + \sqrt{2} \) is irrational.
In simple words: For each case, we assume the given number is rational, meaning it can be written as a fraction. Then, we rearrange the equation to show that a known irrational number (like \( \sqrt{2} \) or \( \sqrt{5} \)) would have to be rational too. Since this is a contradiction, our initial assumption must be wrong, proving that the numbers are indeed irrational.

Exam Tip: For proofs of irrationality, always start by assuming the number is rational. Manipulate the equation to isolate the irrational part. The goal is to show that this leads to a contradiction with a known irrational number, thereby proving the original number is irrational.