NCERT Class 10 Maths Chapter 1 Exercise 1.2: Proving Irrationality of Real Numbers

The Foundation: Primes and Divisibility Theorem

To prove that a number is irrational, we rely on a fundamental theorem in arithmetic. Theorem: 'Let p be a prime number. If p divides a^2, then p divides a, where a is a positive integer.'

Let's understand why this works using YoLearn's conceptual sketchpad. Every positive integer 'a' can be written as a product of its prime factors: a = p1 p2 ... pn (where p1, p2, etc., are primes, not necessarily distinct). Squaring both sides, we get a^2 = (p1 p2 ... pn)^2 = (p1)^2 (p2)^2 ... * (pn)^2. By the Fundamental Theorem of Arithmetic, the prime factorization of a^2 is unique. Therefore, the only prime factors of a^2 are the prime factors of 'a'. If a prime 'p' divides a^2, then p must be one of the prime factors (p1, p2, ... pn) of 'a'. Thus, p must also divide 'a'. This simple yet powerful logical link is the backbone of all irrationality proofs in Exercise 1.2.

The Method of Contradiction: Step-by-Step Proof

1. Step 1: The Opposite Assumption — Assume the contrary. Let us assume that the given number (e.g., √5) is a rational number. This means we can write it in the form of a/b, where 'a' and 'b' are integers, b is not equal to 0, and 'a' and 'b' are co-prime (meaning they share no common factors other than 1).
2. Step 2: Squaring and Rearranging — Write √5 = a/b. Square both sides to eliminate the square root: 5 = a^2 / b^2. Rearranging the equation gives us 5 * b^2 = a^2. This equation implies that 5 divides a^2.
3. Step 3: Applying the Core Lemma — Since 5 is a prime number and divides a^2, it must also divide 'a' (using our Core Theorem). We can therefore express 'a' as a multiple of 5: let a = 5c, where 'c' is some integer.
4. Step 4: Finding the Contradiction — Substitute a = 5c back into our equation: 5 b^2 = (5c)^2, which simplifies to 5 b^2 = 25 c^2, or b^2 = 5 c^2. This implies 5 divides b^2, and thus 5 divides 'b'.
5. Step 5: Rejecting the Assumption — Now, 5 divides both 'a' and 'b'. This means 'a' and 'b' share a common factor of 5, which contradicts our starting assumption that 'a' and 'b' are co-prime. Therefore, our initial assumption that √5 is rational is incorrect. We conclude that √5 is irrational.

Pro Tips for Scoring Full Marks on Board Exams

1. Define Co-prime explicitly: When writing a/b, always write in parentheses: 'where a and b are integers, b ≠ 0, and HCF(a, b) = 1 (i.e., a and b are co-prime)'. Neglecting to write this loses you 1 mark instantly.
2. State the theorem clearly: When moving from 'p divides a^2' to 'p divides a', write: 'According to the theorem, if a prime p divides a^2, then p divides a.'
3. Handle Compound Numbers Efficiently: For numbers like 3 + 2√5, you do not need to prove √5 is irrational from scratch unless specified. Rearrange the equation to isolate √5 on one side: √5 = (a/b - 3)/2. Since the RHS is composed of rational operations on integers, it must be rational, which contradicts the known fact that √5 is irrational.

Practice Questions with Solutions

• Q: Prove that √3 is irrational. A: Step 1: Assume to the contrary that √3 is rational. Therefore, we can find co-prime integers a and b (b ≠ 0) such that √3 = a/b. Step 2: Squaring both sides, we get 3 = a^2 / b^2, which gives 3b^2 = a^2. Since 3 divides a^2, by theorem, 3 must also divide a. Step 3: So, we can write a = 3c for some integer c. Substituting a = 3c in 3b^2 = a^2, we get 3b^2 = 9c^2, which simplifies to b^2 = 3*c^2. Since 3 divides b^2, by theorem, 3 must also divide b. Step 4: Therefore, a and b have at least 3 as a common factor. But this contradicts the fact that a and b are co-prime. Final answer: This contradiction has arisen because of our incorrect assumption that √3 is rational. Hence, √3 is irrational.
• Q: Prove that 7√5 is irrational. A: Step 1: Assume to the contrary that 7√5 is rational. Therefore, we can find co-prime integers a and b (b ≠ 0) such that 7√5 = a/b. Step 2: Rearranging the terms, we get √5 = a / (7b). Step 3: Since a and b are integers, a and 7b are also integers, meaning a / (7b) is rational. This implies √5 is rational. Step 4: But this contradicts the known fact that √5 is irrational. Final answer: This contradiction arises because of our incorrect assumption that 7√5 is rational. Hence, 7√5 is irrational.
• Q: Prove that 3 + 2√5 is irrational. A: Step 1: Assume to the contrary that 3 + 2√5 is rational. Therefore, we can find co-prime integers a and b (b ≠ 0) such that 3 + 2√5 = a/b. Step 2: Subtracting 3 from both sides: 2√5 = (a/b) - 3 = (a - 3b)/b. Step 3: Dividing by 2: √5 = (a - 3b) / (2b). Step 4: Since a and b are integers, the term (a - 3b) / (2b) is rational. This implies that √5 is rational. But this contradicts the fact that √5 is irrational. Final answer: Our assumption was false. Thus, 3 + 2√5 is irrational.
• Q: Prove that 1/√2 is irrational. A: Step 1: Assume to the contrary that 1/√2 is rational. Therefore, we can write 1/√2 = a/b, where a and b are co-prime integers and b ≠ 0. Step 2: Cross-multiplying, we get b = a√2. Squaring both sides gives b^2 = 2 a^2. Step 3: This implies 2 divides b^2, so 2 divides b. We can write b = 2c for some integer c. Substitute b = 2c: (2c)^2 = 2 a^2 => 4 c^2 = 2 a^2 => 2 * c^2 = a^2. This means 2 divides a^2, so 2 divides a. Step 4: Thus, 2 is a common factor of both a and b, contradicting our assumption that they are co-prime. Final answer: Hence, our assumption is incorrect, and 1/√2 is irrational.

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