Exercise 1.4 — Irrationality Proofs

What Does Exercise 1.4 Cover?

Exercise 1.4 of Class 10 Mathematics — Chapter 1: Real Numbers — is dedicated entirely to proving that certain numbers are irrational. All proofs here use the same powerful technique: proof by contradiction. You assume the opposite of what you want to prove, follow the logic until you reach an impossibility, and conclude that the original assumption must be false. This method is a favourite in CBSE, Telangana, and Andhra Pradesh board exams and typically carries 3–4 marks per proof.

What Are Irrational Numbers?

A number is irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0. Unlike rational numbers, irrational numbers have decimal expansions that are non-terminating and non-repeating — they go on forever without settling into any repeating pattern. Common examples include √2, √3, √5, ∛5, π, and any non-perfect-square root.

Two useful properties that this exercise relies on: the sum or difference of a rational and an irrational is always irrational, and the product or quotient of a non-zero rational with an irrational is always irrational.

The Key Supporting Theorem

Before proving any number irrational, the chapter establishes this crucial result: if a prime number p divides a², then p must also divide a. This is what allows the contradiction to work in proofs about √2, √5, and similar numbers. For example, if 2 divides a², then 2 divides a — meaning a is even, so we can write a = 2c for some integer c.

The Standard Proof — Showing √2 Is Irrational

This is the model proof that all others follow. Assume √2 = a/b where a and b are co-prime integers (no common factor other than 1) and b ≠ 0. Squaring both sides gives a² = 2b², which means 2 divides a². By the theorem above, 2 must also divide a, so write a = 2c. Substituting back gives 2b² = 4c², so b² = 2c², meaning 2 also divides b. But now 2 is a common factor of both a and b — this directly contradicts the assumption that a and b are co-prime. The contradiction proves that √2 is irrational.

Question 1 — All Five Proofs Explained

Each sub-question uses the same contradiction framework, but the algebraic manipulation differs depending on the expression being tested.

• 1/√2 is irrational: Assume 1/√2 = a/b. Rearranging gives √2 = b/a, which would make √2 rational — a known contradiction. So 1/√2 is irrational.
• √3 + √5 is irrational: Assume it equals a/b. Isolate √5, square both sides, then rearrange to show that √3 equals a rational expression. Since √3 is known to be irrational, this is a contradiction.
• 6 + √2 is irrational: Assume 6 + √2 = a/b. Then √2 = (a − 6b)/b, which is rational since a and b are integers. This contradicts the irrationality of √2.
• √5 is irrational: Identical structure to the √2 proof — assume √5 = a/b (co-prime), square to get a² = 5b², use the key theorem to show 5 divides both a and b, reaching a contradiction.
• 3 + 2√5 is irrational: Assume 3 + 2√5 = a/b. Then 2√5 = (a − 3b)/b, so √5 = (a − 3b)/(2b), making √5 rational — a contradiction.

Question 2 — Proving √p + √q Is Irrational for Any Two Primes

This is the most general proof in the exercise. Assume √p + √q = a/b (co-prime, b ≠ 0). Isolate √q and square both sides. After careful algebraic simplification, you can show that √p equals a rational expression involving a, b, p, and q. Since p is a prime, √p is irrational — and a rational expression cannot equal an irrational number. This contradiction proves that √p + √q is irrational for any two prime numbers p and q. This result covers pairs like √2 + √3, √5 + √7, and √11 + √13 all at once.

The Proof Template to Memorise

Every proof in this exercise follows the same five-step structure that you can apply to any similar question in board exams:

• Step 1 — Assume the opposite: "Let us assume that [expression] is a rational number."
• Step 2 — Set up the fraction: Write it as a/b where a and b are co-prime integers and b ≠ 0.
• Step 3 — Rearrange: Isolate the known irrational part (√2, √3, √5, etc.) on one side.
• Step 4 — Identify the contradiction: The right-hand side will be rational (ratio of integers), but the left-hand side is a known irrational number.
• Step 5 — Conclude: "This contradicts the fact that [√n] is irrational. Hence our assumption is false. Therefore [expression] is irrational."

What to Study Next

Exercise 1.4 is the final exercise in Chapter 1 for most syllabuses. Now that you have covered Euclid's Algorithm, the Fundamental Theorem of Arithmetic, decimal expansions of rationals, and irrationality proofs, you are fully prepared for any question from Real Numbers in your board exam. Review the connections between all four exercises starting from the Real Numbers Introduction, revisit how prime factorisation supports these proofs in Exercise 1.2, or move on to Chapter 2: Polynomials.