Exercise 1.2 — Real Numbers

1. 13 marksNCERT Cl-10 Maths, Ex 1.2, Q1

   Prove that √5 is irrational.

   Hint. Assume the opposite — that it can be written as a fraction in lowest terms — and show that this forces the fraction not to be in lowest terms after all.

   This is a proof by contradiction, and the whole method rests on one idea: if 5 divides a², then 5 divides a. That comes straight from unique prime factorisation.

   Step 1 — Suppose, for contradiction, that √5 is rational. Then √5 = a/b for some integers a and b with b ≠ 0, and we may take a and b to have no common factor (cancel first if they do).

   Step 2 — Rearrange: a = √5 b, and squaring gives a² = 5b².

   Step 3 — So 5 divides a². Since 5 is prime, 5 must appear in the prime factorisation of a itself. Write a = 5c for some integer c.

   Step 4 — Substitute back: (5c)² = 5b², so 25c² = 5b², which gives b² = 5c².

   Step 5 — By the same reasoning applied to b, 5 divides b² and therefore 5 divides b.

   Step 6 — Now 5 divides both a and b. That contradicts Step 1, where we arranged for them to share no common factor.

   Step 7 — The only assumption we made was that √5 is rational, so that assumption must be false.

   ✦ √5 is irrational.

   Where students slip. Skipping the phrase 'where a and b have no common factor' in Step 1. Without it there is no contradiction at the end, and the proof earns almost nothing.

   Another way. The same six steps prove √2, √3 or √p irrational for any prime p — only the number changes. The step that needs p to be prime is Step 3.

2. 23 marksNCERT Cl-10 Maths, Ex 1.2, Q2

   Prove that 3 + 2√5 is irrational.

   Hint. You may now use the result of Q1. Assume the number is rational and rearrange until √5 sits alone on one side.

   The strategy changes here. You are not proving irrationality from scratch — you are reducing to Q1.

   Step 1 — Suppose, for contradiction, that 3 + 2√5 is rational. Call it a/b, with a, b integers and b ≠ 0.

   Step 2 — Isolate the surd: 2√5 = a/b − 3, so √5 = (a/b − 3) ÷ 2 = (a − 3b) / 2b.

   Step 3 — Look hard at the right-hand side. a, b and 3 are integers, so a − 3b is an integer and 2b is a non-zero integer. A ratio of two integers is by definition rational.

   Step 4 — So this says √5 is rational. But Q1 proved √5 is irrational — a direct contradiction.

   Step 5 — The assumption in Step 1 is therefore false.

   ✦ 3 + 2√5 is irrational.

   Where students slip. Arguing 'rational + irrational = irrational' as a one-line answer. It is a true fact, but in a proof question the marks are for deriving the contradiction, not for quoting the slogan.

   Another way. Same shape for any a + b√5 with rational a and b (b ≠ 0): isolate the surd and the same contradiction appears.

3. 34 marksNCERT Cl-10 Maths, Ex 1.2, Q3

   Prove that each of these is irrational: (i) 1/√2, (ii) 7√5, (iii) 6 + √2.

   Hint. All three reduce to the known irrationality of √2 or √5. Rearrange until the surd is alone, then read off the contradiction.

   Each part uses the same two moves: assume rational, then isolate the surd.

   (i) 1/√2.

   Step 1 — Suppose 1/√2 = a/b with a, b integers, b ≠ 0. Note a ≠ 0, since 1/√2 is not 0.

   Step 2 — Take reciprocals: √2 = b/a.

   Step 3 — b/a is a ratio of integers with a ≠ 0, so it is rational — meaning √2 would be rational. But √2 is irrational (proved by the Q1 method with 2 in place of 5). Contradiction.

   ✦ 1/√2 is irrational.

   (ii) 7√5.

   Step 1 — Suppose 7√5 = a/b with a, b integers, b ≠ 0.

   Step 2 — Divide by 7: √5 = a/7b.

   Step 3 — a and 7b are integers and 7b ≠ 0, so the right side is rational, making √5 rational. That contradicts Q1.

   ✦ 7√5 is irrational.

   (iii) 6 + √2.

   Step 1 — Suppose 6 + √2 = a/b with a, b integers, b ≠ 0.

   Step 2 — Subtract 6: √2 = a/b − 6 = (a − 6b) / b.

   Step 3 — a − 6b and b are integers with b ≠ 0, so √2 would be rational. Contradiction again.

   ✦ 6 + √2 is irrational.

   Where students slip. In (i), forgetting to note a ≠ 0 before taking reciprocals. It is a small point, but dividing by a is only legal once you have said why a is not zero.

   Another way. Once √2 and √5 are known irrational, the general fact behind all three parts is that multiplying an irrational by a non-zero rational, or adding a rational to it, always leaves it irrational.

Solutions written by the tuition.in editorial team and checked against the rationalised NCERT Class 10 Mathematics textbook and the CBSE 2026-27 syllabus (Euclid's division lemma and decimal expansions are no longer part of this chapter). Questions are referenced from the NCERT textbook for identification.