Exercise 1.2 answers: Real Numbers

Open NCERT p. 9One way to think about it

1. Suppose, to the contrary, that √5 is rational. Then √5 = a/b for some integers a and b with no common factor other than 1, and b ≠ 0.
2. So a = √5 × b. Squaring both sides: a² = 5b².
3. This means 5 divides a². Since 5 is prime, 5 must divide a.
4. So a = 5c for some integer c. Substituting: (5c)² = 5b², so 25c² = 5b², so b² = 5c².
5. This means 5 divides b² as well, and since 5 is prime, 5 divides b.
6. So both a and b are divisible by 5 — but we assumed a and b have no common factor. This is a contradiction.
7. So our assumption was wrong: √5 is irrational.

In short√5 is irrational (proved by contradiction).

Watch this explained “The slot filled with 5”, 2:32 into Re-running the same contradiction on other prime square roots · हिंदी में देखें

Open NCERT p. 9One way to think about it

1. Suppose, to the contrary, that 3 + 2√5 is rational. Call it r, so 3 + 2√5 = r.
2. Rearrange: 2√5 = r − 3, so √5 = (r − 3)/2.
3. r is rational and 3 is rational, so r − 3 is rational, and a rational number divided by 2 is still rational. So this would make √5 rational.
4. But √5 is irrational (proved earlier), so this is a contradiction.
5. So our assumption was wrong: 3 + 2√5 is irrational.

In short3 + 2√5 is irrational (proved by contradiction).

Watch this explained “Four to sort”, 11:38 into Re-running the same contradiction on other prime square roots · हिंदी में देखें

Open NCERT p. 9One way to think about it

(i) 1/√2

1. Suppose, to the contrary, that 1/√2 is rational. Call it a/b, where a and b are integers and b ≠ 0.
2. Then √2 = b/a (flip both sides; a ≠ 0 because 1/√2 is not zero).
3. b/a is a ratio of integers, so this would make √2 rational.
4. But √2 is irrational (proved earlier), so this is a contradiction.
5. So 1/√2 is irrational.

In short1/√2 is irrational (proved by contradiction).

(ii) 7√5

1. Suppose, to the contrary, that 7√5 is rational. Call it a/b, where a and b are integers and b ≠ 0.
2. Then √5 = a/(7b), a ratio of integers, since 7b ≠ 0.
3. This would make √5 rational.
4. But √5 is irrational (proved earlier), so this is a contradiction.
5. So 7√5 is irrational.

In short7√5 is irrational (proved by contradiction).

(iii) 6 + √2

1. Suppose, to the contrary, that 6 + √2 is rational. Call it r.
2. Rearrange: √2 = r − 6.
3. r is rational and 6 is rational, so r − 6 is rational. This would make √2 rational.
4. But √2 is irrational (proved earlier), so this is a contradiction.
5. So 6 + √2 is irrational.

In short6 + √2 is irrational (proved by contradiction).

Watch this explained “Four to sort”, 11:38 into Re-running the same contradiction on other prime square roots · हिंदी में देखें