Proof by contradiction:

Assume √5 is rational. Then it can be expressed as a ratio of two coprime integers:

√5 = p/q, where q ≠ 0 and HCF(p,q) = 1

Squaring both sides:

5 = p²/q² ⇒ p² = 5q² …(1)

This implies p² is divisible by 5 ⇒ p is divisible by 5 (by the theorem: if a prime divides a², then it divides a)

Let p = 5k for some integer k

Substitute into (1):

(5k)² = 5q² ⇒ 25k² = 5q² ⇒ q² = 5k²

This implies q² is divisible by 5 ⇒ q is divisible by 5

Contradiction: Both p and q are divisible by 5, but we assumed they are coprime (HCF=1)

∴ Our initial assumption is false, and √5 must be irrational. ∎

Proof by contradiction:

Assume 3 + 2√5 is rational. Then:

3 + 2√5 = a/b, where b ≠ 0 and HCF(a,b) = 1

Rearranging:

2√5 = (a/b) – 3 = (a – 3b)/b

√5 = (a – 3b)/(2b)

Since a and b are integers, (a – 3b)/(2b) is rational ⇒ √5 is rational

Contradiction: We know from Problem 1 that √5 is irrational

∴ Our assumption is false, and 3 + 2√5 must be irrational. ∎

Proof by contradiction:

Assume 1/√2 is rational. Then:

1/√2 = p/q, where q ≠ 0 and HCF(p,q) = 1

Rearranging:

√2 = q/p

This implies √2 is rational (as ratio of two integers)

Contradiction: It’s known that √2 is irrational (similar proof as √5)

∴ 1/√2 must be irrational. ∎

Proof by contradiction:

Assume 7√5 is rational. Then:

7√5 = p/q, where q ≠ 0 and HCF(p,q) = 1

√5 = p/(7q)

Right side is rational (as p and 7q are integers) ⇒ √5 is rational

Contradiction: We proved √5 is irrational in Problem 1

∴ 7√5 must be irrational. ∎

Proof by contradiction:

Assume 6 + √2 is rational. Then:

6 + √2 = a/b, where b ≠ 0 and HCF(a,b) = 1

Rearranging:

√2 = (a/b) – 6 = (a – 6b)/b

Right side is rational ⇒ √2 is rational

Contradiction: It’s known that √2 is irrational

∴ 6 + √2 must be irrational. ∎

Proof by contradiction:

Assume √3 is rational. Then:

√3 = p/q, where q ≠ 0 and HCF(p,q) = 1

Squaring: 3 = p²/q² ⇒ p² = 3q² …(1)

⇒ p² is divisible by 3 ⇒ p is divisible by 3

Let p = 3k

Substitute in (1): 9k² = 3q² ⇒ q² = 3k²

⇒ q² is divisible by 3 ⇒ q is divisible by 3

Contradiction: Both p and q divisible by 3, but HCF(p,q)=1

∴ √3 is irrational. ∎

Proof by contradiction:

Assume 5 – √3 is rational. Then:

5 – √3 = a/b, where b ≠ 0 and HCF(a,b) = 1

Rearranging: √3 = 5 – (a/b) = (5b – a)/b

Right side is rational ⇒ √3 is rational

Contradiction: We know √3 is irrational (from Problem 4)

∴ 5 – √3 is irrational. ∎