Malay K. Ganai

Proof by Induction

Note that 5 cannot be represented in the form 4a + 3b/ Let P (k) be the proposition: for every m with 5 < m ≤ k there exist a and b such that 4a + 3b = m. Proof. We will prove by induction on n ≥ 8 that P (n) holds. We need P (8). Claim 1: P (13) holds. Check Claim 1: 6 = 2 · 3, 7 = 4 + 3, 8 = 2 · 4. Claim 2: P(k) implies P(k+1). Proof of Claim 2. Let m = (k+1)−3. By induction there exist there exist a′ and b′ such that 4a′+3b′ = m. That is, 4a′+3b′ = k−2. So, 4a′+3b′+3 = k+1. So if we set a = a′ and b = b′ + 1, 4a + 3b = k + 1. Since for any k, we have shown P (k) implies P (k + 1) by the mathematical induction we have shown: for all n, P (n).