Induction Practice CS 1050 Prove that is a

Induction Practice CS 1050

Prove that is a positive integer. Proof: Basis Case: Let n = 1, then whenever n

Prove that whenever n is a positive integer. Inductive Case: Assume that the expression is true for n, i. e. , that Then we must show that:

If S is a finite set with n elements, then S has 2 n subsets (1 of 2) We will do a proof by induction. Basis Step: Assume that the set is empty, i. e. , that it has 0 elements, then it has 20 = 1 subsets, which is Ø. Inductive Step: Assume that a set with n elements has 2 n subsets. Then we must show that this implies that a set with n+1 elements has 2 n+1 subsets.

(2 of 2) Let S be the set with n elements and T be the set with n+1 elements. Let a be the element that is added to S to get T. So T = S {a} or S = T-{a}. For every subset, X, of S there must be exactly two distinct subsets of T: X and X {a}. 2*2 n = 2 n+1

Let p(n) be the proposition that there exists t, e N such that 3 t + 8 e = n. Prove n 14 N, p(n) is true. We will do a proof by induction. Basis Step: Prove p(14) Let t = 2 and e =1, then 3(2) + 8(1) = 14. Inductive Step: Assume p(k) is true. Thus there exist t, e N such that 3 t+8 e = k. We need to show that there exist u, f N such that 3 u + 8 f = k+1. There are two cases e = 0 (there are no 8’s) or e 1 (there is at least one 8).

Case 1: e = 0 Since k 14 and e = 0, then t 5. Let u = t-5 and f = 2. Since t 5, u N and clearly f N. Now 3 u + 8 f = 3(t-5) + 8(2) = 3 t+1 = k + 1. (Effectively, take out three five’s and replace with 2 eight’s. )

Case 2: e 1 Let u = t+3 and f = e-1. Since t, e N and e 1, u, f N. Now 3 u + 8 f = 3(t+3) + 8(e-1) = 3 t + 8 e + 1 = k+1. (Effectively what we have done is remove one 8 and replace it with 3 threes to get a net increase of 1)

Generalization of De Morgan’s Law = (p q) ( p q) n 2 De Morgan’s Laws

Generalization of De Morgan’s Law = (p q) ( p q) n 2 De Morgan’s Laws