University:
Gonzaga University
Course:
Math 413 | Real Analysis I
Academic year:

2021
Views:

79

Pages:

3
Author:

Audrey F.

1. Prove (Commutative Law of Multiplication) For all x, y ∈ N, x * y = y * x. Proof: Fix x ∈ N and let y = 1. By Induction: I. Base Step: If y = 1, then x * 1 = 1 * x follows from Lemma B. II. Inductive Step: Assume y ∈ S, then x * y = y * x. We must show that y' ∈ S, in which case x * y' = y' * x. x * y' = x * (y + 1) (Inductive Hypothesis) = x * y + x * 1 (From 2) = y * x + x * 1 (Base case) = y * x + 1 * x (From 2) = (y + 1) * x (Factor out x) = y' * x (From 2) This shows that x * y' = y' * x, and it follows that y' ∈ S. Therefore, S = N. 2. Prove (Associative Law of Multiplication) For all x, y, z ∈ N, (x * y) * z = x * (y * z). Proof: Fix x, y ∈ N, and let S = {z | (x * y) * z = x * (y * z)}. By Induction: Base Step: If z = 1, then (x * y) * 1 = x * (y * 1) (x * y) * 1 = x * y (From 2) and x * (y * 1) = x * y (From 2) So, 1 ∈ S. Step II (Inductive Step): Assume z ∈ S, then (x * y) * z = x * (y * z). We want to show that z' ∈ S. Consider: (x * y) * z' = (x * y) * (z + 1) (eq. 8) = (x * y) * z + (x * y) * 1 (Inductive hypothesis) = x * (y * z) + (x * y) (Distributive property of multiplication) = x * (y * z + y) (eq. 8) = x * (y * z') (eq. 8) z' ∈ S ⇒ S = N 3) Prove: The relation ~ on NxN given by (a, b) ~ (c, d) ⇔ ad = bc. Proof: Case 1: Reflexive: For all (a, b) ∈ NxN, the relation holds: ab = ba (commutative property of addition) The relation (a, b) ~ (a, b) holds for all (a, b). The relation is reflexive. Case 2: Symmetric: Show that if (a, b) ~ (c, d), then (c, d) ~ (a, b). Assume (a, b) ~ (c, d): ad = bc. We must show that (c, d) ~ (a, b): cd = da. ad = da (commutative) bc = cb (commutative) Thus, if ad = bc, then cb = da, showing that this relation is symmetric. Case 3: Transitive: We need to show that if (a, b) ~ (c, d) and (c, d) ~ (e, f), then (a, b) ~ (e, f). Assume (a, b) ~ (c, d) ⇔ ad = bc and (c, d) ~ (e, f) ⇔ cf = de. Show that: (a, b) ~ (c, d) <=> ad = bc and ad = bc and cf = de Substituting d, we get: cf = a(b + c - d) atf = bc (simplify) Thus, (a, b) ~ (e, f) holds and relation is transitive. ~ is an equivalence relation on NxN

Follows from Lemma B.

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