Homework 2 2-24
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This is a detailed solution to problem 2.2.24 from Discrete Mathematics With Graph Theory, 3rd Ed, by Goodaire and Parmenter.
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Homework 2 2-24
- 1. Homework Problem 2.2.24 February 20, 2014 I chose this problem to show you because it demonstrates a very good principle for writing proofs and how we sometimes have to stop and think about what’s going on and what the implications of our situation are. 2.2.24 Let A, B, C be subsets of some universal set U . Prove that A(BC) = (AB) ∪ (AC c ) Note that this problem is asking us to prove that one thing is equal to another. That means we have to prove this in two directions. The first is pretty straightforward, so let’s look at that one. First, unpack what you have. Do this before the proof as scratch work. This will help you see what you have and what you want. You don’t need to include this in your proof. First look at the left-hand side of what we are working with. A(BC) means A ∩ (BC)c . This means A ∩ (B ∩ C c )c . Unpacking further, we get A ∩ (B c ∩ C). Now look at the right-hand side. c (AB) ∪ (AC c ) means (A ∩ B c ) ∪ (A ∩ C c ). Cancelling the double complements we get (A ∩ B c ) ∪ (A ∩ C). So we really want to prove that A ∩ (B c ∩ C) = (A ∩ B c ) ∪ (A ∩ C). Proof. (⇒) Let x ∈ A(BC) be true. This means that x ∈ A ∩ (BC)c . This also means that x ∈ A ∩ (B ∩ C c )c . So x ∈ A ∩ (B c ∩ C). 1
- 2. So x ∈ A and x ∈ B c ∩ C . So x ∈ A and x ∈ B c and x ∈ C. So x ∈ A and x ∈ B. / So x ∈ A ∩ B c . Also, x ∈ A and x ∈ C. So x ∈ A ∩ C. So x ∈ (A ∩ B c )orx ∈ (A ∩ C) is true. (note that we know it is in both. But we want to show that it is in the union, which is an ”or” statement. This is one of those creative steps in realizing what you want versus what you have.) So x ∈ (A ∩ B c ) ∪ (A ∩ C), which is equivalent to (AB) ∪ (AC c ). ————————————————————————————————————This is the first direction. It only took a little bit of out of the box thinking to get where we needed to go. Now let’s prove the other direction and see what happens. ————————————————————————————————————– Proof. (⇐) Let x ∈ (AB) ∪ (AC c ). c This means that x ∈ (A ∩ B c ) ∪ (A ∩ C c ). Simplifying, this means that x ∈ (A ∩ B c ) ∪ (A ∩ C). So x ∈ (A ∩ B c ) or x ∈ (A ∩ C). Now we have two cases. Case 1: Assume x ∈ A ∩ B c . So x ∈ A and x ∈ B c . x ∈ A ∩ Bc. x ∈ A and x ∈ B c . x ∈ A and x ∈ B. x ∈ A and x ∈ B ∩ C c . (Think about why this is true and why we can just add the C c here.) x ∈ A and x ∈ BC. x ∈ A and x ∈ (BC)c . x ∈ A ∩ (BC)c . x ∈ A(BC). Case 2: Assume x ∈ A ∩ C. So, x ∈ A and x ∈ C. So, x ∈ A and x ∈ C c . / So, x ∈ A and x ∈ (C c ∩ B). / So, x ∈ A and x ∈ (B ∩ C c ). / So, x ∈ A and x ∈ (BC). / 2
- 3. So, x ∈ A and x ∈ (BC)c . So, x ∈ A ∩ (BC)c . So, x ∈ A(BC). Some things to notice: 1. How to structure a proof. 2. When you get stuck, think about what you want to get to and ask yourself how you can bring things into your proof that will help you get there. 3. Pay attention to all situations. Most proofs will have more than one case to deal with. 4. All proofs where you show that one thing is equal to another has two directions. 3