Proof writing tips

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Published on March 4, 2014

Author: tylerisaacmurphy

Source: slideshare.net

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proof writing tips for Math 189, BSU, spring 2014

Proof Writing Tips Tyler Murphy February 3, 2014 Here’s a few helpful ideas to remember as you guys are writing proofs for this class. These are based on some observations I’ve made about your work as the semester has gone on. 1. Remember the rules of logical equivalence 2. Use deﬁnitions 3. Put new ideas on new lines. 4. USE MORE PAPER Let me expand on these a little. 1 Remember the rules of logical equivalence What I mean by this is that you have a lot of tools in your belt. We spent the ﬁrst few weeks of the class going over proof styles. We can do things directly, by contrapositive, or by contradiction. Most of the problems we had in chapter 2 we solved directly. That often left a lot of cases to be dealt with. Let’s look at a problem and go through it a few ways. Remember that if you are working through something and you get stuck, try a diﬀerent proof method. Sometimes it works out even better. A lot of times it’ll also help you understand the ﬁrst way you were trying to go. Example: Prove that If A ⊆ B and B ⊆ C, then A ⊆ C. Proof. Directly: Assume A ⊆ B and B ⊆ C is true. WTS: AsubseteqC. Let x ∈ A. Since A ⊆ B by assumption, then x ∈ B. Since B ⊆ C by assumption, then x ∈ C. 1

So x ∈ C. So A ⊆ C. Pretty straight forward. Now let’s try contrapositive. Proof. By contrpositive. First consider what the contrapositive of the statement is. We have ¬Q → ¬P. So we have A C → A B or B C. (Ask yourself why this is now an ”or” statement) Since A C, then ∃x ∈ A such that x ∈ C. / Choose this x. Now consider the relationship x has with B. Either x ∈ B or x ∈ B c . Case 1: Suppose x ∈ B. Then A ∈ B. However, since x ∈ C, then B C. / c . Then x ∈ A and x ∈ B. So A Case 2: Suppose x ∈ B / B. Since either case satisﬁes one of the things we need, then we are done. So, If A ⊆ B and B ⊆ C, then A ⊆ C by contrapositive. Now let’s consider this problem by contradiction. Proof. Suppose that if A ⊆ B and B ⊆ C, then A C. Assume A ⊆ B and B ⊆ C is true. If A C, then existsx | x ∈ A and x ∈ C. / Choose this x. Since A ⊆ B, then x ∈ B. Since B ⊆ C, then x ∈ C. However, we chose the x that is not in C. So we have a contradiction. So, if A ⊆ B and B ⊆ C, then A ⊆ C. So you can see how these proofs can be done by any of these methods. 2 Use deﬁnitions Often times problems can be unclear what is being asked right away. The best way to analyze what is being asked is to use deﬁnitions. For example, if you were given a problem that says ”Prove that (A∩((B ⊕C)A))∪A = A. First, you need to unpack what you have. Like we have always done, use the order of operations. Start inside the smallest parenthesis. 2

B ⊕ C = B ∪ C − B ∩ C. This means that we have everything in B and everything in C, but not the things in both. (So we have unpacked what the symbols means and we can understand what is going on.) So we can put this substitute this into our original statement. (A ∩ ((B ⊕ C)A)) ∪ A = (A ∩ ((B ∪ C − B ∩ C)A)) ∪ A. Now we can unpack the next thing. (B ∪ C − B ∩ C)A) means (B ∪ C − B ∩ C) ∩ Ac ). So we have everything in B and everything in C, but not the things in both intersected with all the things not in A. So if we were to describe the elements we are dealing with, we would say: {x | x ∈ A / and (x ∈ B or x ∈ C, but x ∈ B ∩ C.)} / Now if we were to put this in conjunction with our next requirement, we would have A ∩ ((B ∪ C − B ∩ C) ∩ Ac ). So now we can see that we have the intersection of all the things in A with a set containing only things not in A. So we don’t have any elements that are in A and not in A. So we have the empty set ∅. So (A ∩ ((B ⊕ C)A)) = ∅. So we can substitute this equivalence into the original equation and we get (∅) ∪ A), which equals A. This is what we sought out to prove. Note that we did this all without having to deal with a speciﬁc element and didn’t need to chase it around. It all came from unpacking deﬁnitions. It is a very handy and vital tool to have. 3 Put new ideas on new lines and be clear that you are onto a new thought. This may seem overly simple, but it makes everything so much easier. Proofs are easier to grade, logic is easy to follow, and mistakes can be easily identiﬁed. Let me show you what I mean. Consider the following proofs for the problem ”If A ⊆ B and B ⊆ C, then A ⊂ C.” Proof (incorrect): Proof (correct): Assume A ⊆ B and B ⊆ C. Assume A ⊆ B and B ⊆ C. Let x ∈ A, x ∈ B, x ∈ C. Let x ∈ A. So A ⊆ C. Since A ⊂ B by assumption, x ∈ B. Since B ⊆ C by assumption, x ∈ C. So A ⊆ C. 3

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