1. Proof Writing Tips
Tyler Murphy
March 17, 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: A ⊆ C.
Let x ∈ A.
Since A ⊆ B by assumption, then x ∈ B.
Since B ⊆ C by assumption, then x ∈ C.
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2. So x ∈ C.
So A ⊆ C.
Pretty straight forward. Now let’s try contrapositive.
Proof. By contrapositive.
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 ∈ Bc.
Case 1: Suppose x ∈ B. Then A ⊆ B. However, since x /∈ C, then B C.
Case 2: Suppose x ∈ Bc. Then x ∈ A and x /∈ B. So A 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 ∃x | 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.
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3. 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.
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4. 4 USE MORE PAPER
This is the best advice I can give to any student of mathematics. The paper is dead. Use
it. You can recycle stuff later. For now, use the paper. This goes in hand with the last
section. Take your time, use the space.
There are a number of reasons for this. First, when your papers are being graded, it
makes it a lot easier to provide feedback if you leave a visible space between problems, put
ideas on new lines, and take up space. If you are trying to cram your entire proof onto two
lines, there’s no room for Jason to add comments to your work.
Second, if you take up space, it’s much easier for you to follow your reasoning as you
look back over your problems if you get stuck. If you are cramming everything onto a
couple of lines, you’re going to find it much more difficult to follow your own reasoning
later, especially if you’re using your work to study from and it’s got some comments from
Jason crammed in there somehow, making one big cluttered mess of work.
Finally, if you aren’t worried about taking up space, then that’s one less thing you’re
worried about. That’s always a win.
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