Jonathan Love - Thursday, September 27, 2012
Proof Techniques
Direct, Contrapositive, and Contradiction
Say you are given the following:
Definition 1. A jumble is a collection of blickers and snarks satisfying the following
four properties:
(1) Any two blickers are common to at least one snark.
(2) Any two snarks are common to at least one blicker.
(3) There are at most four blickers.
(4) No snark is common to every blicker.
And you are asked to prove:
Proposition 1. If a jumble contains exactly three blickers, then every snark is common
to exactly two blickers.
How can you expect to be able to figure that question out? You may come across
something like this quite often in math - you’re just given a list of statements, and
expected to prove things about them. This would normally be extremely difficult.
However, there are two crucial aspects to any mathematical problem - intuition and
logical proof. These two work together in fascinating ways, but it’s important to be
able to keep the two separate. Most questions you encounter will just want a logical
proof; but very often, you will need some intuition about what the question is even
asking before you can see where to begin. We can teach you proof techniques, but the
only way to learn how to get an intuition for problems is to practice, practice, practice.
In the case of this problem, a bit of intuition might help a little bit. I attached
some figures at the end of this document: the circles are blickers, and the squiggles are
snarks. Suddenly, it becomes incredibly easy to figure out what is a jumble and what
isn’t.
In fact, the theorem becomes almost obvious now; if you have three blickers, and
try to connect them using snarks in a way that fits the properties of a jumble, each
jumble has to connect to exactly two blickers. But unfortunately, that won’t cut it for
a proof. Here is what a proof of our theorem might look like
Proof. Given a jumble with exactly three blickers, by property (4), no snark can be
common to all three; therefore each snark is common to at most two blickers.
We now show that at least three snarks must exist. By property (1), any two blickers
are common to at least one snark. There are three pairs of blickers; but since each snark
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is common to at most two blickers, every pair requires a distinct snark. Thus there are
at least three snarks.
Finally, we show that each snark is common to at least two blickers. Given any
snark S1 , choose another snark (call it S2 ); since there are at least three snarks, we
know that we can find S1 and S2 . By property (2), these are common to at least one
blicker (choose one and call it B1 ). By property (1), the other two blickers (besides
B1 ) are common to at least one snark (choose one and call it S3 ). This snark is not
common to B1 since each snark is common to at most two blickers. Now by property
(2), S3 and S1 are common to at least one blicker, and this blicker is not B1 ; therefore,
S1 is common to at least two blickers: B1 and one other.
Since any snark is common to at most two blickers, and any snark is common to at
least two blickers, we conclude that any snark is common to exactly two blickers.
(my apologies to anyone who attended the talk - I messed up the proof as I was
presenting it. The proof given here is correct, as far as I can tell).
Note that the “logical proof” isn’t just a string of truth tables - there is a lot of
prose writing spread through. Just because intuition doesn’t count as proof doesn’t
mean you should make your explanation as hard to understand as possible!
This was an example of a “direct proof.” We started with certain assumptions, and
just followed the logical implications until we got what we wanted. Note how this was
divided up into steps - (a) each snark is common to at most two blickers, (b) there are
at least three snarks, and (c) each snark is common to at least two blickers. (a) was
used to prove (b), (a) and (b) were used to prove (c), and (a) and (c) together proved
the proposition. This is a good practice to follow: break down your proof into smaller,
easier parts and prove each individually. If you can’t follow the proof, try going through
each step on some examples (such as the ones at the back of these notes).
Direct proof can be useful quite often, but there are some questions it doesn’t work
for, such as the following:
Proposition 2. For any integer n, if n2 is even, then n is even.
If you try to start with assuming n2 is even, it’s almost impossible to learn anything
about n. Many students are then tempted to flip it around: they start with n being
even, and try to prove that n2 is even. DO NOT DO THIS.
If you have a statement “If A then B,” the statement “If B then A” given by flipping
A and B is called the converse of the original statement. And in general, a statement
and its converse are not the same. Consider the statement “If it is raining, then the
ground outside is wet.” Even if this statement is true, its converse, the statement “If
the ground outside is wet, then it is raining” is not true. Perhaps someone had a water
balloon fight, or there was dew, or there was snow and it melted. A statement and its
converse are very different things.
However, a statement, “If A then B,” and its contrapositive, “If not B then not A”
are identical. That is, saying “If it is raining, then the ground outside is wet” and
saying “If the ground outside isn’t wet, then it’s not raining” are logically equivalent 2
if one is true, the other must be. (Try proving this using the “If A then B” and “not”
truth tables!)
So in this case, we can prove the theorem by looking at the contrapositive, and it
suddenly becomes very simple to prove.
Proof. Assume n is odd; then it can be written as 2k + 1 for some integer k. Then
n2 = (2k + 1)2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1,
showing that n2 is odd. Thus, if n2 is even, then n must be even.
To illustrate the last proof technique, consider the following proposition:
Proposition 3. There is no smallest positive rational number.
This problem is tricky because it asks a question about something that (supposedly)
doesn’t exist; how are you supposed to prove anything about that?
Well, one way to prove that a statement A is true is to prove that the statement
“A is false” is false (double negation). So, in a technique called proof by contradiction,
we actually assume the opposite of what we’re trying to prove. This may seem like a
strange thing to do, but if we can then use that assumption to get some contradictory
result, then we know the assumption couldn’t have been true to begin with - that is,
our original statement was true all along.
The best way to understand this technique is to see it in action. Let’s prove Proposition 3 by assuming the opposite of what we want.
Proof. Assume that a smallest positive rational number exists; we will show that this
gives a contradiction.
Let r ∈ Q be the smallest positive rational number. Consider 2r ; this is rational, since
r ∈ Q, 2 ∈ Q, and the quotient of two rational numbers is rational. It is positive, since
r > 0, 2 > 0, and the quotient of two positive numbers is positive. But we have 2r < r,
contradicting the assumption that r is the smallest positive rational number. Since our
assumption that a smallest positive rational number exists results in a contradiction,
no smallest positive rational number can exist.
As a brief summary, we covered three basic proof techniques:
• Direct Proof: Start with what you know, and proceed logically until you get what
you need.
• Proof by Contrapositive: To prove “If A then B,” prove “If not B then not A.”
• Proof by Contradiction: To prove “A,” assume “not A,” and show that this results
in a logical contradiction.
It takes a bit of practice to be able to realize quickly which proof techniques will
work for which problems. This is all a part of building intuition; the more proofs you
do, the better you will get at discovering exactly what you need to do.
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