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Proof by contrapositive, contradiction
Margaret M. Fleck
8 September 2010
This lecture covers proof by contradiction and proof by contrapositive (section 1.6 of Rosen).
1 Announcements
The second quiz will be a week from today (September 15th). It’s based on material through the end of this week. Study materials are posted on the 173 Exams web page. If you need extra time or other special arrangements for the quiz, please contact Margaret ASAP.
2 Direct proof: example with two variables
Last class, we saw examples of direct proof. Let’s do another example. First, let’s define
Definition 1 An integer n is a perfect square if n = k^2 for some integer k.
And now consider the claim:
Claim 1 For any integers m and n, if m and n are perfect squares, then so is mn.
Proof: Let m and n be integers and suppose that m and n are perfect squares. By the definition of “perfect square”, we know that m = k^2 and n = j^2 , for some integers k and j. So then mn is k^2 j^2 , which is equal to (kj)^2. Since k and j are integers, so is kj. Since mn is the square of the integer kj, mn is a perfect square, which is what we needed to show.
Notice that we used a different variable name in the two uses of the definition of perfect square: k the first time and j the second time. It’s important to use a fresh variable name each time you expand a definition like this. Otherwise, you could end up forcing two variables (m and n in this case) to be equal when that isn’t (or might not be) true.
Notice that the phrase “which is what we needed to show” helps tell the reader that we’re done with the proof. It’s polite to indicate the end in one way or another. In typed notes, it may be clear from the indentation. Sometimes, especially in handwritten proofs, we put a box or triangle of dots or Q.E.D. at the end. Q.E.D. is short for Latin “Quod erat demonstrandum,” which is just a translation of “what we needed to show.”
3 Another example with two variables
Here’s another example of direct proof that I didn’t do in class. Suppose that we want to prove the following
Claim 2 For all integers j and k, if j and k are odd, then jk is odd.
The proof might look like:
Proof: Let j and k be integers and suppose they are both odd. Because j is odd, there is an integer p such that j = 2p + 1. Similarly, there is an integer q such that k = 2q + 1. So then jk = (2p+1)(2q+1) = 4pq+2p+2q+1 = 2(2pq+p+q)+1. Since p and q are both integers, so is 2pq + p + q. Let’s call it m. Then jk = 2m + 1 and therefore jk is odd, which is what we needed to show.
5 Proof by contrapositive
A particularly common sort of rephrasing is to replace a claim by its contra- positive. If the original claim was ∀x, P (x) → Q(x) then its contrapositive is ∀x, ¬Q(x) → ¬P (x). Remember from last week that any if/then statement is logically equivalent to its contrapositive.
Remember that contructing the hypothesis requires swapping the hypoth- esis with the conclusion AND negating both of them. If you do only half of this transformation, you get a statement that isn’t equivalent to the original. For example, the converse ∀x, Q(x) → P (x) is not equivalent to the original claim.
For example, suppose that we want to prove
Claim 8 For any integer k, if 3 k + 1 is even, then k is odd.
This is hard to prove in its original form, because we’re trying to use information about a derived quantity to prove something about a more basic one. If we rephrase as the contrapositive, we get
Claim 9 For any integer k, if k is not odd, then 3 k + 1 is not even.
which is equivalent to:
Claim 10 For any integer k, if k is even, 3 k + 1 is odd.
When you do this kind of rephrasing, your proof should start by explain- ing to the reader how you rephrased the claim. It’s technically enough to say that you’re proving the contrapositive. But, for a beginning proof writer, it’s better to actually write out the contrapositive of the claim. This gives you a chance to make sure you have constructed the contrapositive correctly. And, while you are writing the rest of the proof, it helps remind you of exactly what is given and what you need to show.
So the proof of our original claim might look like:
Proof: We will prove the contrapositive of this claim, i.e. that for any integer k, if k is even, 3k + 1 is odd. So, suppose that k is an integer and k is even. Then, k = 2m for some integer m. Then 3k + 1 = 3(2m) + 1 = 2(3m) + 1. Since m is an integer, so is 3m. So 3k + 1 must be odd, which is what we needed to show.
There is no hard-and-fast rule about when to switch to the contrapositive of a claim. If you are stuck trying to write a direct proof, write out the contrapositive of the claim and see whether that version seems easier to prove.
6 Another example
Here’s another claim where proof by contrapositive is helpful.
Claim 11 For any integers a and b, a + b ≥ 15 implies that a ≥ 8 or b ≥ 8.
A proof by contrapositive would look like:
Proof: We’ll prove the contrapositive of this statement. That is, for any integers a and b, a < 8 and b < 8 implies that a + b < 15. So, suppose that a and b are integers such that a < 8 and b < 8. Since they are integers (not e.g. real numbers), this implies that a ≤ 7 and b ≤ 7. Adding these two equations together, we find that a + b ≤ 14. But this implies that a + b < 15.
Notice that when we negated the conclusion of the original statement, we needed to change the “or” into an “and” (DeMorgan’s Law).
7 Proof by contradiction
Another way to prove a claim P is to show that its negation ¬P leads to a contradiction. If ¬P leads to a contradiction, then ¬P can’t be true, and
Also, we proved last class that, for any integer k, if k is odd then k^2 is odd. So the contrapositive of this statement must also be true: (*) if k^2 is even then k is even.
Now, we can prove our claim:
Suppose not. That is, suppose that
2 were rational. Then we can write
2 as a fraction a b where a and b are integers with no common factors. Since
2 = a b , 2 = a 2 b^2. So 2b
(^2) = a (^2).
By the definition of even, this means a^2 is even. But then a must be even, by (*) above. So a = 2n for some integer n. If a = 2n and 2b^2 = a^2 , then 2b^2 = 4n^2. So b^2 = 2n^2. This means that b^2 is even, so b must be even. We now have a contradiction. a and b were chosen not to have any common factors. But they are both even, i.e. they are both divisible by 2. Because assuming that
2 was rational led to a contradiction, it must be the case that
2 is irrational.
