Introduction to Basic Proof Techniques Mathew A. Johnson Throughout this class, you will be asked to rigorously prove various mathematical statements. Since there is no prerequisite of a formal proof class, I think it is wise to take this time and introduce a few of the basic methods of mathematical proof. The four we shall discuss here are mathematical induction, direct proof, proof by contradiction, and proof by contrapositive. Also, we will briefly discuss the structure of proofs of if and only if statements. For tips on how to write clear mathematical proofs, see an accompanying handout on the course website. 1 Mathematical Induction We have discussed this method in detail in class. Your text also has a nice readable summary of this method. For the sake of completeness, we state the method here and give an example. Principle of Mathematical Induction: Let m Z and let P m, P m+1, P m+2,... represent statements which may be true or false. Then all statements {P j } j=m are true if P m is true and if P k+1 is true whenever P k is true. Notice this statement contains a slight generalization of the principle of mathematical induction described on the first day of class. Mainly, the base case need not always correspond to m = 1. Example 1. Prove that 3 divides n 3 + 2n for all n N. Solution: We are asked to prove a list of statements which can be indexed by the natural numbers, and hence it is natural to use mathematical induction. The base case here corresponds to n = 1, and hence we must verify that 3 divides n 3 +2n = 1+2 = 3, which is true. Now for the induction step: assume that 3 divides n 3 + 2n for some n N. We want to show this implies that 3 divides (n + 1) 3 + 2(n + 1). Well (n + 1) 3 + 2(n + 1) = n 3 + 3n 2 + 3n + 1 + 2n + 2 = (n 3 + 2n) + 3 ( n 2 + n + 1 ). Since 3 divides n 3 + 2n by the induction hypothesis and since 3 clearly divides 3 (n 2 + n + 1), it follows that 3 divides (n + 1) 3 + 2(n + 1) as claimed. Thus, the induction step holds and hence the proof is complete by mathematical induction. 1
2 Direct Proof This is probably the most used form of proof. The idea is that if you want to show a statement P is true, then you start from known results and use the rules of inference to show that P is true. No logical tricks or gimmicks, just a straight forward brute force attack! Lets illustrate with an example. Example 2. Prove that if n is odd, then n 2 is odd. Solution: If n is odd, then by definition there must be a k Z such that n = 2k + 1 (definition of an odd integer). Substituting this into the expression n 2 gives n 2 = (2k + 1) 2 = 4k 2 + 4k + 1 = 2(2k 2 + 2) + 1. Setting l = 2k 2 + 2 and noting that l Z, it follows that n 2 = 2l + 1 and hence n 2 must also be odd, as claimed. Notice in the special case where the statement you want to prove is an implication of the form if P, then Q, written in logic symbols as P Q, then you are essentially proving that the combination that P is true and Q is false never happens!! 3 Proof by Contradiction The method of proof by contradiction works as follows. Suppose you have a particular statement, say P, that you want to show is true. In order to prove P by contradiction, one assumes on the contrary that P is false and tries to prove this assumption leads to some logical contradiction. This method is particularly useful in trying to prove statements of the form if P then Q, i.e. P Q. The method of proof by contradiction would approach this by assuming that the desired result is actually false (here, meaning that P is true but that Q is false) and showing this leads to some sort of logical contradiction. This method is probably best illustrated by an example. Example 3. Prove that 3n + 2 is odd whenever n is odd. Solution: This can be proved by a direct proof similar to that given in the previous section. Here, we wish to provide a proof by contradiction. To this end, we assume that the statement 3n+2 is odd whenever n is odd is false. Well, if this statement is false then there must exist some odd number n for which 3n+2 is not odd, and hence is even 1. Since n is odd (by assumption), there exists a k Z such that n = 2k + 1. 1 Indeed, notice this statement can be rephrased as if n is odd, then 3n + 2 is odd, which is in the P = Q form with P meaning n is odd and Q meaning 3n + 2 is odd. 2
Substituting this into the expression 3n + 2 we see 3n + 2 = 3(2k + 1) + 2 = 6k + 5 Since 3n + 2 is even (by assumption), there exists a l Z such that 3n + 2 = 2l (definition of an even integer). From above, it follows that 6k + 5 = 2l, which can be rewritten as 2(l 3k) = 5. This latter equality implies that 5 is divisible by 2 and hence that 5 is an even integer, which is clearly false! Thus, our assumption that 3n + 2 was even for some odd n must be false, and hence 3n + 2 must be odd for all odd n, as claimed. Remark 1. Notice that my first step in the above proof was to simply write down the definitions of what means for n to be odd and for 3n + 2 to be even. It is often good practice to begin a proof with simply writing down the definitions of the information that you are given... if nothing else, this always gives you a place to start! Example 4. Prove that if the integer n is odd, then n 2 is odd. Solution: We have already seen a direct proof of this fact. Here, for comparison purposes, we proceed with a proof by contradiction. To this end, we begin by assuming that the implication if n is odd, then n 2 is odd is false. What does this mean? It means that there is some n Z such that n is odd and n 2 is even. If this is true then since n is odd and n 2 is even, there exists k, l Z such that n = 2k + 1 and n 2 = 2l. Substituting n = 2k + 1 into the formula for n 2, it follows that (2k + 1) 2 = 2l which can be expanded to read as 2l = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1. Setting j = 2k 2 + 2k and noting that j Z, it follows that 2l = 2j + 1, which is impossible (the left hand side is clearly even, while the right hand side is clearly odd). Thus, our assumption that n 2 is even for some odd n must be false. Therefore, it must be the case that if n is an odd integer, then n 2 is also odd, as claimed. Remark 2. Notice that proofs by contradiction sometimes seem quite circular and difficult to follow. This is essentially because when using a proof by contradiction, you have no clear set goal in mind: you just assume what you want to show is false and then try to show that this leads to some sort of logical contradiction. Thus, when trying to prove an implication if P, then Q, as a rule of thumb (to some 2 ) proofs by contradiction are discouraged. A much better (again, in the opinion of some) technique in this case is to use a proof by contrapositive, which we discuss below. In fact, a common mistake is to write a proof by contradiction that is really a proof by contrapositive in disguise... see Remark 3 below. 2 Not necessarily in the opinion of your professor. 3
4 Proof by Contrapositive (or Indirect Proof) We now proceed to consider indirect proofs, which are better known as proofs by contrapositive. To prove an implication if P, then Q, where P and Q are statements, by contrapositive, one assumes that Q is false and tries to prove that P is also false. Notice that if this is true, then it must be that if P is true then Q is true, which is what you originally wanted to prove! Said another way, the statement if P, then Q and the statement if not Q, then not P are logically equivalent!! The benefit of this method over a proof by contradiction is that it provides a very clear goal: while in a proof by contradiction one would assume that P is true and Q is false try to derive some kind of logical contradiction (again, no clear cut goal in mind... you just know that you are trying to derive something which can t possibly be true), when using a proof by contrapositive you assume that Q is false and your goal is to prove that P must also be false. This difference is very subtle, but is an important one. Lets illustrate this method with a few examples. Example 5. Prove that 3n + 2 is odd whenever n is odd. Solution: We saw above how to prove this using contradiction and we now proceed with a proof by contrapositive. To this end, notice that the statement we are trying to prove is if n is odd, then 3n + 2 is odd. Thus, for a proof by contrapositive we begin by assuming that 3n + 2 is not odd, and is hence even. Our goal is then to prove that this implies that n is also not odd, and hence is also even. Since we are assuming that 3n + 2 is even, there exists a k Z such that 3n + 2 = 2k. Thus, 3n = 2(k 1) and hence the product of the numbers 3 and n must be even. Since 3 is odd (it can be written as 2 1 + 1), it follows that n must be even as claimed. Thus, whenever the conclusion of the statement if n is odd, then 3n+2 is odd is false the hypothesis must also be false. Therefore, whenever the hypothesis (n is odd) is true the conclusion (3n + 2 is odd) must also be true as claimed. Example 6. Prove that if n Z and n 3 + 5 is odd, then n must be even. Solution 1: We will prove this statement first by contradiction and then by contrapositive. We begin with contradiction. To this end, we assume that there exists a n Z such that both n and n 3 + 5 are odd. By definition, it follows that there exists k, l Z such that n = 2k + 1 and n 3 + 5 = 2l + 1. Substituting n = 2k + 1 into the 4
formula for n 3 + 5, it follows that the integers k and l satisfy 2l + 1 = n 3 + 5 = (2k + 1) 3 + 5 = 8k 3 + 12k 2 + 6k + 6 = 2(4k 3 + 6k 2 + 3k + 3). Setting j = 4k 3 + 6k 2 + 3k + 3 and noting that j Z, it follows that l and j satisfy 2l + 1 = 2j, which is impossible (the left hand side is odd while the right hand side is even). This contradiction implies that our original assumption that n is odd implies n 3 + 5 is also odd must be false. Hence, it must be that if n 3 + 5 is odd then n must be even, as claimed. Solution 2: Here, we prove the statement by contrapositive. To this end, we assume only that the conclusion that n is even is false, and hence we assume that n is odd, and our goal is to prove this implies that n 3 + 5 is even. Since n was assume to be odd, there exists a k Z such that n = 2k + 1 and hence n 3 + 5 = (2k + 1) 3 + 5 = 8k 3 + 12k 2 + 6k + 6 = 2 ( 4k 3 + 6k 2 + 3k + 3 ). Thus, we see that n 3 + 5 must be even and hence whenever n 3 + 5 is odd it must be the case that n is even, as claimed. Remark 3. It is often the case that students (and professors!) claim to write a proof by contradiction when in fact what they have written would be better interpreted as a proof by contrapositive. Here is an example where this could arise... Example 7. Prove that if n Z and n 3 + 5 is odd, then n must be even. Solution: To illustrate the above remark, I want to provide a proof by contradiction that is actually a proof by contrapositive in disguise. In an effort to form a contradiction, we assume that n is odd and that n 3 + 5 is also odd. Since n is assumed to be odd there exists a k Z such that n = 2k + 1 and hence n 3 + 5 = (2k + 1) 3 + 5 = 8k 3 + 12k 2 + 6k + 6 = 2 ( 4k 3 + 6k 2 + 3k + 3 ). It follows that n 3 + 5 is even, which contradicts our assumption that n 3 + 5 was odd. This contradiction implies that our original assumption that n is odd implies n 3 + 5 5
is also odd must be false. Hence, if n 3 +5 is ever odd it must be the case that n is even. Now, notice that while the above is technically a completely correct proof by contradiction 3, it is really a proof by contrapositive. Why? Let P denote the statement n 3 +5 is odd and let Q denote the statement n is even so that what we are trying to prove is P Q. In the above proof by contradiction, we began by assuming (as one should in any proof by contradiction) that P is true and that Q is false. However, what we did from this point was to prove that Q being false implies that P can not be true 4. Said differently, we proved that if Q is false then P is false. But this is EXACTLY the strategy of a proof by contrapositive! 5 A Note on If and Only If Proofs Throughout this course, there will be several times when you need to prove a statement of the form P is true if and only if 5 Q is true. Essentially, the statement P Q means that the only way P can be true is if Q is true, and vice versa, and hence you can see why such statements are so important in mathematics. The main observation is that the statement P iff Q is equivalent with both of the statements if P, then Q and if Q, then P being true. Thus, in order to prove P Q it is necessary to prove both of the statements P Q and Q P. Lets illustrate this with an example. Example 8. Prove that the integer n is odd if and only if n 2 is odd. Solution: We must prove both of the statements if n is odd, then n 2 is odd and if n 2 is odd, then n is odd. We begin with the former. If n is odd then there exists a k N such that n = 2k + 1 and hence n 2 = (2k + 1) 2 = 4k 2 + 4k + 1 = 2 ( 2k 2 + 2 ) + 1 and hence n is odd as claimed. Now, we must prove the implication if n 2 is odd then n is odd. To do this, we use an indirect proof, i.e. a proof by contrapositive. To this end, we assume that n is even and try to prove that n 2 is also even. Well, if n is even then there exists a m Z such that n = 2m and hence n 2 = (2m) 2 = 4m 2 = 2 ( 2m 2). 3 In particular, students would never lose points for submitting such a proof. 4 Indeed, the assumption that P is true was only used to provide the contradiction, and not in the steps leading up to the contradiction. 5 The phrase if and only if is used so much in mathematics that it is given its own special abbreviations: it is written either as iff or as the symbol. 6
Thus, n 2 is also even and hence we have proven the implication if n 2 is odd then n is odd by contrapositive. Remark 4. The above example shows that when proving iff type statements, one often uses different methods of proofs for each implication. Thus, when confronted with and iff type statement, don t be afraid to just start trying various proof techniques until you find one that works for you! 7