Exercise 37/40, p. 94/93

Prove or disprove: the sum of an integer and its cube is even.

1. Decide whether it's true or not. You might want to sketch some ideas on some scratch paper (what a concept!).
2. If false, give a counterexample;
3. if true, map out a proof strategy. Pick out P and Q, and then choose from amongst your favorite proof strategies. In this case,
   • P: x an integer
   • Q: the sum of x and its cube is even

To illustrate the ideas of the section, we'll do the proof of this one in three different ways:

1. directly,
2. by contradiction, and
3. contrapositively

Direct

This illustrates the use of cases in an argument. The cases must exhaust all possibilities!

Let x be an integer.

We consider , and hope to show that it is even. Now we can factor, , so we consider two cases:

• Let x be even: then x=2n for some integer n, and then is even, as 2 is a factor.
• Let x be odd: then is even, as is odd (as the product of two odds (x and x!)). We can write for some integer m, hence is even, as 2 is a factor.

In either event, is even. Q.E.D.

You may object that we need to prove that the product of odd integers is odd. If so, then this is called a lemma (a theorem proven to help prove another theorem).

Lemma: the product of odd integers is odd.

Proof: given two odd integers, a and b. Then we can write a=2m+1 and b=2n+1, where m and n are integers. Consider the product: ab=(2m+1)(2n+1)=2(2mn+m+n)+1, which is odd as the sum of an even integer and one. Q.E.D.

Contradiction

Note that the contradiction occurs not in P or Q, but in a sub-wff. Any contradiction will do!

Let x be an integer, and assume that is odd. A product ab is odd if and only if both a and b are odd. , so x must be odd, and must be odd. But is odd, as a product of odd integers; hence is even. But this is a contradiction. Therefore, is not odd, but rather is even. Q.E.D.

Contrapositive

Restated: ``If is not even, then x is not an integer.''

Notes:

• Be careful in this case, as the negation of ``even'' is not ``odd''! (Note also the importance of punctuation in the previous sentence for the logic!)
• We're trying to show that x is not an integer, and even and oddness doesn't make any sense for general real numbers!
• Since this proof invokes the proof by contradiction, it is clearly more painful than that proof.
• Having restated it, would you want to prove it? I argue that this form is far less appealing than the others!

Let be not even, so that is not even. If x is an integer, then so is , and so it is either even or odd. Since we are assuming that is not even, it must be odd. [Now goto the proof by contradiction, to show that cannot be odd.] Since it cannot be even, by hypothesis, and cannot be odd, by contradiction, we conclude that x is not an integer. Q.E.D.

More elaborate proof

Direct

  
Table: Direct Proof, gory detail