Geometry / Mr. Hansen 11/8/2004 [rev. 11/10/2004]

Name: _________________________

Some Examples of Proof by Contradiction

	Note: The abbreviation bwoc means “by way of contradiction.” We always begin by negating the conclusion. In other words, we assume that the conclusion is false. We then proceed to show that this leads to an absurdity, i.e., a contradiction of known mathematical facts or of one of the givens. From this, we conclude that to have assumed the conclusion to be false was invalid. Therefore, the conclusion must have been true all along.
Example I: “If a St. Albans student earns a high grade on a test, then he has studied.”
	To prove this theorem, you need an additional given, namely that failure to study leads inevitably to a mediocre (or worse) grade on a test. If you are clever, you will notice that this second given is logically equivalent to the theorem, but what the heck; this is only an example for learning purposes. Let H = a St. Albans student earns a high grade on a test Let ~H = a St. Albans student earns a not-high grade on a test (i.e., low or mediocre) Let S = a St. Albans student has studied Let ~S = a St. Albans student has not studied Given: H             ~S Þ ~H Prove: S Proof: Assume (bwoc) that S is false. Then by the second given, we must have ~H. But that would contradict the first given, namely that H is true. Since our assumption that S is false leads to a contradiction, it cannot be the case that S is false. (Q.E.D.)
Example II: “There is no such thing as a largest real number.”
	Proof: Assume (bwoc) that a largest real number exists. Although it might be extraordinarily difficult to write the digits of the number down, we will not even attempt to do that. Instead, we will call our number B for “biggest.” Take the inequality 1 > 0, which is certainly true, and add B to both sides. This is valid by the Addition Property of Inequality, which is true for any real number. However, that gives us the inequality B + 1 > B, which proves that the number B + 1 is greater than B. That contradicts our original assumption that a largest real number exists. (Q.E.D.)
Example III: “Mama ain’t happy, ain’t nobody happy.”
	This may be rephrased as “If Mama is not happy, then nobody in the family is happy.” Let M = the event that Mama is happy, and let S be the event that somebody in the family is happy. The theorem that we wish to prove can therefore be written symbolically as ~M Þ ~S. We will take ~M as a given (i.e., hypothesis) and will establish that the ~S follows as a logical conclusion. We are to assume that Mama is never in a “blah” mood; she is always either happy (M) or unhappy (~M), but not both. (In math, we indicate that one or the other is true, but not both, by using the word “xor.”) A crucial additional given is that if a person is a member of the family, then Mama’s mood rubs off on that person. Given: ~M M xor ~M             If X is a member of the family, then Mama’s mood rubs off on X. Prove: ~S Proof: Assume (bwoc) that S is true. That means that someone in the family is happy. In other words, there exists a family member (temporarily named A) such that A is happy (*). By applying the third given, we know that Mama’s mood rubs off on A. By the second given, we have only two cases to consider: M xor ~M. (*) I wanted to have a simple way of referring to the intermediate conclusion that A is happy. That is why I marked it with a star. Case I: If M is true, then we have a contradication of the first given. Therefore, the assumption that S is true (made at the beginning) cannot be true. Case II: If ~M is true, then Mama is unhappy. Again applying the second given, we conclude that A is unhappy. However, that contradicts the “*” statement above. Therefore, the assumption that S is true (made at the beginning) cannot be true. Because both Case I and Case II lead to a contradiction, we must conclude that ~S is true. (Q.E.D.)
Example IV: “If you want an omelet, someone’s gotta break some eggs.”
	Given: You are making an omelet.             Omelets require eggs, i.e., if you are making an omelet, then you must use eggs in it.             An omelet is a conventional dish.             If eggshells are not broken, then the eggs cannot be used in any conventional dish. Prove: Eggs must be broken. Proof: Assume (bwoc) that eggs need not be broken. The second given says, in effect “making omelet Þ using eggs in the omelet,” which is equivalent to the contrapositive statement, “not using eggs in the omelet Þ not making an omelet.” By our initial assumption that eggs need not be broken, together with the fourth given, we can conclude that the eggs cannot be used in any conventional dish. Furthermore, since an omelet is a conventional dish, the eggs cannot be used in an omelet, i.e., we are not using the eggs in an omelet. But since earlier we proved that “not using eggs in the omelet Þ not making an omelet,” we conclude that we are not making an omelet. That, of course, contradicts the first given. (Q.E.D.)