Then, n = k, If k mod(4) = 1, then k = 4q + 1, for some integer q. Therefore, instead of proving p ⇒ q, we may prove its contrapositive ¯ q ⇒ ¯ p. Since it is an implication, we could use a direct proof: Assume ¯ q is true (hence, assume q is false). I will assume that x is odd and y is even without loss of generality, since x and y are commutative. What we have learned then is that if we have a statement if A, then B, we can equally use the statement if not B, then not A. There are four cases to consider.
If n is a positive integer such that n mod(3) = 2, then n is not a perfect square. This statement is true because it is a restatement of a definition. The method of contrapositive comes from the broader field of logic and is a method of rewriting any statement that looks like “If A is true, then B is true.” Method of Contradiction: Assume P and Not Q and prove some sort of contradiction.

Contrapositive proof: Assume that x and y have different parity (~Q). This method can have the convenience that working backwards has in problem-solving. So the ‘if-then’ versions must also be saying the same thing. Then, n = k. If x and y are two integers whose product is even, then at least one of the two must be even. The method of proof by contraposition is based on the logical equivalence between a statement and its contrapositive. If A gives you plenty of information, you are working forwards towards B. Since all odd numbers have the shape for some other whole number , we can calculate using substitution: Since is a whole number, this means that is odd. The latter can be proved by contradiction. The underlying reasoning is that since a conditional statement is logically equivalent to its contrapositive, if the contrapositive is true, then the statement must also be So, here is the proof – using contrapositive. The rule I have just described is what goes by the title of ‘contrapositive.’. ( Log Out /  Change ), You are commenting using your Facebook account. 20 0 obj <> endobj In other words, you begin to realize that the inside/outside statements are actually just different ways of saying the same thing. Methods of Proof – Exam Worksheet & Theory Guides Now then, we compute the sum x+y = 2k + 2m + 1 = 2(k+m) + 1, which is an odd integer by definition. Proof by contrapositive is based on the fact that an implication is equivalent to its contrapositive. Prove by contrapositive: Let x 2Z. We will prove the contrapositive version: "If n is a perfect square then n mod(4) must be 0 or 1." (Hint: To begin with is even and end with is even, we know since is even that for some whole number . � �R ��PF�¦B9'AԩB% Note: When made to match the Venn diagram earlier in the article, the green region is to be thought of as ‘ is an even number’ and the red region is to be thought of as ‘ is also even.’. If x and y are two integers for which x+y is even, then x and y have the same parity.

About Proof by Contrapositive In this new visual situation, it is equally easy to think about outsides of shapes rather than their insides. Translating this into ‘if-then’ language as we did earlier with A and B, we can say that ‘If G, then R‘ (where by G we mean ‘inside of the green circle, and similarly with R). In case this isn’t familiar, a Venn diagram is a collection of circles used to represent some kind of situation happening. The method of contrapositive comes from the broader field of logic and is a method of rewriting any statement that looks like “If A is true, then B is true.” To see how this works, we will use visuals using what are normally called Venn diagrams.
Proof: We use the method of contrapositive. The outside of the green circle G are the regions colored blue and red. Then, n = k, If k mod(4) = 3, then k = 4q + 3, for some integer q. Suppose that x is even. So, the outside of R is literally contained inside of the outside of G. Since we have already learned that this ‘contained in’ relationship can be translated into if-then, we can use this to express the idea. aH��/����t^��(����uT������|V��P9\ q��5q���M��s3���4 Proof by Contrapositive 1. One piece of good advice in problem-solving is to try to work backwards. And this one doesn’t involve the very strange idea of intentionally saying something false. Change ), You are commenting using your Google account. ( Log Out /  This contrapositive, like the original statement, is also true. Write x = 2a for some a 2Z, and plug in: x26x+ 5 = (2a)26(2a) + 5 = 4a212a+ 5 = 2(2a26a+ 2) + 1: Thus x26x+ 5 is odd.

endstream endobj startxref You can solve this problem in the ‘regular way’ if you use some information about prime numbers – but if you do contrapositive, the only thing involved is the distributive property and the definition of even and odd numbers – which are simpler than the ideas you have to use to go directly from A to B. A proof by contraposition (contrapositive) is a direct proof of the contrapositive of a statement. hޤY�SG�W�c�Rhޯ�U��@p����I^-6�������.`�]ۚ������c�

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Since one of these integers is even and the other odd, there is no loss of generality to suppose x is even and y is odd. Suppose n = k2. If n is a positive integer such that n mod(4) is 2 or 3, then n is not a perfect square.

Theorem: If is an even number, then is also even. Proof.


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