Aproof by contrapositiveproves a conditional theorem of the form p→c by showing that thecontrapositive ¬c→¬p is true. In other words, ¬c is assumed to be true and ¬p is proven as aresult of ¬c.Many theorems are conditional statements that also have a universal quanti±er such as:The domain of variable n is the set of all integers. If D(n) is the predicate that says that n is odd,then the statement is equivalent to the logical expression:∀n (D(n)→D(n)). A proof bycontrapositive of the theorem starts with n, an arbitrary integer, assumes that D(n) is false, andthen proves that D(n) is false.The animation below gives a proof by contrapositive that for every integer n, if 3n + 7 is odd thenn is even. The theorem is a conditional statement in which the hypothesis is that 3n + 7 is oddand the conclusion is that n is even. The proof implicitly uses the fact that every integer is evenor odd, so if an integer is not even, then it is odd. A contrapositive proof assumes the negation ofthe conclusion (n is odd) and uses the assumption to prove the negation of the hypothesis(3n + 7 is even). Why use a contrapositive proof for this proof? One reason, is that the negationof the assumption (n is odd) is a little simpler and therefore easier to work with than thehypothesis (3n + 7 is odd).The statements "n is odd" and "3n + 7 is even" need to be translated from English into amathematical expression to which the prover can apply the standard rules of algebra. Anevenintegercan be expressed as 2k for some integer k. Anodd integercan be expressed as 2k + 1for some integer k. The (2k + 1) expression for n can be plugged into 3n + 7. The goal of thealgebra is then to show the resulting expression is 2 times some integer.PARTICIPATIONACTIVITY2.3.1: Proof by contrapositive example.Deciding whether to prove a conditional statement using a direct proof or a proof bycontrapositive often involves some trial and error. The decision should be based on whether the222Animation captions:1. A proof by contrapositive starts with the negation of the conclusion and derives the negation othe hypothesis.2. The negation of the conclusion is thatnis an odd integer. Therefore,n= 2k+ 1for some integek.3. The next step is to plug the expression forninto3n+ 7and show that the result is equal to2(3k+ 5), which is 2 times an integer.4. Therefore,3n+ 7is even, which is the negation of the hypothesis.©zyBooks 02/19/18 18:48 156189Timothy NgoIITCS330BistriceanuSpring2018©zyBooks 02/19/18 18:48 156189Timothy NgoIITCS330BistriceanuSpring2018