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# Intro To Abstract Math MA 315

Purdue University Calumet

GPA 3.93

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This 0 page Class Notes was uploaded by Vergie Ankunding on Sunday November 1, 2015. The Class Notes belongs to MA 315 at Purdue University Calumet taught by Peter Turbek in Fall. Since its upload, it has received 14 views. For similar materials see /class/232684/ma-315-purdue-university-calumet in Mathematics (M) at Purdue University Calumet.

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Date Created: 11/01/15

MA 3157 Spring 2009 Implications And An Implied For All77 Assume A and B are conditional statements We de ned the truth value of the conditional statement If A then B77 as follows A B A x B Tr ue Tr ue Tr ue True False False False True True False False True In the last class we considered the following conditional statement If n is even7 then 71 is divisible by 3 Applying the truth table to this conditional statement yields n is even 71 is divisible by 3 n is even i n is divisible by 3 True True True True False False False True True False False True We determined integers n which apply to each row of the truth table For example7 1 n 6 yields the values in the rst row 2 n 4 yields the values in the second row 3 n 9 yields the values in the third row 4 n 11 yields the values in the fourth row This yields that the conditional statement7 If n is even7 then 71 is divisible by 3 is true or false depending on the value of 71 On the other hand7 a mathematician would say that the satement If n is even7 then 71 is divisible by 3 is false This is because7 in mathematics7 implications almost always have an implied7 For all in front of them The statement7 For all 717 if n is even7 then 71 is divisible by 3 is a false statement7 because there are values of 717 for example 71 4 for which the statement is not true Consider the statement7 If n is odd7 then 71 1 is even Since there is an implied for all in front of it7 the statement really is7 For all 717 if n is odd7 then n1 is even Notice that7 in deciding the truth of this statement7 the truth table still comes into play7 in fact7 the de nitions it contain allow us to conclude that this is a true statement The for all forces us to consider the case that 71 may be even the truth table allows us to conclude that the implication is true in this case since the hypothesis is false Note that the statement For all 717 if n is even7 then 71 is divisible by 3 is false However7 there are values of 717 for example7 n 6 for which this statement is true This is expressed by the statement There exists an n with the property that n is even and n is divisible by 3 Notice that this statement is not an implication it contains a conjunction of the hypothesis and conclusion If you try to express this idea by using an implication you will probably not obtain the statement you desire For example the statement There exists an n such that if n is even then 71 is divisible by 3 is true if we can nd one integer n that has the property if n is even then 71 is divisible by 3 It is true that n 6 has that property It is also true that n 11 has that property since 71 11 is not even the hypothesis of if n is even then 71 is divisible by 3 is false and so the implication is true in this case MA 3157 Spring 2009 More on sets and logic Exercise Let A7 B7 and C be sets How can you rewrite each of the following sets PI OVS your answers are correct 1 A m B u 0 2 A o B m 0 Exercise Let A and B be subsets of a set D How can you rewrite each of the following sets Prove your answers are correct 1 A U B0 2 A Bc Exercise Assume A and B are integers Assume C is a subset of both A and B Prove or disprove 0 must be a subset of A B We can de ne intersections and unions for more than two sets Suppose I is a set7 and for each 239 E I we can associate a set Ai We de ne UAZ xl 6 Al for some 2396 I ieI Ai 6 Ai for all 2396 I 61 In the above de nitions7 I can be an in nite set Notation Let a and b denote real numbers with a lt b We de ne 1 11 Ma lt z and z lt b 2 119 Ma S x and z lt b 3 ab Ma lt z and x S b 4 11 Ma S x and x S b Exercise Let NJr denote the set of positive integers What are each of the sets equal to Prove your answers are correct 1 1 1 1 0177 7 w 7 39l H Z1 11071 171ll071 Z1mm I01Z 16N 16N 16N 16N

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