CMPSC 360 Lecture Notes - Lecture 8: Reflexive Relation, Inverse Relation, If And Only If
26 Feb 2017
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Assume the first proposition, but negate the conclusion. Assume that the fraction is reduced, so a and b have no common factor a = 3m + 1 where m is an int then a2 = 9m2 + 6m + 1 so, 3 does not divide a2. Proof: if 3|a2 then 3|a when a is an integer (contrapositive) Suppose 3 does not divide a there is a remainder 1 or 2 when a is divided by 3. So, a = 3d where d is an int. There exists (cid:1853),(cid:1854) (cid:1852) (cid:1871)(cid:1873)(cid:1855) (cid:1872) (cid:1853)(cid:1872) (cid:883)+(cid:885) (cid:884)=(cid:3028)(cid:3029)and b is not zero so (cid:885) (cid:884)=(cid:3028)(cid:3029) (cid:883)=(cid:3028)(cid:3029) (cid:3029)(cid:3029) But (cid:884)is irrational, so the above is a contradiction. P q = p q ^ q p. Prove each of the conditionals on the right by any of the previously mentioned methods try to avoid proof by contradiction, as it can lead to ambiguous results. A relation is a set of ordered pairs.
