CSE 215 Lecture 11: Midterm Review
3 Mar 2017
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If the proof begins with forall x and does not give us a p(x), we cannot do the step where we assume. The difference of the squares of two consecutive integers is odd. Gupta believes we will make a mistake here (formalizing) Aka take any x, if we have two consecutive numbers, Since there is no p(x), we cannot assume anything. [(x+1)2 - x2] = (x2 + 2x + 1) - x2 = 2x + 1. (2x+1) is odd, since x is an integer. For all integers a, b, and c, if a|b and a|c then a|(2b-3c) Forall a,b,c, if (a|b and a|c) then a|(2b-3c). We know that a divides b and a divides c are both true. By definition, a|b implies that b = ar for some integer r. By definition, a|c implies that c = as for some integer s.
