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# 7 - LECTURE 7 CASES AND COMMON ERRORS(3.4-3.5 1 Proof by...

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LECTURE 7: CASES AND COMMON ERRORS ( § 3.4-3.5) 1. Proof by Cases Occasionally it is useful to break a proof into smaller pieces that can each be dealt with. Example 1. If the domain is N we might consider x = 1 as one case, and x > 1 as another case. If the domain is Z we might consider the two cases: x is odd, x is even. If the domain is R we might consider x = 0, x > 0, and x < 0 as three cases. Result. If n Z then n 2 + 3 n + 5 is odd. Proof. We proceed by cases, according to the parity of n . Case 1: n is even. Then n = 2 x for some x Z . Thus n 2 +3 n +5 = 4 x 2 +6 x +5 = 2(2 x 2 +3 x +2)+1 is odd. Case 2: n is odd. Then n = 2 y + 1 for some y Z . Thus n 2 + 3 n + 5 = (2 y + 1) 2 + 3(2 y + 1) + 5 = 4 y 2 + 4 y + 1 + 6 y + 3 + 5 = 2(2 y 2 + 5 y + 4) + 1 is odd. We say that two integers have the same parity if they are both even, or both odd. Otherwise, we say they have opposite parity . Result. Let x, y Z . Then x and y are of the same parity if and only if x + y is even. Proof. First, assume that x and y are of the same parity. We have two cases.

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7 - LECTURE 7 CASES AND COMMON ERRORS(3.4-3.5 1 Proof by...

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