# Lecture_7 - MA1100 Lecture 7 Mathematical Proofs Proof by...

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1 MA1100 Lecture 7 Mathematical Proofs Proof by Cases Existence Statements

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MA1100 Lecture 7 2 Using Cases in Proofs (P ¤ Q) Ø R logically equivalent to (P Ø R) (Q Ø R) Proposition (from lecture 6) For m, n œ Z , if 3 | m or 3 | n, then 3 | mn. If 3 | m, then 3 | mn If 3 | n, then 3 | mn same meaning as
MA1100 Lecture 7 3 Using Cases in Proofs (P ¤ Q) Ø R logically equivalent to (P Ø R) (Q Ø R) Proposition If n is an integer, then n 2 + n is an even integer. If n is an even integer , then n 2 + n is an even integer If n is an odd integer , then n 2 + n is an even integer same meaning as

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MA1100 Lecture 7 4 Using Cases in Proofs Proposition If n is an integer, then n 2 + n is an even integer. (i) If n is even , then n = 2k for some integer k. Consider two cases: (i) n even and (ii) n odd Proof So n 2 + n = i.e. n 2 + n is even. (ii) If n is odd , then n = 2k + 1 for some integer k. So n 2 + n = i.e. n 2 + n is even. So for any integer n, we have n 2 + n an even integer.
MA1100 Lecture 7 5 Using Cases in Proofs (P ¤ Q) Ø R logically equivalent to (P Ø R) (Q Ø R) Proposition If n is an integer, then 3 divides n 3 –n. If n is an even integer , then 3 | n 2 –n. If n is an odd integer , then 3 | n 2 –n. same meaning as doesn’t work

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MA1100 Lecture 7 6 Using Cases in Proofs n is an integer 2 cases: n is even or n is odd same as: n = 2k or n = 2k + 1 for some integer k same as: n ª 0 or n ª 1 mod 2 remainder 0 remainder 1 divided by 2 When n is divided by 3, 3 cases: remainder 0 or 1 or 2 same as: n = 3k or n = 3k + 1 or n = 3k + 2 for some integer k same as: n ª 0 or n ª 1 or n ª 2 mod 3
MA1100 Lecture 7 7 Using Cases in Proofs (P ¤ Q ¤ R) Ø S equivalent to (P Ø S) (Q Ø S) (R Ø S) Proposition If n is an integer, then 3 divides n 3 –n. If n ª 0 mod 3, then 3 | n 2 –n. If n ª 1 mod 3, then 3 | n 2 –n. same meaning as If n ª 2 mod 3, then 3 | n 2 –n.

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MA1100 Lecture 7 8 Using Cases in Proofs Proposition If n is an integer, then 3 divides n 3 –n. Case 1: n ª 0 mod 3 Case 2: n ª 1 mod 3 Case 3: n ª 2 mod 3 Consider three cases: Proof So n = 3k for some integer k Then n 3 –n= So 3 | n 3 –n.
MA1100 Lecture 7 9 Congruence and Remainder Proposition Let a, n be integers with n > 0. 2. a is

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## This note was uploaded on 03/19/2012 for the course SCIENCE MA1100 taught by Professor Forgot during the Fall '08 term at National University of Singapore.

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Lecture_7 - MA1100 Lecture 7 Mathematical Proofs Proof by...

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