Example Proofs (easy)

Example Proofs (easy) - Proof. Suppose that x and y are two...

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Math 100 In Class Proofs Martin H. Weissman On October 8, students were asked to write proofs of three (or four) theorems. Here are some “model proofs”. Theorem 1. The product of two odd numbers is odd. Proof. Suppose that x and y are two odd numbers. Choose integers a and b such that x = 2 a + 1 and y = 2 b + 1. Then x · y = (2 a + 1)(2 b + 1). Hence x · y = 4 ab + 2 a + 2 b + 1. Factoring yields: x · y = 2(2 ab + a + b ) + 1 . Therefore x · y is odd. / Theorem 2. If x 1 ,x 2 ,y 1 ,y 2 Z , and x 1 y 1 ( mod n ) and x 2 y 2 ( mod n ) , then x 1 + x 2 y 1 + y 2 ( mod n ) . Proof. Suppose that x 1 ,x 2 ,y 1 ,y 2 Z , and x 1 y 1 (mod n ) and x 2 y 2 (mod n ). Choose integers m 1 and m 2 such that x 1 - y 1 = m 1 n and x 2 - y 2 = m 2 n . It follows that: x 1 - y 1 + x 2 - y 2 = m 1 n + m 2 n. It follows that: ( x 1 + x 2 ) - ( y 1 + y 2 ) = ( m 1 + m 2 ) n. Hence x 1 + x 2 y 1 + y 2 (mod n ). / Theorem 3. If x 1 ,x 2 ,y 1 ,y 2 Z , and x 1 y 1 ( mod n ) and x 2 y 2 ( mod n ) , then x 1 · x 2 y 1 · y 2 ( mod n ) . Proof. Suppose that x 1 ,x 2 ,y 1 ,y 2 Z , and x 1 y 1 (mod n ) and x 2 y 2 (mod n ). Choose integers m 1 and m 2 such that x 1 - y 1 = m 1 n and x 2 - y 2 = m 2 n . It follows that x 1 = m 1 n + y 1 and x 2 = m 2 n + y 2 . Therefore, x 1 x 2 = m 1 m 2 n 2 + m 1 y 2 n + m 2 y 1 n + y 1 y 2 . Rearranging terms yields x 1 x 2 - y 1 y 2 = ( m 1 m 2 n + m 1 y 2 + m 2 y 1 ) n. Hence x 1 x 2 y 1 y 2 (mod n ). / Theorem 4. The product of two consecutive integers is even.
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Unformatted text preview: Proof. Suppose that x and y are two consecutive integers. We may assume that x is the smaller of the two numbers. Thus y = x + 1. It follows that the product of x and y is equal to x 2 + x . Now, we consider two cases: when x is even, and when x is odd. If x is even, then choose a Z such that x = 2 a . It follows that: x 2 + x = (2 a ) 2 + 2 a = 4 a 2 + 2 a = 2(2 a 2 + a ) . Hence x 2 + x is even. In the other case, x is odd. Choose b Z such that x = 2 b + 1. It follows that: x 2 + x = (2 b + 1) 2 + (2 b + 1) = 4 b 2 + 4 b + 1 + 2 b + 1 = 2(2 b 2 + 3 b + 1) . Hence x 2 + x is even. In both cases, we nd that the product of consecutive integers is even. / 1...
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This note was uploaded on 04/17/2008 for the course MATH 100 taught by Professor Weissman during the Fall '08 term at UCSC.

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