My Math Forum Mathematical Proof Help!

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 October 23rd, 2013, 10:57 PM #1 Newbie   Joined: Oct 2013 Posts: 4 Thanks: 0 Mathematical Proof Help! Hey, I'm super stuck on these two questions and I need help! Theorem: The sum of three even integers is divisible by 6. Proof: Suppose that x, y and z are even integers. Then x = 2k for some integer k. Similarly, y = 2k for some integer k, and z = 2k for some integer k. Thus x + y + z = 2k + 2k + 2k = 6k, which is indeed divisible by 6. Theorem: The product of an even integer and an odd integer is even. Proof: Suppose m is an even integer and n is an odd integer. If mn is even, then by de finition of even there exists an integer r such that mn = 2r. Also, since m is even, there exists an integer p such that m = 2p, and since n is odd there exists an integer q such that n = 2q + 1. Thus mn = (2p)(2q + 1) = 2r where r is an integer. By definition of even, then mn is even as required. Essentially both proofs are wrong and I'm supposed to find out why but I'm super stuck. It's the last question I have to do! It'd be awesome if someone could help and explain it to me. Thank you!
 October 23rd, 2013, 11:42 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Mathematical Proof Help! For the first part, the problem is that the three 'k's don't have to be the same number.
October 24th, 2013, 04:16 AM   #3
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Re: Mathematical Proof Help!

Quote:
 Originally Posted by Rket12 Hey, I'm super stuck on these two questions and I need help! Theorem: The sum of three even integers is divisible by 6. Proof: Suppose that x, y and z are even integers. Then x = 2k for some integer k. Similarly, y = 2k for some integer k, and z = 2k for some integer k. Thus x + y + z = 2k + 2k + 2k = 6k, which is indeed divisible by 6.
So, let x, y and z be even integers. Then x = y = z = 2k. So, x, y and z must all be the same, so there is only one even integer? That can't be right!

The theorem is nonsense in any case:

2 + 2 + 4 = 8
2 + 4 + 8 = 14

Are you sure it's not three consecutive even integers? It would be true then.

Quote:
 Theorem: The product of an even integer and an odd integer is even. Proof: Suppose m is an even integer and n is an odd integer. If mn is even, then by de finition of even there exists an integer r such that mn = 2r. Also, since m is even, there exists an integer p such that m = 2p, and since n is odd there exists an integer q such that n = 2q + 1. Thus mn = (2p)(2q + 1) = 2r where r is an integer. By definition of even, then mn is even as required.
You can't prove something (mn = 2r) by assuming that in the first place! You've assumed mn is even and then proved it is even (under that assumption). A correct proof is:

Let m be even = 2r and n be odd, then mn = (2r)n = 2(rn), which is even.

 October 24th, 2013, 08:35 AM #4 Newbie   Joined: Oct 2013 Posts: 4 Thanks: 0 Re: Mathematical Proof Help! Yeah, I'm certain it's not three consecutive even integers. It's kind of why I'm so confused, his theorems are kind of off but I think the point is just to find what's false in the proof. Thanks for your help though, it's clearer now!

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