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October 23rd, 2013, 10:57 PM   #1
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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!
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October 23rd, 2013, 11:42 PM   #2
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Re: Mathematical Proof Help!

For the first part, the problem is that the three 'k's don't have to be the same number.
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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.
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October 24th, 2013, 08:35 AM   #4
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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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