My Math Forum Mathematical Proof Help!

 Applied Math Applied Math Forum

 October 23rd, 2013, 09:57 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, 10:42 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, 03:16 AM   #3
Senior Member

Joined: Jun 2013
From: London, England

Posts: 1,316
Thanks: 116

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, 07:35 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!

 Tags mathematical, proof

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post arun Algebra 5 November 23rd, 2011 07:23 AM doleary22 Applied Math 6 October 4th, 2011 09:12 PM Dave New Users 19 November 23rd, 2010 11:49 PM Jesusdpm Algebra 2 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top