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April 8th, 2014, 03:16 AM  #1 
Senior Member Joined: Oct 2013 From: Far far away Posts: 422 Thanks: 18  Prove: mn odd iff m odd & n odd
Let m, n denote any two natural numbers. prove that mn i odd IFF m and n are odd My attempt: 1) If m odd and n odd then mn odd Let m = 2a + 1 and n = 2b + 1 mn = (2a + 1)(2b + 1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1 = odd 2) If mn odd then m = odd and n = odd There are three possibilities: i) If mn odd then m = even and n = even Let mn = odd and m = 2x and n = 2y mn = (2x)(2y) = 2(2xy) = even Contradiction so m = even and n = even is false ii) If mn odd then m = even and n = odd Let mn = odd and m = 2x and n = 2y + 1 mn = (2x)(2y + 1) = 2(2xy + x) = even Contradiction so m = odd and n = even is false iii) If mn odd then m = odd and n = odd Let m = 2x + 1 and n = 2y + 1 mn = (2x + 1)(2y + 1) = 4xy + 2x + 2y + 1 = 2(2xy + x + y) + 1 = odd No contradiction So if mn is odd then m = odd and n = odd Therefore mn is odd IFF m is odd and n is odd Is this proof acceptable? Is there a better proof? Thanks Question 2) With reference to the previous question, is it true that mn is even iff m and n are even? Prove your answer My attempt: 1) If m even and n even then mn is even Let m = 2a and n = 2b then mn = (2a)(2b) = 2(2ab) = even 2) If mn even then m is even and n is even There are three possibilities: i) if mn even then m is odd and n is odd mn is even and let m = 2a + 1 and n = 2b + 1 mn = (2a + 1)(2b + 1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1 = odd Contradiction so not possible ii) If mn even then m is even and n is even Let m = 2a and n = 2b mn = (2a)(2b) = 2(2ab) = even So this is possible iii) if mn even then m is odd and n is even let m = 2a and n = 2b + 1 mn = (2a)(2b + 1) = 2(2ab + a) = even This is possible too So if mn is even then m is even and n is even is false Therefore it is false that mn is even IFF m is even and n is even Is my proof correct? Is there a simpler better way? Thanks Last edited by shunya; April 8th, 2014 at 03:22 AM. 
April 8th, 2014, 03:23 AM  #2 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 155 Math Focus: Abstract algebra 
Your proofs are fine, though they can be shortened a little. For example to prove that (2) is false you only have to give one counterexample: Let $m=2$, $n=3$; then $mn=6$ is even even though one of $m$ and $n$ is odd.
Last edited by Olinguito; April 8th, 2014 at 03:31 AM. 

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