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April 8th, 2014, 03:16 AM   #1
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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.
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April 8th, 2014, 03:23 AM   #2
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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.
Thanks from shunya

Last edited by Olinguito; April 8th, 2014 at 03:31 AM.
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