Can this be proved directly?
If $mn$ is odd, then both $m$ and $n$ are odd.
My homework is telling me to prove this directly, but I believe a direct proof doesn't exist. Obviously this is an easy proof by contrapositive.
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Bill Dubuque
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ddd
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3prime decomposition? – James S. Cook Sep 02 '16 at 22:50
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2Is presenting the $4$ cases and showing that "$mn$ odd" only occurs in one case a direct proof ? – Peter Sep 02 '16 at 22:55
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2$\mathbf Z/2\mathbf Z$ is field. – Bernard Sep 02 '16 at 22:56
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Don't know what it is meant by "directly". Possibly, that if you write $n=2k_n+j_n$ with $j_n=0,1$, and analogously for $m$ and multiply, you will get that for the result to be odd it must be $j_n j_m=1$ . – G Cab Sep 02 '16 at 22:58
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Both $m$ and $n$ divide $mn$. Since $mn$ is odd, it only has odd primes in its factorization, which means only products of odd primes can divide it. – Vincenzo Oliva Sep 02 '16 at 23:57
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By Division Algorithm, if we divide $m$ and $n$ by $2$, we have $m=2p+r$ and $n=2q+s$ with $0\leq r,s <2$. Then $r$ and $s$ are $0$ or $1$, so $rs = 0$ or $1$. Then $mn = (2p+r)(2q+s) = 2(2pr +ps+qr) + rs$. Since $mn$ is odd, $rs$=1. So neither $r$ nor $s$ is $0$. Therefore $r=s=1$ and so $m$ and $n$ are odd.
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2I don't really know what the setter of the OP's homework meant by "prove directly", but how are you "proving directly" that $rs = 1$ implies neither $r$ nor $s$ is $0$? – Rob Arthan Sep 02 '16 at 23:02
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1If $rs=1$, then $r$ and $s$ are both units in the ring $Z$. Since they're positive, they're both equal to one. – B. Goddard Sep 02 '16 at 23:27
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1@B.Goddard That method presumably depends on a lemma that $,0,$ is not a unit (in a nonzero ring), which probably doesn't count as "direct" (i.e. hiding the indirect part in a lemma does not eliminate it). – Bill Dubuque Sep 03 '16 at 14:06
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If by contrapositive you mean contradiction you can simply state:
Every multiple of 2 is divisible by 2.
If you mean that you can only work with odd numbers I'd consider the fundamental theorem of arithmetic. With this you can notice that multiplication carries over the original prime factors thus yielding an odd number.
Lorenzo
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No, a proof by contrapositive (Wikipedia) and a proof by contradiction (Wikipedia) are different. http://math.stackexchange.com/questions/262828/proof-by-contradiction-vs-prove-the-contrapositive – user236182 Sep 03 '16 at 00:16
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Oh ok, right, I only saw it called modus tollens. nice to refresh that ol' stuff. :-) – Lorenzo Sep 03 '16 at 00:49