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Answer to Question #74361 in Discrete Mathematics for Marquise
Question #74361
Proposed proof: Suppose that n or m are even. Then n = 2k and m = 2j for some integers k and j. This shows that n*m = (2k)*(2j) = 4k*j. Therefore, n*m is even
Expert's answer
Suppose that m is even and n is odd.
So, m=2k,n=2i+1.
Thus, mn=2k(2i+1)=2l is even.
So, m=2k,n=2i+1.
Thus, mn=2k(2i+1)=2l is even.
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