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Stuck on this proof...any help is appreciated. Here is what I have thus far.

Suppose n is odd, then n = 2j + 1 for some integers j, k, and m. then m(2j + 1) = 2jm + m. Since the product of two integers is an integer, let k= jm. => 2k + m

I'm sure I'm just not seeing some algebraic trickery here. Thx!

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If there are no restrictions on $k$, then this assertion can fail. In particular, if $k = 2$, then $kn = 2n$ is always even. – Austin Mohr Nov 11 '12 at 4:27
Ah, good point! Thx! – electr0hed Nov 11 '12 at 4:30

2 Answers 2

up vote 1 down vote accepted

This is false, as written. If $k$ is even, then so is $kn$, regardless of whether $n$ is odd or not.

Now, suppose that $k,n$ are both odd. Then $k=2j+1$ and $n=2m+1$ for some integers $j,m$. Therefore $kn=2(\text{stuff})+1$. (I leave the determination of "stuff" to you, though I will tell you it's an integer.)

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Would it be fair to say, using that counterexample k = 2 would prove the assertion as it is written false then? The "fix" would be to restrict k to being odd? – electr0hed Nov 11 '12 at 4:37
Absolutely right. – Cameron Buie Nov 11 '12 at 6:27

$n=2m+1 \Rightarrow kn=2mk+k=k(2m+1)$. Note that the latter expression is the product of an integer and an odd integer, which is not always odd.

For example, Even*Odd: $(2a)\cdot (2b+1)=4ab+2a=2a(2b+1)$, which is even.

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