I'm practicing for a test on Monday and I'm trying to do some proofs - but I'm not entirely sure if this is sufficient enough for the question.

"Prove that for all integers, m and n, if m is odd and n is even, then $m\cdot n$ is even."

These are the steps I've gone through - but I'm a little unsure on the result.

  1. $m = 2k_1+1$ and $n = 2k_2$
  2. $m\cdot n= (2k_1 + 1)(2k_2)$
  3. $= 4k_1k_2+2k_2$
  4. $=2(2k_1k_2+k_2)$

The part I'm unsure on is the last step, I wasn't sure if I was supposed to factor out the $k_2$ as well.

  • $\begingroup$ That's fine. You could have jumped directly from step 2 to step 4. $\endgroup$ – Adriano Sep 20 '14 at 18:42
  • $\begingroup$ Apart from the fact that you used equations, with no explanatory words, everything is fine. We can even skip part of the multiplication step. We have $m=2a+1$ and $n=2b$ for some integers $a$ and $b$. Thus $mn=nm=(2b)(2a+1)=(2)\left[(b)(2a+1)\right]$, so $mn$ is even. $\endgroup$ – André Nicolas Sep 20 '14 at 18:45

You've done all the work, and it's correct. Let me try to demonstrate how improve the style of your proof. This can be very important—graders are invariably tired and overworked, and lack of clarity is one of the most frequent causes of correct proofs being marked as incorrect on exams.

Additionally, it feels good for both reader and writer when the math is crisp and clear—it is an often-overlooked fact that the entire purpose of mathematical writing is to communicate ideas to other people.

Claim: If $m$ and $n$ are integers, and $n$ is even, then $mn$ is even.

Proof: Because $n$ is even, we can write $n=2l$, where $l$ is an integer. Then $mn=m(2l)=2(ml)$ is a multiple of $2$. In other words, $mn$ is even.

Notice that each step is justified—though if we had an extremely strict teacher, we might want to point out that we got $mn=m(2l)$ by multiplying both sides of $n=2l$ by $m$, and that $m(2k)=2(ml)$ follows from the associativity and commutativity of multiplication. But you're probably safe in omitting these details unless you've been told otherwise.

Also note that I didn't assume that $m$ is odd, because this makes the equations a little more complicated. Here is a more faithful recreation of your proof:

Claim: If $m$ is an odd integer and $n$ is an even integer, then $mn$ is even.

Proof: Because $m$ is odd and $n$ is even, we can write $m=2k+1$ and $n=2l$, where $k$ and $l$ are integers. Then $mn = (2k+1)(2l) = 2\cdot(l(2k+1))$ is a multiple of $2$, hence even.


Define $2k_1k_2+k_2=k_3$ as an integer number.


You could further formalize your text, if you want (it is always handy to write down the definitions or the theorems you use, it also makes it easier for others to read and make it look more complete): By definition: $$m \text{ even }\iff \exists k \in \mathbb Z \text{ such that }2k = m $$


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