I am given the following proposition:

If $m$ and $n$ are even integers, then $mn$ is also an even integer.

This is my strategy:

An integer $m$ is said to be even if it is divisible by 2 (integer). When $m$ and $n$ are integers, $m$ is divisible by $n$ if there exists $j\in\mathbb Z$ such that $m = jn$. Thus, an even number $m$ could be written as $m = j2$. If $m$ and $n$ are even, then they can be defined as:

$m = j2$ and $n = k2$ where $j,k \in\mathbb Z$

\begin{align*} m\times n &= m\times n\\ m\times n &= j2k2\\ m\times n &= (2jk)2 \end{align*}

Please correct me if I am wrong, but at this point, I am finished because $2jk \in\mathbb Z$, and I have derived the 2. I will perform two multiplications in the parentheses; each multiplication will generate an integer because the multiplication is a binary operation. By definition, a binary operation takes two elements from a set $S$ and generates another element of set $S$ as output. In this case, $S$ is $Z$.

  • $\begingroup$ It is correct. Note that the proof does not depend on the hypothesis that $\,m\,$ is even, so that can be omitted. Also $\,2j\,$ vs. $\,j2\,$ is more customary. $\endgroup$ – Bill Dubuque Feb 5 '15 at 0:36
  • $\begingroup$ oups, I meant "when m and n are even integers" $\endgroup$ – Johnathan Feb 5 '15 at 0:39
  • $\begingroup$ What I meant by the above remark is that your proof shows more generally that $\,(2j)n = 2(jn),\,$ i.e. it shows that an even integer times any integer remains even. if this is a course in number theory (vs. say foundations or proof theory), then you can omit explicit mention of basic arithmetical laws, such as commutativity and associativity of multiplication, that integers are closed under multiplication, etc. $\endgroup$ – Bill Dubuque Feb 5 '15 at 0:46

Proof: Let $m=2k$ and $n= 2l$ for some $k$ , $l$ $\in Z$. We now multiply. $m \cdot n$ = $(2k)(2l)$ = $(4kl)$ = $2(2kl)$. Since $k$ and $l$ are integers therefore $kl$ is also an integer. Thus $m \cdot n$ is an even number. This is more formal and simpler to understand. Note after solving a proof try to clean it up and make it more finer. Hope this advise help. Good job by the way.

  • $\begingroup$ Thank you very much! I am learning proofs, and I guess that at a certain level, I don't need to define everything (i.e the reader has at least basic knowledge like what is an even integer). :) $\endgroup$ – Johnathan Feb 5 '15 at 0:56
  • $\begingroup$ Same here but always highlight the main idea of your proof, meaning the essential element that makes the proofs work. The idea of solving difficult problem is to find the tricks behind it, once you find that then that is the heart of the proof. $\endgroup$ – user146269 Feb 5 '15 at 1:01

I assume that you're asked to prove the preposition, and that is what you are asking for help with.

Though correct, your proof is verbose. If you are told that an integer is even, then you already know what even numbers are. If you know what even numbers are, you already know about the properties of multiplication of integers.

Try to get the proof down to two or three lines. You only need to write $m \times n$ once.

  • $\begingroup$ How about: mn = 2j2k mn = 2(2jk) where (2jk) belong to the set of integers. $\endgroup$ – Johnathan Feb 5 '15 at 0:41

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