TITLE: How can my proof be improved? "Let $n$ be an integer. If $3n$ is odd then so is $n$."
QUESTION [8 upvotes]: I am attempting to self-study proof techniques and your criticism of my following proof would be greatly appreciated. Feel free to nitpick minor/trivial things; that's how I'll learn!
Edit: I have appended a revised proof with the criticisms received so that other rookies can learn from my progress too.
Edit 2: And another, third time lucky?

Theorem: Let $n$ be an integer. If $3n$ is odd then $n$ is odd.
Let $P$ be the sentence "$3n$ is odd" and $Q$ be the sentence "$n$ is odd." We therefore have;
$$P \rightarrow Q$$
By the law of the contrapositive, we may obtain;
$$\lnot Q \rightarrow \lnot P$$
Which is translates to "If $n$ is not odd then $3n$ is not odd", or put another way; "If $n$ is even then $3n$ is even."
If an integer $n$ is even then there exists some integer $m$ such that;
$$n = 2m$$
By multiplying this by $3$ we may obtain;
$$3n = 6m \equiv n = 2m$$
It has therefore been shown that if $n$ is even so is $3n$ and that this is equivalent to showing that if $3n$ is odd then so is $n$.
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Attempt 2, taking into consideration previous criticism
Theorem: Let $n$ be an integer. If $3n$ is odd then $n$ is odd.
Let $P$ be the sentence "$3n$ is odd" and $Q$ be the sentence "$n$ is odd." We want to show that $$P \rightarrow Q$$ and that by taking contrapositives this is equivalent to showing $$\lnot Q \rightarrow \lnot P$$ which translates to;
"If $n$ is not odd then $3n$ is not odd", or put another way; "If $n$ is even then $3n$ is even."
If an integer $n$ is even then there exists some integer $m$ such that;
$$n = 2m$$
By multiplying this by $3$ we may obtain;
$$3n = 6m$$
This must still be even as an even integer multiplied by an odd integer produces an even one.
We must now show that $3n$ is even. If this is so, then $6m = 2k$ for some integer $k$.
As $3n=6m=2k$ we have;
$3n=2k$
It has therefore been shown that if $n$ is even so is $3n$ and that this is equivalent to showing that if $3n$ is odd then so is $n$.
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Attempt 3
Theorem: Let $n$ be an integer. If $3n$ is odd then $n$ is odd.
Let $P$ be the sentence "$3n$ is odd" and $Q$ be the sentence "$n$ is odd." We want to show that $$P \rightarrow Q$$ and that by taking contrapositives this is equivalent to showing $$\lnot Q \rightarrow \lnot P$$ which translates to;
"If $n$ is not odd then $3n$ is not odd", or put another way; "If $n$ is even then $3n$ is even."
If an integer $n$ is even then there exists some integer $m$ such that;
$$n = 2m$$
By multiplying this by $3$ we may obtain;
$$3n = 6m$$
Which can be rewritten as;
$$3n = 2(3m)$$
Thus we have shown that $3n$ is even, as it is equal to $2(3m)$ which, as an integer multiplied by 2, must be even.
It has therefore been shown that if $n$ is even so is $3n$ and that this is equivalent to showing that if $3n$ is odd then so is $n$.
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REPLY [5 votes]: Your argument is structured as if you were arguing from your conclusion.  The correct way to state this argument is "we want to show that $P \Rightarrow Q$, and by taking contrapositives this is equivalent (you also did not say this) to showing that $\neg Q \Rightarrow \neg P$."  
I guess you are using $\equiv$ to mean "is equivalent to," but this is usually reserved for an equivalence relation on mathematical objects, not on mathematical statements.  For statements you should use $\Leftrightarrow$.  
And, of course, Arturo's comments are spot-on.