Prove: if $n-5$ is odd then $3n+2$ is even (make a true table on the statement that shows that is always true) This problem consist in two parts. first: prove the statement. Second: create a true table showing that the statement is always true ($n-5$ is odd if and only if $3n+2$ is even).
This is my reasoning for both parts, and I will be grateful for your insights.
Part 1: 
If $n-5$ is odd, then $n-5=2k+1$ for some integer $k$, and
$3(n-5)=3(2k+1)$
$3n-15=6k+3$
$3n-15+17=6k+3+17$
$3n+2=6k+20=2(3k+10)$
Proving that $3n+2=2m$ is even for some integer $m=3k+10$
Part 2: I really don´t know how to show that. Should I use $n-5$ is odd as $P$ and $3n+2$ as $Q$? Because that's the normal table for $P\leftrightarrow Q$ and doesn't show that it's always true.
Thanks in advance for your help.
 A: Your proof for the first part is fine.
For the reverse direction showing that $3n+2$ is even implies that $n-5$ is odd you could do a very similar process, beginning with $3n+2=2k$, subtracting an appropriate even number from both sides and then adding one to both sides, i.e. $n-5 = (3n+2)-2n-8+1=2(k-n-4)+1$ is odd.
Both of these can probably be cleaned up considerably though by using an inbetween step, showing that $n-5$ is odd iff $n$ is even iff $3n+2$ is even.
As for using a "truth table" to show this... even I don't quite understand what they want.  Something that might be useful that might be similar to what they want would be something like this:
$$\begin{array}{|c|c|c|} n&n-5&3n+2\\\text{odd}&\text{even}&\text{odd}\\\text{even}&\text{odd}&\text{even}\\\end{array}$$
After all, the content of the truth table is proven in step 1, and I don't necessarily think that a visual representation of what we've learned in step 1 is necessary... we already learned it no need to put it into a table.  The truth table doesn't show that it is true in this scenario, it just organizes our results.
A: The sum of two numbers is odd when they are of opposite parity, that is, when one of them is even and the other is odd.
Since $(n-5)+(3n+2)=4n-3=2(2n-2)+1$, we see that $n-5$ and $3n+2$ are of opposite parity.
A: $$\underbrace{n-5}_{odd}+\{?\}=\underbrace{3n+2}_{even}\\
\underbrace{n-5}_{odd}+\bf{\underbrace{2n+7}_{odd}}=\underbrace{3n+2}_{even}$$ if you look to $2n+7=2n+6+1=2(n+3)+1=\underbrace{2q+1}_{odd}$ there is no condition ,It is always true
