Why can't a direct proof be made backwards? Say we have the following implication:
$$\textit{Let $x\in \mathbb{Z}$. If  $5x-7$ is even, then x is odd. }$$
The method used by my book to prove this implication is by means of a proof by contrapositive. But what prevents me from saying:
$$ \textit{Assume that x is odd. Then: } x=2k+1, \textit{and we have}: 5(2x+1)-7 =10x-2=2(5x-1).$$
$$\textit{Since  $5x-1$ is an integer, $2(5x-1)$ is even. Ergo, $5x-7$ is even when x is odd.}$$
 A: You've shown that if $x$ is odd, then $5x-7$ is even, but that is not what you were supposed to show.  You were supposed to show that if $5x-7$ is even, then $x$ is odd.
To see the problem with your method, let's use your method to prove this very similar statement:

Let $x\in \Bbb Z$.  If $4x-2$ is even, then $x$ is odd.

Your method now goes like this:

Assume that $x$ is odd.  Then: $x = 2k+1$, and we have $4(2k+1) -2 = 8k+2 = 2(4k+1)$.
Since $4k+1$ is an integer, $2(4k+1) $ is even, Ergo, $4x-2$ is even when $x$ is odd.

So it looks okay, right?  But “If $4x-2$ is even, then $x$ is odd.” is false, because $4x-2$ is even whether or not $x$ is odd.
A: But then you are not proving the original implication: you are proving its converse.
A: You have shown that $5x-7$ is even when $x$ is odd, but this does not show that it is only even when $x$ is odd - it could also be even when $x$ is even (as for example $6x-8$ would be).
Your "proof" would go through for $6x-8$ but it would not be true that if $6x-8$ is even then $x$ is odd.
