I have a few problems proving stuff with integers, basically the most basic stuff is driving mad right now.
Basically, there's two exercices.
1) If 3x is even, then x is even
Which is true (I think, at least). The method of resolution that I have here assumes x is even, and then proves that 3x is also even. But that's completely the opposite of what I need. Using some methods that I saw in other questions, the only way I could think of is doing something like this:
"Assuming 3x is even, then 3x = 2a and x = $\frac{{2}}{3}a$, so x = $2(\frac{{1}}{3}a)$ and $(\frac{{1}}{3}a) = b$. Finally, x = 2b proves that x is even.
Naturally, both a and b are integers.
1) If 4x is even, then x is even
This is clearly false, and it's pretty easy to find a counterexample.
Except that if I use the method above, I can easily prove a FALSE statement.
Doing a quick copy paste:
"Assuming 4x is even, then 4x = 2a and x = $\frac{{1}}{2}a$, so x = $2(\frac{{1}}{4}a)$ and $(\frac{{1}}{4}a) = b$. Finally, x = 2b proves that x is even.
I am of course doing something very wrong somewhere, but I can't find a recipe for this.
Thank you very much!
Since the mistake in your reasoning has been pointed out in the comments, let me provide you with the correct argument.
1) Assume $3x$ is even. For contradiction assume that $x$ is odd. Then $x=2n+1$ for some $n \in \mathbb{Z}$. Then $$3x=3(2n+1)=6n+3=2(3n+1)+1$$ where $3n+1 \in \mathbb{Z}$. Thus, we expressed $3x$ as $2m+1$ for some $m \in \mathbb{Z}$ (namely, $m=3n+1$). Thus, $3x$ is odd, contrary to the assumption that $3x$ is even. Thus, by contradiction, $x$ has to be even, as desired.