Proving $\forall x \in R$: $x = \frac{2k}{m}$ $\iff$ $x \in Q$

Question

Proving $\forall x \in R$, $\exists k \in Z, \exists m \in Z$: ($x = \frac{2k}{m}$ $\iff$ $x \in Q$)

I am trying to prove or disprove this statement and I've tried couple examples and I think it is True.

However, I'm completely lost and do not know where to start

Answer 1

Suppose $x=\frac{2k}{m}$ for some integers $k$ and $m$. Remember that a real number is rational if you can write it as a fraction of integers. Both $2k$ and $m$ are integers, so $x$ is rational.

Now assume $x$ is rational. This means there exists integers $a$ and $b$ such that $x=\frac{a}{b}=\frac{2a}{2b}$. Take $k=a$ and $m=2b$ so $x=\frac{2a}{2b}=\frac{2k}{m}$, which is what we wanted to show.

Answer 2

$x =\frac{2k}{m} \implies x \in Q$ because $Q = ${$r=\frac{p}{q}$ with $n,m \in Z$}

$x in Q$ then $x=\frac{p}{q}$ for some $p,q \in Z$

if p is even, then $2k=p$ and $m=q$ if p is odd, then $2k = 2p$ and $m=2q$

then x can be written as $\frac{2k}{m}$ then $x \in Q \implies x=\frac{2k}{m}$ for some $k,m \in Z$