$\forall x \in \Bbb N\left(x > 1 \to \exists k \in \Bbb N \exists m \in \Bbb N\left(m \equiv 1\pmod 2 \land x = 2^k \cdot m\right)\right)$
So far I have written it as:
For all $x$ in the set Natural numbers where $x$ is larger than $1,$ there exists $k$ in Natural numbers existing $m$ in Natural numbers.
That is how far I could go and not sure if it is correct.
I need help for the $\left(m \equiv 1\pmod 2 \land x = 2^k \cdot m\right)$ part.
Any edits of my existing statement and any help in converting to plain English is welcomed.
Thank you in advance.