# Discrete math into plain English help $1$.

Given Statement:
$\forall x \in \Bbb N\left(x > 1 \to \exists k \in \Bbb N \exists m \in \Bbb N\left(m \equiv 1\pmod 2 \land x = 2^k \cdot m\right)\right)$

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.

• It says that any natural number greater than $2$ can be expressed as the product of an odd number and a power of $2.$ For example, $360=2^3\cdot45.$ By the way, do the natural numbers start with $0$ or $1?$ – bof Apr 7 '17 at 0:53
• @Bof, most conventions start the naturals at $1$, but perfectly valid conventions exist starting them at $0$. In my experience, without further explanation, people assume $\mathbb{N}=\{1,2,3,...\}$ unless otherwise explicitly stated. – The Count Apr 7 '17 at 0:59
• @TheCount Thank you. I'm aware that both conventions exist. Not sure about "most". In this case, assuming the given statement is supposed to be true, one would have to allow $k=0$ in case $x$ is an odd number. – bof Apr 7 '17 at 1:04
• Yeah, in this one they seem to be using $0\in\mathbb{N}$, and when I say "most" I mean "most of the time I use or see used the symbol $\mathbb{N}$", not necessarily a rigrous survey of all uses of it, haha. – The Count Apr 7 '17 at 1:07
• @bof I JUST realized you were asking the OP, not in general. Sorry! Ha. – The Count Apr 7 '17 at 2:37

• Where did the $n$ in this statement "then there exist natural numbers $m$ and $n$" come from? – user430574 Apr 7 '17 at 1:06
• Right. There seems to be no particular reason for specifying $x\gt1$ instead of $x\gt0$ but I guess it does no harm. – bof Apr 7 '17 at 0:58