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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.
Thank you in advance.

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  • $\begingroup$ 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 $1?$ $\endgroup$ – bof Apr 7 '17 at 0:53
  • $\begingroup$ @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. $\endgroup$ – The Count Apr 7 '17 at 0:59
  • $\begingroup$ @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. $\endgroup$ – bof Apr 7 '17 at 1:04
  • $\begingroup$ 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. $\endgroup$ – The Count Apr 7 '17 at 1:07
  • $\begingroup$ @bof I JUST realized you were asking the OP, not in general. Sorry! Ha. $\endgroup$ – The Count Apr 7 '17 at 2:37
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The first part of your translation seems correct, although you might have a hard time giving a clear interpretation of the second half if you don't rephrase it.

I would say "For all x in the natural numbers, if x is greater than one, then there exist natural numbers m and k such that m is congruent to 1 modulus 2, and x equals 2 to the power of k times m".

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  • $\begingroup$ Where did the $n$ in this statement "then there exist natural numbers $m$ and $n$" come from? $\endgroup$ – user430574 Apr 7 '17 at 1:06
  • $\begingroup$ That was meant to be a k, I've edited it to make sense. $\endgroup$ – Yehuda Apr 16 '17 at 19:05
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I believe what it is saying is that every natural number greater than 1 is the product of a pure power of 2 and an odd number. This is only true if you accept 0 as a natural number.

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  • $\begingroup$ Right. There seems to be no particular reason for specifying $x\gt1$ instead of $x\gt0$ but I guess it does no harm. $\endgroup$ – bof Apr 7 '17 at 0:58

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