Discrete math statement $\rightarrow$ plain English. $\left(Longer\right)$

Question

I found this one more challenging, I think the length of it confused me the most.

Given Statement:
$\forall x \in \Bbb N \left[\left(\exists y \in \Bbb N \left(2 \le y \land y \lt x \land y \mid x \right)\right) \to \left(\exists z \in \Bbb N \left(2 \le z \land z \le \sqrt x \land z \mid x\right)\right)\right]$

So far I have written it as:
For all $x$ in the natural numbers and there exists $y$ for all natural numbers where $z$ is smaller or equal to $y$ if and only if $y$ is smaller than $x$ if and only if $y$ is a divisible of $x$ then, $\cdot \cdot \cdot$

I only wrote it up to the $\rightarrow$, but not sure if it makes sense and furthermore if it's correct. Thus, can I continue like this for the next part.

Any edits of my existing statement and any help in converting to plain English is welcomed. Thank you in advance.

Are you sure that you want $z\leq z$ or should it be $2\leq z$? — martin.koeberl, Apr 07 '17 at 02:41
Going the other way around: how would you quantify the statement "any positive integer $x$ which is not a prime has a proper divisor $z$ no larger than $\sqrt{x},$". — dxiv, Apr 07 '17 at 02:55
@martin.koeberl it was supposed to be $2 \le z$, edited the question. — , Apr 07 '17 at 02:57
We have it already : forall-x-in-mathbbn-exists-y-in-mathbbn2-leq-y-land-yx-land-y-x — Mauro ALLEGRANZA, Apr 07 '17 at 06:44

score 0 · Accepted Answer · answered Apr 07 '17 at 02:51

0

Take any natural number $x$. If we can find a natural number $y$ between 2 and $x-1$ (inclusive) such that $y$ divides $x$, then there exists another natural number $z$ between 2 and $\sqrt(x)$ (inclusive) that divides $x$ as well.

answered Apr 07 '17 at 02:51

wto

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Discrete math statement $\rightarrow$ plain English. $\left(Longer\right)$

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