Proof by contrapositive, what should I be assuming? I am attempting to prove the following statement by proving the contrapositive. 

For all natural numbers n, IF n = k * l and (k less than or equal to l) THEN k less than or equal to square root of n

I have found the contrapositive to be:

For all natural numbers n, IF k greater than the square root of n THEN n does not equal k * l OR k is greater than l

I know to prove this we must first assume the first the part to be true (that k is greater than the square root of n) and that can lead me to say that k squared is greater than n. 
I am used to approaching the proof by expressing one variable in terms of another such that I can substitute the expression in for the variable in another expression. I would then solve for another variable and prove it is an integer or something like that. In this situation, how can I take the expression from one statement to sub into another if there are no equal signs involved? I do not know how to solve for an expression being not equal to another. Am I even on the right track? If I express k squared to be greater than n, how can I even move that into another equation and still keep the idea that it is greater than.
I don't understand how I can say anything about l without it being on the left side of the if/then statement. In the current iteration of the proof, I am only assuming k to be greater than the square root of n. I don't see how I can assume anything more? 
Any hints on what direction I can go from here would be very appreciated. 
 A: I find this easier to do when you do it in symbols.
So you have:
$$\forall n,k,l \in \mathbb{N} ((n = k\cdot l \land k \le l) \rightarrow k \le \sqrt{n})$$
Taking the contrapositive (or at least the contrapositive of the main body):
$$\forall n,k,l \in \mathbb{N} (\neg k \le \sqrt{n} \rightarrow \neg (n = k\cdot l \land k \le l))$$
which works out to:
$$\forall n,k,l \in \mathbb{N} (k > \sqrt{n} \rightarrow (\neg n = k\cdot l \lor k > l))$$
Also, since $\neg P \lor Q \Leftrightarrow P \rightarrow Q$, it might be nice to rewrite this to:
$$\forall n,k,l \in \mathbb{N} (k > \sqrt{n} \rightarrow (n = k\cdot l \rightarrow k > l))$$
And since $P \rightarrow (Q \rightarrow R) \Leftrightarrow (P \land Q) \rightarrow R$, we can also do:
$$\forall n,k,l \in \mathbb{N} ((k > \sqrt{n} \land n = k\cdot l) \rightarrow k > l)$$
which is probably the most useful format if you have to actually prove it.
A: What you have stated is wrong,
so it will be hard to prove.
You stated
"For all natural numbers n, IF n = k * l and (k greater than or equal to l) THEN k less than or equal to square root of n".
However,
you can always write
$n = n\times 1$
and this $k$
(with $k=n$)
does not satisfy the conclusion.
What is correct is
"For all natural numbers n, IF n = k * l and (k less than or equal to l) THEN k less than or equal to square root of n".
The contrapositive is
If $k > \sqrt{n}$ and
$k \le l$
then
$n \ne k \times l$.
The proof is easy:
Since
$l \ge k > \sqrt{n}$,
then
$k \times l
\gt \sqrt{n}\times \sqrt{n}
=n
$
so
$k \times l \ne n$.
Note that this holds
for any positive reals,
not just integers.
A: "For all natural numbers n, IF k greater than the square root of n THEN n does not equal k * l OR k is greater than l"
Assume $k > \sqrt {n}$.
If $k \le l$ then $k*l > k*k >{\sqrt{n}}^2 =n$.
If $k*l = n$ then $l = \frac {n}k < \frac {n}{\sqrt n} = \sqrt n < k$.
So either $n \ne kl$ or $k> l$.
