How to find the first same-proportion rectangle(s) whose both sides are also whole numbers? Let's elaborate the issue with an example:
I have an arbitrary rectangle. Let's choose the rectangle to be $5.6 \times 4.2$. The width-to-height ratio is $\frac 4 3$. The rectangle is in between the two whole-numbered rectangles of $4 \times 3$ and $8 \times 6$, both having the same ratio.

So, given an arbitrary rectangle, how do I find (if any) the closest* rectangle(s) whose sides are both whole numbers and also has (have) the same ratio?

 * By closest, I mean the rectangle(s) whose sides have the same ratio, and both sides have a whole-numbered length that are the first possible whole numbers that form such a ratio.

General algorithm
Let's say we start from a rectangle with side lengths $w$ and $h$ (in your case $w=5.6$ and $h=4.2$). Now you try to express $w/h$ as a rational number $p/q$ with $\gcd(p,q)=1$. In your case we have
$$\frac{5.6}{4.2}=\frac{56}{42}=\frac{\color{red}2\cdot 2\cdot 2\cdot \color{blue}7}{\color{red}2\cdot 3\cdot \color{blue}7}=\frac{2\cdot 2}{3}=\frac 43.$$ 
So we have $p=4$ and $q=3$. Define $\alpha=w/p$. In your case we find $\alpha=5.6/4=1.4$. Then compute
$$\check w
= p\cdot \lfloor \alpha\rfloor
= 4\cdot \lfloor1.4\rfloor
= 4\cdot1
= 4,
\qquad
\check h
= q\cdot \lfloor \alpha\rfloor
= 3\cdot \lfloor1.4\rfloor
= 3\cdot1
= 3,
$$
$$\hat w
= p\cdot \lceil\alpha\rceil
= 4\cdot \lceil 1.4\rceil
= 4\cdot 2
= 8,
\qquad
\hat h
= q\cdot \lceil \alpha\rceil
= 3\cdot \lceil 1.4\rceil
= 3\cdot 2
= 6.
$$
and so we found your nearest rectangles $(\check w,\check h)=(4,3)$ and $(\hat w, \hat h)=(8,6)$. And this works in general.
