How do you find a common multiple (which could be irrational) of a two numbers? I wasn't really sure how to phrase the question nor know how I should tag it.
Lets say I have a piece of paper (42x25) that I need to cut into squares (the size and number of the squares do not matter).  How many pieces can I cut out with having little to no remainder?
I did this problem by following an art book's instructions with the dimensions of 21x12.5 cm instead of 42x25 cm and doing these steps.


*

*lw = R ; 21 - 12.5 = 8.5

*R/2 = length of segment ; 8.5/2 = 4.25

*Then I measured out 4.25 cm along the edges of a paper until I reached 5 segments on the length and 3 segments on the width.

*Then I divided the length and width by their respective segments.  And found out their lengths were around 4.2 cm


What I am asking:
What is the formula to find the common multiple when multiplied with a whole number it roughly equals to given lengths?  The only thing I can this of is that if:
x = the segment length
y = the number of segments of the length of an edge1
z = the number of segments of the length of an edge2
Then
xy ~= Edge1
xz ~= Edge2
But I can't figure out how to actually solve it since there are two variables.
 A: You want to find two small integers $y$ and $z$ such that 
$${\text{Edge}_1\over\text{Edge}_2}\approx {y\over z}.$$
The standard way is that of expressing ${\text{Edge}_1\over\text{Edge}_2}$ as a continued fraction: its convergents will then give the best rational approximations.
Example: for ${\text{Edge}_1\over\text{Edge}_2}={42\over25}$, the continued fraction representation is
$$
{42\over25}=1+{1\over1+\displaystyle{1\over2+\displaystyle{1\over8}}}
$$
and its  convergents are
$$
1,\quad 2,\quad{5\over3},\quad{42\over25}.
$$
The best one is probably ${5\over3}$, which gives your solution: divide $42$ into $5$ parts and $25$ into $3$ parts.
A: Leat $a$ and $b$ be the sides in question. Their ratio $\frac{a}{b}$ is either rational or irrational number. If it is rational, let say $\frac{p}{q}$, the "unit" length will be
$$
u=\frac{a}{p}=\frac{b}{q}.
$$
If the ratio is irrational you can take any rational approximation for it, e.g.
$$
\sqrt2\sim\frac{141}{100} 
$$
and proceed as before.
A: From your examples, it looks like you're dealing with rational lengths for the paper, $l$ and $w$.  It even looks like you're using a finite decimal expansion (the decimal doesn't go on forever). 
From this, I suggest that you multiply $l$ and $w$ by powers of $10$ until you have two whole numbers, I'll call them $L$ and $W$.  Let $G$ be the greatest common divisor of $L$ and $W$.
Let $g$ be $G$ divided by the power of $10$ that you multiplied in the previous step.  This is the length of the side of a square.  Now, $l/g$ and $w/g$ will tell you the number of squares along an edge of the paper.
If $l$ and $w$ are rational, but aren't decimals with finite expansions, multiply them by the least common multiple of their denominators, and mimic the steps above.
If all your side lengths are rational, then you will never have an irrational side length for your squares.  If your side lengths are irrational, then the only way to get squares without anything left over is that the ratio of the side lengths is rational.
