Let the set of all inch measures $\bigl( \mathbb{IMS} \bigr)$ be the set of unitized real numbers constructed from the set of strings $\begin{Bmatrix} f(x) + “ \,\,\,\,\mathtt{inch(es)}” : x \in \dfrac{n}{2^p} \\ \text{ where } n \leq 2^p \text{ and } n, p \in \mathbb{N} \bigcup \{0\} \end{Bmatrix} $.
Specifically, we define a unitized decimal number to be class having two fields:
a value field taken from $\mathbb{R}$
a unit field, which is a string such as “inches” or “feet”
For example, if $\mathtt{obj}$ is an element of the set of all inch measures, then maybe we have:
$\mathtt{obj.unit} = “inch(es)"$
$\mathtt{obj.value} = “5.5"$
$f$ is a function.
When you put a number $x$ inside of function $f$ and turn the crank, then out pops a string.
For example, $f(0.25) = “ 1/4 inch(es)"$
One example of an inch measure is “1/16 inch(es)"
Most questions without a little LaTeX an mathematical maturity demonstrated get deleted. Thus, we introduced a the notation and a few pseudo-formalisms.
My primary question is, what is the set of $\mathbb{IM} \in \mathbb{IMS}$ such that after converting inches to millimeters, the measurment in millimeters is almost a whole number?
More formally, let $g$ be a function which accepts unitized real numbers in inches as input and outputs a unitized real number in millimeters.
When is g(x) less the floor of g(x) less than 0.1 millimeters?
After converting inches to millimeters the result is less than one tenth of a millimeters away from a whole number of millimeters.
This is really just a comment, but it's a bit too long to fit in the comment box.
You might find the theory of continued fractions to be interesting, with particular emphasis on the convergents of the continued fraction expansion of a real number.
The idea is that for each real number $r$ there is a special sequence of rational numbers $p_n/q_n$ called the convergents of $r$, such that for each of these numbers we have $\left| r - \frac{p_n}{q_n}\right| < \frac{c}{q_n^2}$ (for some constant $c$). Intuitively this says that $p_n/q_n$ is an unusually good approximation to $r$: if one fixes an integer $q$ one would expect to find an integer $p$ such that $\left| r - \frac{p}{q}\right| < \frac{1/2}{q}$, so to get that denominator to be $q^2$ would be unusual.
Now take $r$ to be the ratio of 1 millimeter to 1 inch. Then for each convergent $p_n/q_n$, $p_n$ inches is unusually close to $q_n$ millimeters.