Consider the following general system of linear equations
$$
\begin{pmatrix} a & -b\\ -c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} v \\ w \end{pmatrix}
$$
where $a, b, c, d \in \mathbb{N}^+$, $x, y, v, w \in \mathbb{R}^+$ so that the following constraints are also satisfied: $0 \leq v, w <e$, $0 < x,y < e$ where $e$ is the exponential constant.
I am trying to understand what the general solution set to such a system would look like. That is, for all $a, b, c, d \in \mathbb{N}^+$ and $ v, w \in \mathbb{R}^+$, what is the corresponding set of solutions $(x, y)^T$?
For example, I am unsure if every $x, y$ pair satisfying $0 < x,y < e$ is a solution to one of these general systems satisfying the constraints, or if only a subset of such $x, y$ values create the solution set to such an arbitrary system. If this is indeed the case, what would that (solution) subset be?
Best Answer
According to your last paragraph it seems you are interested in the problem:
Given $0 < x, y, < \mathrm{e}$ find $a, b, c, d \in ℕ+$ and $\mathrm{e} > v, w \geq 0$ such that the above system of equations is true. Especially you want to know whether there is a solution for arbitrary $0 < x, y < \mathrm{e}$.
And the answer is that there is.
The problem can be reformulated in a following way: for given $x, y$ find $a, b, c, d \in ℕ+$: $$ \mathrm{e} > ax - by \geq 0\\ \mathrm{e} > -cx + dy \geq 0. $$
Let's take the first inequality. We can transform it -- add a term $by$ first and then divide by $bx$ (recall that $x, b > 0$ thus the inequalities don't switch the direction) to $$ \frac{y}{x} + \frac{\rm e}{bx} > \frac{a}{b} \geq \frac{y}{x}. \tag{1}\label{1} $$
Let's continue with the left hand side inequality. If we managed to find $a/b$ such that $$ \frac{y}{x} + \inf_{0 < x < \mathrm{e}} \frac{\rm e}{bx} > \frac{a}{b}\tag{2}\label{2} $$ then the LHS inequality of \eqref{1} would be fulfilled for free.
Well, let's try it. The infimum equals to $\frac{\rm e}{b{\rm e}}$. Cancel those $\rm e$ in it and replace the left inequality of \eqref{1} by \eqref{2} $$ \frac{y}{x} + \frac{1}{b} > \frac{a}{b} \geq \frac{y}{x}. $$
Now take the positive real number $y/x$ and round it upwards to the first decimal place. The error is clearly less than 1/10 and since it is rounded upwards it is at least 1/10. In other words: $$ \frac{y}{x} + \frac{1}{10} > \frac{a}{10} \geq \frac{y}{x}, $$ where $a/10$ is the rounded value. Clearly, $a \geq 1$ as the value is at least $1/10$. Thus, you have found $a, b \in ℕ+$ and it suffices to set $v = ax - by$.
The second inequality can be solved in the same way.