If $\frac{a}{c}<\frac{a'}{c'}<\frac{b}{d}$ and $bc-ad=1$, Then $a' = \lambda_1 a + \lambda_2 b$ and $c' = \lambda_1 c + \lambda_2 d$

Question

with $a,b,c,d>0$, and $c'>0$

I feel like this result follows from the denseness of the reals, but I am unsure how to prove it. Also, it is not obvious to me why $c'$ and $a'$ should have the same $\lambda_1$ and $\lambda_2$

Answer 1

The existence of a solution is a mere linear algebra problem: you want to solve $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \lambda_1 \\ \lambda_2 \end{pmatrix} = \begin{pmatrix} a' \\ c' \end{pmatrix}, $$ which has a unique solution since the $ad - bc \ne 0$. Furthermore, using the given inequalities and the assumption that $bc - ad = 1$, you can prove that both $\lambda_1$ and $\lambda_2$ are strictly positive.

Answer 2

let us solve ( determinant is $-1$) $$ \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} a' \\ c' \end{array} \right) $$ Then $$ \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{cc} -d & b \\ c & -a \end{array} \right) \left( \begin{array}{c} a' \\ c' \end{array} \right) $$ so $$ xa+yb = a' \; \; , \; \; \; xc +yd = c' \; \; , $$ also $$ x = b c' - d a' > 0 \; \; , \; \; y = ca' - a c' > 0 $$ because $$\frac{a}{c}<\frac{a'}{c'}<\frac{b}{d} \; \; .$$

These are integers, $$ x = b c' - d a' \geq 1 \; \; , \; \; y = ca' - a c' \geq 1 $$ so $$ a' \geq a+b \; \; , \; \; c' \geq c+d $$

This is the same information as at Largest Fraction Smaller than a Given Fraction a bit cleaned up this time.