Suppose $a,b,c,d\geq 0$ and that $ad-bc=1$
on This Wiki page it says
$$ \frac{\lambda_1 a+\lambda_2 b}{\lambda_1 c +\lambda_2 d} -\frac{a}{c} = \lambda_2 \frac{bc-ad}{c(\lambda_1 c +\lambda_2 d)} $$ and $$ \frac{b}{d} -\frac{\lambda_1a+\lambda_2 b}{\lambda_1 c +\lambda_2 d} = \lambda_1 \frac{bc-ad}{d(\lambda_1c +\lambda_2 d)} $$ must be positive (OK),
Then it says, because $ad-bc=1$, it must be that
- $\lambda_1,\lambda_2$ are integers
- solving the system of linear equations $$ a'=\lambda_1a+\lambda_2b\\ c'=\lambda_1c+\lambda_2d $$
- Therefore $c'\geq c+d$
I am confused on 1), 2), and 3)
- specifically, why must $\lambda_1,\lambda_2$ be integers (that solve that system of linear equations), and why can they not be negative integers?
First, it's $bc-ad$ which equals 1, not $ad-bc$, so the reason $\lambda_2>0$ is because
$$0< a'/c'-a/c= \frac{\lambda_1a+\lambda_2b}{\lambda_1c+\lambda_2d}-\frac{a}{c}$$ which simplifies to $\frac{\lambda_2}{cc'}>0$ when you use the fact that $bc-ad=1$ and $c'=\lambda_1c+\lambda_2d$. Since $c,c'$ are assumed to be $>0$, you must have $\lambda_2>0$. The argument for $\lambda_1$ is similar.
Now solve $$ a'=\lambda_1a+\lambda_2b\\ c'=\lambda_1c+\lambda_2d $$ to get $\lambda_1$ in terms of $a',a,b,c',c,d$. You will get an expression that makes it clear $\lambda_1$ is an integer (when you use the fact that $bc-ad=1$).
Finally you have $c'=\lambda_1 c+\lambda_2d$ where $\lambda_1\geq 1$ and $\lambda_2 \geq 1$, which gives the inequality you want.