If $(a,b)$ is a multiple of $(c,d)$, show that $(a,c)$ is a multiple of $(b,d)$ I need help with this problem:
If $(a,b)$ is a multiple of $(c,d)$ with $abcd\neq0$, show that $(a,c)$ is a multiple of $(b,d)$. This is suprisingly important: call it a challenge question. You could use numbers first to see how $a,b,c$ and $d$ are related. The question will lead to:
If A = $\left[\begin{array}{l}a&b\\c&d\end{array}\right]$ has dependent rows the it has dependent columns.
I tried to do it this way:
$(a,b)=x(c,d)\Rightarrow(a,b)=(xc,xd)$
$(a,c)=(xc,c)\Rightarrow c(x,1)$ and $(b,d)=(xd,d)\Rightarrow d(x,1)$
I don't know what to do after that, what should I do next?

First note the following: $(a,b) = (a,(\frac{b}{a}) a)$. [As $abcd \not =0$ we can assume that $\frac{b}{a}$ exists]
So for some scalar $x$ we note: $(c,d) = x(a,b) = (xa,xb)$ $=(xa,x(\frac{b}{a}) a)$. Thus $c$ can be written $c=xa$ and $d$ can be written $d=x(\frac{b}{a}) a$.
Thus $(b,d) = (\frac{b}{a} a, x(\frac{b}{a})a) = \frac{b}{a}(a,xa) = \frac{b}{a}(a,c)$. So $(b,d) = y(a,c)$ where $y = \frac{b}{a}$.
