Showing $x-a>y-b$ implies $\frac{x}{y}>\frac{b}{a}$ and $x>y$?

Question

Edit 2: I think the claim is not true Let $x=9; y=10, a=.1, b=1.2$ Then $x-a=8.9>y-b=8.8$ and $\frac{x}{y} = .9<\frac{b}{a}$

Edited because there was a typo

I'm trying to show this ($x-a>y-b$ implies $\frac{x}{y}>\frac{b}{a}$ and $x>y$) but cannot seem to figure it out.

I can show that the reverse direction is true, but reversing my steps didn't seem to help me. perhaps I should try "assuming $x<y$", show contradiction, then assume fraction not true, and contradiction?

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Edit: Here is the full statement I am trying to show: $$ u+x-a > v+y-b \ IFF\quad u+x>v+y\ \text{ and } \frac{u+x}{v+y}>\frac{b}{a}\\ u,v,a,b\geq 0 $$ In the question I am looking for a proof of a (simplified version of) the $\rightarrow$ direction.

If someone wants to prove the forward direction of this full version I am fine with that too.

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Answer 1

Let $x=5$ and $y=3$,$a=2$ and $b=1$

We have $x>y$, $x-a =3$, $y-b =2$, $x/y = 5/3$, $a/b =2$

Thus $$x-a>y-b\iff \frac{x}{y}>\frac{a}{b}$$

is not true.