Proof of the ratio increased and decreased theorem

Question

Let $$\frac{a}{b}$$ be a fraction. Suppose I decrease the numerator and the denominator both by the same number. $$\frac{a-x}{b-x}$$ Then, $$\tag{Equation 1} \frac{a}{b}-\frac{a-x}{b-x}=\frac{x}{b}\frac{b-a}{b-x}$$ If $$a>b$$ Then,

Equation 1 is negative

This implies $$\tag{Inequality 1} \frac{a}{b}<\frac{a-x}{b-x}$$ But from previous question, here on this site , I got to know that equation 1 is not necessarily negative.

I read this proof but I think it is only true if $$a>b>x>0$$ So how can we prove that $$\frac{a}{b}<\frac{a-x}{b-x}$$if $$a>b$$

Answer 1

If $c,d,e,f$ are all $> 0$, then
$\frac{c}{d} > \frac{e}{f} \iff (cf > de).$

Therefore
$a > b > 0$ and $(a-x) > (b-x) > 0$ implies that $\frac{a}{b} < \frac{a-x}{b-x}$.

This is because $(-ax) < (-bx) \implies (ab - ax) < (ab - bx).$

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Assume that $a > b > 0$, and $(0 > [b-x])$.

If $([a-x] > 0)$ then $\frac{a-x}{b-x}$ is a negative number, and so $\frac{a}{b} > 0 > \frac{a-x}{b-x}.$

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If $([a-x] < 0$ then $\frac{a-x}{b-x} = \frac{x-a}{x-b}$, where $(x-b) > (x - a) > 0.$

This allows $\frac{a}{b}$ to be compared against $\frac{x-a}{x-b}$
by comparing $(ax - ab)$ against $(bx - ab).$

Since $a > b,$ you have that $(ax - ab) > (bx - ab),$
and so, in this situation, $\frac{a}{b} > \frac{x - a}{x - b}.$

Edit
Shortcut to the last situation is to realize that since $a > b$, $(x - b) > (x - a).$

Therefore, in the last situation, you have that $\frac{a}{b} > 1 > \frac{x-a}{x - b} > 0.$

Answer 2

Let $0<b<a$. Define for $x\neq b$ the function $$f(x)=\frac{a}{b}-\frac{a-x}{b-x}.$$ We have $f(x)=0\iff x =0$. As $f$ is continuous it may change sign at its zeroes or where it's undefined, that is in $x=0$ or $x=b$. We easily find that $f$ is negative, which means that $$\frac{a}{b}<\frac{a-x}{b-x},$$ iff $0<x<b$.