Prove that for any $x$ and $y:0<x<y< 1,\; \frac xy > x$

Question

How can I prove that for any $x$ and $y$, given that $0 < x < y < 1$, $\frac{x}{y} > x$?

The way I think about it is the following:

$\frac{x}{y} > x \implies \frac{1}{y} > 1$ which is true, because as $0 < y < 1$, we know that $\frac{1}{y}$ must be greater than $0$, as we need many $y$ to end up summing up to $1$.

Whereas:

$\frac{x}{y} < x \implies \frac{1}{y} < 1$ which is false, because of the above statement $0 < y < 1$.

Are my assumptions right and is there a more formal way to state this relation between these numbers?

Thank you for the attention.

Answer 1

From $$\frac{x}{y}>x$$ we get by dividing by $$x>0$$ $$\frac{1}{y}>1$$ or $$1>y$$ which is true, since $1>y>0$

Answer 2

Your way is correct, but here's an alternative way:

$\dfrac{x}{y}>x\Leftrightarrow x>xy$ (because $y>0$) $\Leftrightarrow x(1-y)>0$, also true because $x>0;y<1$.

In fact, you don't need $x<y$ for the inequality to be true, only $x>0$ and $0<y<1$ is enough.

Answer 3

$\frac xy-x=\frac{x(1-y)}{y}$ which is $>0$ because all the three terms in the product are.