# when the numerator is less than the denominator

when the numerator is less than the denominator the result is always between 0 and 1? for example if I have a number like x/y where x<y then the result will be between 0 and 1 always? Is there a proof for this?

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If the numerator is negative it could be between -1 and 0, e.g. $\displaystyle-\frac{3}{4}$ –  manthanomen May 9 '13 at 6:27
Assuming both $x$ and $y$ are positive, and we have existence of inverses, then $x < y \implies x \cdot \frac{1}{y} < y \cdot \frac{1}{y} \implies \frac{x}{y} < 1$. –  AWertheim May 9 '13 at 6:28
so basically depending on it's sign it would be either -1->0 or 0->1? –  themhz May 9 '13 at 6:28

Assuming that $x$ and $y$ are positive, you have $0<x<y$, so $\frac1y>0$, and $$0\cdot\frac1y<x\cdot\frac1y<y\cdot\frac1y\;,$$ which on simplification becomes
$$0<\frac{x}y<1\;.$$
If $0 < x < y$, then by dividing all three numbers by the positive quantity $y$, you have $$0 < \frac{x}{y} < 1.$$