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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
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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

2 Answers 2

up vote 5 down vote accepted

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

$$0<\frac{x}y<1\;.$$

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so that is also the proof. Looks great thank you. –  themhz May 9 '13 at 6:35
    
@themhz: You’re welcome. –  Brian M. Scott May 9 '13 at 9:03

If $0 < x < y$, then by dividing all three numbers by the positive quantity $y$, you have $$ 0 < \frac{x}{y} < 1. $$

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