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Because technically, the numerator is smaller than the denominator as $-2 < -1$

I know it's an extremely stupid question.

I mean I know that I can just multiply $-1$ to the numerator and the denominator and I'll get $\frac 2 1$ which is greater than one.

But what exactly is happening here?

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    $\begingroup$ I think what you mean is "smaller than $1$", judging from your $-2 < -1$ argument. If that's the case, one could argue that the rule should be: If $|a| < |b|$ then $|a/b| < 1$. $\endgroup$ – Darth Geek Jun 20 '15 at 9:06
  • $\begingroup$ $a/b$ = how many times $b$ can be entered in $a$. so, in this case $-1$ can be entered twice in $-2$. $\endgroup$ – d_e Jun 20 '15 at 9:08
  • $\begingroup$ $\frac{-2}{-1} = 2\times \frac{-1}{-1}= 2 \times 1= 2$ $\endgroup$ – Bhaskar Vashishth Jun 20 '15 at 9:15
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    $\begingroup$ That's not a stupid question at all. It's a great question! Noticing a pattern, then noticing that it doesn't always apply, and then trying to understand the true pattern that does always apply--this is what mathematicians do. $\endgroup$ – MJD Jun 20 '15 at 10:13
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    $\begingroup$ Very good question! $\endgroup$ – Kartik Jun 20 '15 at 10:36
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If you regard $\dfrac82=4$ as true because $8 = 2+2+2+2$, i.e. you can write $8$ as the sum of four $2$s,

then you can regard $\dfrac{-2}{-1}=2$ as being equally true because $-2 = (-1)+(-1)$.

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What happens is that what you were taught (or found out) does not apply to negative numbers. If you multiply a negative number $a$ with a number $b > 1$, the result is not larger than $a$, but smaller. Essentially, multiplying $a$ by $b > 1$ "amplifies" $a$: if $a$ is negative, then the result is "more negative" than $a$ (i.e., smaller). Now, $a/b$ can be said to be the "amplification factor" you need to apply to $b$ in order to obtain $a$. In your case of $\frac{-2}{-1}$, $-2$ is "more negative" than $-1$, so you need to apply a factor which is larger than $1$.

The correct rule to remember if you want to account for negative numbers is that if the numerator of a fraction is smaller than its denominator in absolute value, then the fraction is smaller than $1$ again in absolute value. (The sign of the fraction is then determined by the usual sign rules.)

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    $\begingroup$ This answer is the only one that addresses the OP's intended question. $\endgroup$ – Ryan Jun 20 '15 at 17:49
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The fact that the numerator is smaller than the denominator does not imply that the fraction is smaller than $0$. Consider for example $\frac{1}{2 }$, where the exact same situation occurs, while obviously $0 < \frac{1}{2}$.

Rather, a fraction $\frac{a}{b}$ (with $b\neq 0$) should be interpreted as: what number do I have to multiply with $b$ to obtain $a$? In this case: which number should I multiply with $-1$ to obtain $-2$? Clearly, this number equals $2$. Therefore, $\frac{-2}{-1} = 2$.

Edit: Since OP has edited the question, the first paragraph of my answer is not relevant anymore. The second paragraph still is.

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    $\begingroup$ It might imply that the fraction is smaller than 1 – if the denominator is positive· $\endgroup$ – Bernard Jun 20 '15 at 9:31
  • $\begingroup$ OP has revised the question. $\endgroup$ – MJD Jun 20 '15 at 10:14
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Whether the fraction is less than zero or not is independent of the magnitude of the numerator and denominator. It only depends on their signs.(unlike signs = less than zero)

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In my early days I used to interpret it as how many of the denominators can you fit in the numerator. I need 2 '-1's to get one '-2'. So 2 is your answer. Also your premise is an incorrect way of defining division. Leave this idea. Good luck!

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  • $\begingroup$ well it would be good if the guy who downvoted this gives an explanation...thanks $\endgroup$ – Zero Jun 27 '15 at 21:04
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When you multiply by a negative number the inequality sign switches.

For example

$$-1>-a \implies 1 < a$$

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