good morning guys,i have such question,when i was reading GRE book,there was such kind of property related to rational number,in shortly if we are trying to determine if $a/b$ is more then $c/d$ or vice verse,there was explained very short method,namely

$a*d>c*b$ but why it is so?i have tried following thing,let us least common multiple,which would be $d*b$; and on numerator place we have multiplies $b*c$ and $a*d$,it is clear for positive numbers,but for negative number are same?i means method for comparing rational numbers

  • $\begingroup$ What you've written ($c/d\gt a/b$ if $ad\gt cb$) is wrong, even if everything is positive, as any example will show. You are correct in surmising that you need something different for negative numbers than for positive. The main thing you need is that if $r\gt s$ then $ar\gt as$ if $a\gt0$, $ar\lt as$ if $a\lt0$. $\endgroup$ – Gerry Myerson Nov 14 '12 at 6:34

If $a,b,c,d>0$ then the inequality $a/b<c/d$ is equivalent, on multiplying both sides by the positive number $bd$, to the inequality $ad<bc.$ So you just got things flipped around in your question, and the GRE book is right. (Hopefully; look again at exactly what inequalities were said to be equivalent.)

Of course no such simple rule will work in all cases where $a,b,c,d$ can have arbitrary signs, since you have to know the sign of $bd$ so as to know whether to reverse the inequality at the "multiply by common denominator" step. One could make a list of all four cases on the signs of $b,d$ and in each case get an equivalent inequality with no fractions, but this might be more trouble than it's worth.

ADDED NOTE: In the case of fractions (numerator and denominator are integers), it is usual to express a fraction as having a positive denominator. For example one wouldn't write $(-2)/(-3)$ but instead $2/3$, and one woudn't write $5/(-7)$ but rather $(-5)/7$. Under this convention, if $a/b$ and $c/d$ are fractions (with positive denominators) then we can definitely say that $a/b<c/d$ is equivalent to $ad<bc$ [evan if the numerators of either or both fractions are negative]. This is because in multiplying through by $bd$ to get to the integer inequality, we know that $bd>0$, so the inequality is not reversed.


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