Prove $rs<r's'$ given $0\lt r<r'$ and $0\lt s<s'$

Question

Suppose I have the rational numbers r,r',s,s', for which r,s>0 and

$ r < r' $

and

$ s < s' $

From these assumptions, how would you then prove

$ rs<r's' $

I don't know why I can't figure these out but whatever, I can't.

Answer 1

Use the rule, $$\text{if } a<b \text{ and } b<c \text{ then } a<c.$$

First, multiply $r<r'$ by $s$, to get $rs<r's$. Note that the inequality won't reverse because $s>0$. Then multiply $s<s'$ by $r'$ to get $r's<s'r'$. Again, because $r>0\rightarrow r'>0$ the inequality doesn't reverse. All you have to do is put it all together with the above rule. Because $rs<r's$, and $r's<s'r'$, we have $rs<r's'$.

Note that order doesn't matter. That is, $r's'=s'r'$.

Answer 2

Hint: $rs-r's'=(rs-rs')+(rs'-r's')=r(s-s') + s'(r-r') \lt 0\,$.

Answer 3

Order axioms include

(i) if $x>y$ then $x+z> y+z$ and

(ii) if $x>z$ and $y>0$ then $xy>zy.$

From (ii) with $z=0$ we have : If $x>0$ and $y>0$ then $xy>0.$ From (i) with $z=-y$ we have: $x>y \iff x-y>0.$ From (i) we also have: If $x,y,z$ are positive then so is $x+y+z.$

If $0<r<r'$ and $0<s<s',$ let $r'=r+d$ and $s'=s+e.$ Then $d$ and $e$ are positive. We have $$r's'>rs\iff r's'-rs>0\iff (r+d)(s+e)-rs>0\iff re+ds +de>0.$$ Now $re,ds,$ and $de$ are positive because $r,e,d,s$ are all positive. So $re+ds+de$ is indeed positive.