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Given two inequalities the rules of multiplication are said to be (check link for the source):

1.) If $0<a<b$ and $0<c≤d$, then $0<ac<bd$

2.) If $0<a<b$ and $c≤d<0$, then $bd<ac<0$

3.) If $0<a<b$ and $c<0<d$, then $ac<0<bd$

4.) If $a<b<0$ and $c<0<d$, then $bd<0<ac$

5.) If $a<0<b$ and $c<0<d$, then no conclusion may be drawn about the relative positions of $ac$ and $bd$ on the number line

How do I prove the second and fourth rules "If $0<a<b$ and $c≤d<0$, then $bd<ac<0$" and "If $a<b<0$ and $c<0<d$, then $bd<0<ac$" ?

My Attempts:

1.) $$ 0<a<b \quad\&\quad 0<c<d\\0<ac<bc \quad\&\quad 0<bc<bd \implies 0<ac<bc<bd\implies 0<ac<bd $$

3.) $$ 0<a<b \quad\&\quad c<0<d\\ 0<ad<bd \quad\&\quad ac<0<ad\implies ac<0<ad<bd\implies ac<0<bd $$ Similarly,

2.) $$ 0<a<b\quad\&\quad c<d<0\\ 0>ac>bc \quad\&\quad bc<bd<0\implies ? $$

4.) $$ a<b<0 \quad\&\quad c<0<d\\ad<bd<0 \quad\&\quad ac>0>ad\implies ? $$

Sooraj S
  • 7,573

2 Answers2

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For 2), you can use 1): $$d\le c < 0\quad\Rightarrow\quad 0 < -c \le -d,$$ so by 1), $0 < -ac < -bd$ and thus $bd < ac <0$.

Edit: Note that item 2) as stated is false. I've left this here under the hypothesis that the conclusion of 2) is actually misstated.

For 4), note that $b<0$ and $d>0$, so that $bd<0$. Further, $a$, $c<0$, so that $ac>0$.

rogerl
  • 22,399
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2) is not true: take $a=1$, $b=2$, $c=-5$, $d=-2$, then the inequality is not satisfied since $2(-2) \not<1(-5)$
If you ment $ac<bd$ then you just need to observe that $|a|=a<b=|b|$ and $-|c|=c<d=-|d|$ hence $ac=-|ac|<-|bd|=bd$

la flaca
  • 2,593
  • thanx. rule 2.) seems to be wrong given your example. Can i confirm it is rather "If $0<a<b$ and $c<d<0$, then $ac<bd<0$". What about the 4th rule ?. Pls check the source link i have edited the post. – Sooraj S Jan 08 '18 at 20:04