# Inequality proof

1. Aug 29, 2013

### phospho

1. The problem statement, all variables and given/known data
Prove

If $0 \leq a < b$ and $0 \leq c < d$ then $ac < bd$

3. The attempt at a solution

not sure how to even start on this,

was thinking if a = 0 or c = 0, then ac = 0, but bd > 0 (which is given) so bd > ac
however this seems like I'm cheating because they give you bd > ac, so I don't think it's valid

2. Aug 29, 2013

### verty

They don't give you ac < bd, that is what you must show.

What you have done so far is correct but you still have to show it is true when 0 < a, 0 < c.

Hint: you were probably given the rule that, for 0 < c, a < b → ac < bc. Use it.

3. Aug 29, 2013

### phospho

Yes we were given that, but I couldn't proceed how to use it. I used ac < bd to get to where I am right now.

4. Aug 29, 2013

### tiny-tim

no, you can't use ac < bd !!

use verty's hint

(second hint: if you could prove ac < bc, what would you still need to get ac < bd ? )

5. Aug 31, 2013

### phospho

I'm sorry, I'm trying but I just don't see how to proceed.

For the proof of c>0 and a<b prove ac<bc it was very simple, all I had to use was (b-a).c > 0, but when there are two inequality signs I don't know how to deal with it. Is multiplying by b or d to each of the inequalities allowed? For instance, if 0<=a<b and 0 <=c<d is it true that 0<=bc < bd?

6. Aug 31, 2013

### tiny-tim

hi phospho!
that's right! … if you can prove ac < bc and bc < bd,

then obviously you have ac < bc < bd, ie ac < bd

sooo … now prove ac < bc and bc < bd (separately)