Proof question from How to Prove It

[paridiso]

proof question from "How to Prove It"

3.2.8 Suppose a & b are nonzero real numbers. Prove that if a < 1/a < b < 1/b then a < -1

I understand intuitively why this is true, but I can't figure out how to prove it. According to the hints at the back of the book it says to prove a < 0, then use to prove a < -1.

When I go through the inequalities I come up with this:

a < 1/b
ab < 1

1/a < b
1 < ab

I know that when you multiply both sides of inequality you have to switch the signs. But if both a and b are negative the signs both switch so I don't really understand how this can be true.

 

[Dick]

Just take it one step at a time. If a<1/a then either a is in (0,1) or (-infinity,-1). You can show that, right? Same for b. Now if b is also in (0,1) and a<b can 1/a be less than 1/b?

 

[paridiso]

Thanks, I understand it now.