Homework Help: How To Prove it Inequality Proof

1. Jul 26, 2011

Sorgen

1. The problem statement, all variables and given/known data
Suppose that a and b are nonzero real numbers. Prove that if a<1/a<b<1/b then a<-1.

3. The attempt at a solution
So after a while I realized that I could prove that a<-1 by contradiction but first I have to prove that a<0. I figured out how to prove it but I'm not sure if my wording is convincing enough and I feel that it might be redundant at points. Anyway, here's the proof:

Proof: Suppose a<1/a<b<1/b. It then follows that 1/a<1/b. Multiplying both sides of the inequality yields b<a, but this contradicts a<b. Therefore, one and only one variable a or b must be negative, and it follows that since b>a then a<0. Now suppose a$\geq$-1. Plugging the value a=-1 into a<1/a yields -1<1/-1 or -1<-1, which contradicts a<1/a. Therefore a<-1.

Now I feel like the wording of that proof is a mess but I'm not sure how I would reword it. Also am I being redundant in saying "one and only one variable a or b must be negative" or does that need to be there? Or am I crazy and the proof is acceptable as is?

2. Jul 26, 2011

SammyS

Staff Emeritus
What is it you multiply by? What if ab is negative?

3. Jul 26, 2011

Sorgen

Oops yeah I multiply both sides by ab. If ab is negative then the statement doesn't contradict because we then get a<b, but didn't I state that by pointing out the contradiction of if ab is positive?

If I modified that part to say:

Suppose ab is a positive real number. Multiplying both sides of the inequality 1/a<1/b by ab yields b<a, but this contradicts a<b therefore ab must be negative and thus because a<b a<0.

Is that better?

4. Jul 26, 2011

SammyS

Staff Emeritus
Yes, that's better.

5. Jul 26, 2011

Sorgen

Awesome, thanks!