# Please check a basic proof: Non-zero $a,b\in\mathbb{R}$, and $a<\frac{1}{a}<b<\frac{1}{b}\implies a<-1$.

Prove that for non-zero $a,b\in\mathbb{R}$, if $a<\frac{1}{a}<b<\frac{1}{b}$, then $a<-1$.

This was challenging for me, so I understand it might be a tad roundabout. I might be more explicit, than necessary, but that's because I'm just learning this stuff and I want to make sure I don't jump steps and miss something.

My strategy is to use proof by contradiction, and break up the negation of $a<-1$, $(a\geq-1)$ into three cases. Also, I've already proved that if $x$ is positive, then $\frac{1}{x}$ is positive (and vice-versa for negative).

• Looks good. There may be easier ways of course, for e.g. if $a > 0$, just multiply everything by $ab$ to get the contradicting $b < a$ – Macavity Sep 25 '14 at 5:09
• If $a>0,$ then $a<\frac 1a$ implies $a^2<1,$ so $a<1$ as you require to finish case 3. If $a<0,$ then $a<\frac 1a$ implies $a^2>1,$ which is the contradiction you need in case 2. But I think what you wrote is also correct. – David K Sep 25 '14 at 5:19
• Thanks for the suggestions. – Marco Sep 25 '14 at 23:13

Case 1: $a=-1$
If $a=-1$, then $-1<\frac{1}{-1}=-1$ which can't be.

Case 2: $-1<a<0$
Clearly $a$, and $\frac{1}{a}$ are negative. Then, since $\frac{-1}{a}$ is positive,
$-1<a\implies \frac{-1}{a}*(-1)<\frac{-1}{a}*a\implies \frac{1}{a}<-1<a$, or that $a>\frac{1}{a}$
But this contradicts the hypothesis (that $a<\frac{1}{a}$) so that can't be.

Case 3: $a>0$
To make this easier, I'm going to prove something else: If $0<a<\frac{1}{a}$, then $a<1$.
(PBC again): Assume $a>1$, then $\frac{1}{a}*a>\frac{1}{a}*1$, or $\frac{1}{a}<1$, but $\frac{1}{a}<1<a$, or $\frac{1}{a}<a$.
As in case 2, this contradicts the hypothesis (that $a<\frac{1}{a}$).

Also, if $0<x<1$, then $\frac{1}{x}*x<\frac{1}{x}*1\implies0<x<1<\frac{1}{x}$.

So, since $0<a<1<\frac{1}{a}$ (and by the same logic, above) $0<b<1<\frac{1}{b}$, then $b<\frac{1}{a}$, which is the final contradiction that shows all three cases are impossible.