0
$\begingroup$

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).

Thanks in advance.

--Proof posted as answer.

$\endgroup$
  • 1
    $\begingroup$ Looks good. There may be easier ways of course, for e.g. if $a > 0$, just multiply everything by $ab$ to get the contradicting $b < a$ $\endgroup$ – Macavity Sep 25 '14 at 5:09
  • 1
    $\begingroup$ 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. $\endgroup$ – David K Sep 25 '14 at 5:19
  • $\begingroup$ Thanks for the suggestions. $\endgroup$ – Marco Sep 25 '14 at 23:13
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.