Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question is from the book How to Prove It and I'm having trouble getting started with it. The book provides the hint "first prove $a \lt 0$". However, I can't figure out how to get that far with the provided givens.

The question is:

Given $a\ne 0$ and $b\ne 0$ and $a \lt \frac{1}{a} \lt b \lt \frac{1}{b}$. Prove $a \lt -1$.

share|cite|improve this question
up vote 3 down vote accepted

Suppose that $a$ were positive. Then, multiplying the string of ineqs. (doesn't change their direction) by $a$ gives $a^2 <1 < ba < a/b$. From here we see that $b>1$ (why?), but then, from the last ineq., $b^2 a < a$ which is impossible (why?). Therefore $a<0$. Now multiply again by $a$ and conclude.

share|cite|improve this answer
1  
Thank you very much for your answer. I ended up (possibly) solving it slightly differently based on your answer. I started the same, by saying a^2 < 1 < ba < a/b and concluded that since a^2 < 1 can reduce to a < 1. Since a < 1 and ab > 1, b > 1. Since b < 1/b can reduce to b < 1 we have a contradiction so a must be less than 0. Is that a valid proof (that a > 0) or did I make a mess of it? – John Sep 29 '10 at 22:58
    
That's a correct proof that a must be < 0, yes. – Weltschmerz Sep 29 '10 at 23:13

HINT $\ \ \ $ Multiplying $\rm\ a < 1/a\ \:$ by $\rm\ a^2\ $ yields $\rm\:\ a^3 < a\ $. By symmetry $\rm\ b^3 < b\ $. So both $\rm\: a\:$ and $\rm \:b\:$ must lie in the intervals where the graph of $\ x^3 - x < 0\:,\ $ i.e. either in $(-\infty,-1)$ or in $(0,1)\:$. But $\rm\ a \in (0,1)\ \Rightarrow\ 1/a > 1\ $ contra $\rm\ 1/a < b < 1\ $.

alt text

share|cite|improve this answer
    
A simpler graph-based approach is to consider the (very familiar) graphs of the rectangular hyperbola $y=1/x$ and the line $y=x$. Then the values of $x$ for which $x<1/x$ correspond to the points of the hyperbola lying above the line. These $x$-values comprise two intervals, and the key to the problem is to show that if $a<1/a<b<1/b$ then $a$ and $b$ cannot lie in the same one of these two intervals. – Robin Chapman Sep 30 '10 at 10:21
    
I don't find that approach to be any simpler. It amounts to precisely the same observations on the same intervals. I chose the above approach since students usually have more intuition about roots of polynomials than they do about the intersection points of arbitrary curves. – Bill Dubuque Sep 30 '10 at 15:53

$\ $ If $\rm\ a > 0\ $ then $\rm\ a < b\ $ times $\rm\ 1/a < 1/b\ \Rightarrow\ 1 < 1$

So $\rm\ a < 0\ \:$ and $\rm\:\ a < 1/a\ \:\Rightarrow\:\ a^2 > 1 \ \:\Rightarrow\:\ a < -1$

share|cite|improve this answer

Hint: Try to prove $a < 0$ by contradiction, i.e. assume $a > 0$ and try to derive a contradiction.

share|cite|improve this answer

Hint: What does a<1/a say about a? What does b<1/b say about b?

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.