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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 $a \lt -1$.

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5 Answers 5

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.

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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\ $.

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

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

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

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