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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 $a \lt -1$.

6 Answers 6

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Suppose $a\ge-1$. Then adding $1$ to each term in the inequalities gives

$$0\le1+a\lt{1+a\over a}\lt1+b\lt{1+b\over b}$$

From $0\le1+a\lt{1+a\over a}$ and $0\lt1+b\lt{1+b\over b}$, it follows that $a$ and $b$ are positive, in which case

$${1+a\over a}\lt{1+b\over b}\implies b+ab\lt a+ab\implies b\lt a$$

But this contradicts the original inequality $a\lt b$. Therefore we must have $a\lt-1$.

  • 0
    Why is $1+a < \frac{1+a}{a}$ when $a \geq -1$?2016-12-05
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    @HarshCurious, $a\lt{1\over a}\implies1+a\lt1+{1\over a}={a+1\over a}$, regardless of the sign of $a$. The assumption $a\ge-1$ only enters into the opening of the string, $0\le1+a$.2016-12-05
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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 $a<1/a$a$ and $b$ cannot lie in the same one of these two intervals. – 2010-09-30
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    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.2010-09-30
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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?2010-09-29
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    That's a correct proof that a must be < 0, yes.2010-09-29
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$\ $ 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?