1

Suppose that a and b are nonzero real numbers. Prove that if $a < \frac{1}{a} < b < \frac{1}{b}$ then $a < -1$.

Here's my attempt:

Suppose that $a < \frac{1}{a} < b < \frac{1}{b}$, and that $a > 0$.

Multiplying the inequality by $a$, we get $0 < a^2 < 1 < ab < \frac{a}{b}$.
Taking the square root of $a^2 < 1$, we get $a < 1$.
*Because $0 < a < 1 < ab$, $b > 1$.
*But if $0 < a < 1 < b$, then $ab > \frac{a}{b}$, which contradicts $0 < a^2 < 1 < ab < \frac{a}{b}$.

Therefore, $a < 0$.

Because $a < 0$, multiplying $a < \frac{1}{a}$ by $a$, we get $a^2 > 1$.
So, $\sqrt{a^2} > 1$, and since it has been proven that $a < 0$, $a < -1$.

Therefore, if $a < \frac{1}{a} < b < \frac{1}{b}$, then $a < -1$.
//end proof

I looked for a solution in the back of the book, and there was just a tip for writing the proof, which was not helpful to me.

Questions in order of importance to me:
Did I prove it?
Were the statements with * in front of them justified?
If I did prove it, is there a much more elegant way of proving this and what is it? (I would bet my life that there is.)
How is my style?
Do you have any general advice for me based on this post?

A Name
  • 11
  • Welcome to the site, and thanks for the submission. I can see you have put work into this! You are not too far off, but I would deduct marks for the line: "Taking the square root of...", as it does not follow that $a > 1$: consider $a = -50$, say. Also, logically, if we don't have $a > 0$, it does not follow that $a < 0\dots$ – DanLewis3264 Feb 10 '20 at 05:22
  • Thank you for the welcome, @bounceback. If I had instead written "Taking the square root of $0 < a^2 < 1$, we get $0 < a <1$" would it then be valid? And if I had specified that $a$ and $b$ are nonzero real numbers, wouldn't it then follow that $a < 0$? – A Name Feb 10 '20 at 05:36
  • I'm afraid not: $0 < a^2 = 0.25 < 1$ but that may be true for $a = -0.5$! I would avoid any mention whatsoever of square roots when dealing with inequalities - split into cases. And yes, that would be fine. Either $a > 0$, $a = 0$ or $a < 0$ for a trichotomy, or $a \geq 0$ plays $a < 0$ (or $\leq, >$ resp.) for a dichotomy – DanLewis3264 Feb 10 '20 at 23:20

1 Answers1

1

We have: $$a<\frac{1}{a}$$ or $$\frac{(a-1)(a+1)}{a}<0$$ gives $$a<-1$$ or $$0<a<1.$$ Similarly, $$b<-1$$ or $$0<b<1$$ and we'll get a contradiction after assuming $0<a<1$ because if so we obtain: $$a<b$$ and $$\frac{1}{a}<\frac{1}{b},$$ which is impossible.

Michael Rozenberg
  • 194,933