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?
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.