Prove that if $a<1/a
Suppose that $a$ and $b$ are nonzero real numbers. Prove that if $a<1/a<b<1/b$ then $a<−1$.
This is a problem from Velleman's "How To Prove It". There is a hint in the back of the book to solve this question: "Assume $a<1/a<b<1/b$. Now prove that $a<1$, then use this fact to prove that $a<0$, and then use this fact to prove that $a<-1$." 
I am getting stuck after starting with 
$$a < \frac{1}{a}$$
$$a^2a < a^2\frac{1}{a}$$
$$a^3 < a$$
At this point, is it valid to divide by a? We do not know whether it is positive or negative, so we could need to switch the less than sign to a greater than. How does one complete this proof from this point? Even if we could divide by $a$, I get stuck again at $(a-1)(a+1)<0$ for the same reason.
This question has been asked on here before, but all of the answers skip this (probably trivial) part that I am confused about. 
 A: For the first part of your provided hint, consider
$$a \lt \frac{1}{a} \tag{1}\label{eq1A}$$
If $a \ge 1$, then it's positive, so you can multiply both sides by $a$ to get
$$a^2 \lt 1 \implies \sqrt{a^2} \lt \sqrt{1} \implies |a| \lt 1 \implies  a \lt 1 \tag{2}\label{eq2A}$$
This contradicts your assumption, so you have that $a \lt 1$.
With the second part of the provided hint, note that what was just shown above gives that since $b \lt \frac{1}{b}$, then $b \lt 1$. If $0 \lt a \lt 1$, then $\frac{1}{a} \gt 1$, but this contracts $b \gt \frac{1}{a}$. Thus, you have that $a \lt 0$.
With the final part of the provided hint, since $a \lt 0$, multiplying both sides of \eqref{eq1A} changes the inequality around, so you get
$$a^2 \gt 1 \implies \sqrt{a^2} \gt \sqrt{1} \implies |a| \gt 1 \implies a \lt -1 \tag{3}\label{eq3A}$$
A: $$a-\frac{1}{a}<0$$ it's $$\frac{(a-1)(a+1)}{a}<0,$$ which gives $$a\in(-\infty,-1)\cup(0,1).$$ 
By the same way $$b\in(-\infty,-1)\cup(0,1).$$
But $$\frac{1}{a}<b,$$ which says that $0<a<1$ is impossible.
A: Not sure I'd do the hints in order but ...
We can't have $a \ge 1$ because that would mean $1 \ge \frac 1a$ and $a \ge 1 \ge \frac 1a$ and we are told that is not so.
So $a < 1$.  If $a > 0$ then $0 < a < 1$ and $\frac 1a > 1$ so $b>1$ and $1 < \frac 1b$ so $\frac 1b < 1 < \frac 1b$.  But we we were told that isn't so.  
So $a \le 0$.
If $a = 0$ then ... oh, get out of here $\frac 10$ is not defined...
So $a < 0$.  That means we flip inequality sign!.
So $a < \frac 1a$ means $a^2 > 1$ so $|a| > 1$.  So $a < -1$.
