Prove $a < 1/a < b < 1/b$ implies $a < -1$

Question

I've solved a simple proof from Velleman's How to Prove It (p. 107, Q. 8) but I think my proof is suboptimal and was wondering if there's a better way I could prove it. The prompt is as follows:

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

I proved it as follows:

Suppose a and b are nonzero real numbers and $a \lt \frac{1}{a} \lt b \lt \frac{1}{b}$. Suppose $a \gt 0$. Then $a \lt 1$ because $a \lt \frac{1}{a}$. Since $b \gt a$, $b$ must also be greater than 0, and since $b \lt \frac{1}{b}$, $b < 1$. But since $\frac{1}{a} > 1$, $\frac{1}{a} > b$ which is a contradiction, therefore $a \lt 0$.

Since $a \lt 0$, then $a^2 \gt 1 \gt ab \gt \frac{a}{b}$. Therefore, $a$ must be greater than $1$ or less than $-1$. But we already know $a \lt 0$, therefore $a \lt -1$.

Thus if $a$ and $b$ are nonzero real numbers and $a \lt \frac{1}{a} \lt b \lt \frac{1}{b}$, then $a \lt -1$.

Answer 1

This is an answer without using the proof by contradiction.

We have, multiplying by $a^2\gt 0$, $$a\lt\frac 1a\implies a^3\lt a\implies a\lt 1\tag1$$ Similarly, $$b\lt 1\tag2$$ So, from $(1)(2)$, $$\frac 1a\lt b\lt 1\implies a\lt a^2\implies a\lt 0$$ It follows that $$a\lt\frac 1a\implies a^2\gt 1\implies a\lt -1$$

Answer 2

Your proof seems perfectly fine to me. I don't really think there's any need for improvement, other than some minor details, such as mentioning where the inequalities $a^2>1>ab>\frac ab$ come from, and how you know that if $a>0$ and $a<\frac1a$, then $a<1$.

But like I said, it's fairly obvious where these come from, and I don't think there's any need to improve the proof any further.

Answer 3

The proof that $a<0$ is fairly good.

Now $$ a<\frac{1}{a} $$ is equivalent to $$ -a>\frac{1}{-a} $$ and you have already observed that this implies $\frac{1}{-a}<1$. Therefore $a<-1$.