Proof of $a<\frac{1}{a}I am currently working on Velleman's How to Prove It section 3.2, question nine (question eight in the second edition).
Why does he hint in the solution section to go through the steps to first prove $a<1$ and $a<0$, rather than just using a proof by contradiction by assuming $a\ge-1$ to conclude the goal in one step? Can someone point out where I have gone wrong in this proof?
Suppose $a$ and $b$ are nonzero real numbers. Suppose $a<\frac{1}{a}<b<\frac{1}{b}$ and $a\ge-1$. Because $a$ is nonzero, either $a=-1$ or $a>0$. Clearly it is not the case that $a>0$ because this contradicts $a<\frac{1}{a}$. It cannot be the case that $a=-1$, as this also contradicts $a<\frac{1}{a}$. Therefore we conclude $a<-1$ as required.
EDIT: Right I can see the problem with my proof. The solution manual has this: "Suppose that $a\ge-1$. Dividing by the positive
number $−a$, we conclude that $−1\ge\frac{1}{a}$. Combining $a\ge-1$ and $−1\ge\frac{1}{a}$ we get $a\ge\frac{1}{a}$,
contradicting the fact that $a<\frac{1}{a}$. Therefore $a<−1$." Why can he divide by the "positive number" $-a$?
 A: Suppose $a$ and $b$ are nonzero real numbers such that $a < \frac{1}{a} < b < \frac{1}{b}$. We want to show that $a < -1$.
To this end, suppose towards a contradiction that $a \geq -1$. Then there are three cases to consider:
Case 1: Suppose that $-1 \leq a < 0$. Then since $a$ is negative, we know that $-a$ is positive, so we can divide it from both sides of the inequality $-1 \leq a$ to obtain $\frac{1}{a} \leq -1$. But since $a < \frac{1}{a}$, it follows that $-1 \leq a < \frac{1}{a} \leq -1$ so that $-1 < -1$, a contradiction.
Case 2: Suppose that $a \geq 1$. Then since $a$ is positive, we can divide it from both sides of the inequality $1 \leq a$ to obtain $\frac{1}{a} \leq 1$. But since $a < \frac{1}{a}$, it follows that $1 \leq a < \frac{1}{a} \leq 1$ so that $1 < 1$, a contradiction.
Case 3: Suppose that $0 < a < 1$. Then since $a$ is positive, we can divide it from both sides of the inequality $a < 1$ to obtain $1 < \frac{1}{a}$. But since $\frac{1}{a} < b$, it follows that $1 < b$. Since $b$ is positive, we can divide it from both sides of this inequality to obtain $\frac{1}{b} < 1$. But since $1 < b < \frac{1}{b} < 1$, we conclude that $1 < 1$, a contradiction.
Thus, we conclude that $a < -1$, as desired. $~~\blacksquare$
