Prove that if a < 1/a < b < 1/b then a < -1.

[jfierro]

Hello again. I recently submitted a thread asking for feedback on a couple of very basic proofs for an exercise from the book "How To Prove It" by Velleman. This is another request for you to help me understand how wrong my proof for a new exercise could be improved.

1. Homework Statement

Exercise 3.2.8. Suppose that a and b are nonzero real numbers. Prove that if a < 1/a < b < 1/b then a < -1.

Homework Equations

My proof uses three other facts, the first one of which is proven in a previous exercise. The second and third ones I didn't bother proving since I consider them to be rather trivial.

• $0 < a < b \to \frac{1}{b} < \frac{1}{a}$
• $a < 0 \to \frac{1}{a} < 0$
• $a < 0 \to -a > 0$

I also gave some background on what Velleman has introduced as far as proof techniques go in my previous thread.

The Attempt at a Solution

"By the contrapositive of $0 < a < b \to \frac{1}{b} < \frac{1}{a}$ we can conclude that $\frac{1}{a} \le \frac{1}{b} \to a \le 0 \vee b \le a$. But then since $b > a$ and $a \ne 0$, it follows that $a < 0$. Now suppose that $-1 < a < 0$. Since $a < 0 \to \frac{1}{a} < 0$, we know that $\frac{1}{a} < 0$. Therefore, $-\frac{1}{a} > 0$. This means that we can multiply both sides of the $-1 < a$ inequality by $-\frac{1}{a}$ to obtain $\frac{1}{a} < -1$. However, we assumed that $-1 < a$. This would mean that $\frac{1}{a} < a$, which is a contradiction. Therefore, it must be the case that $a \le -1$. We can also discard the $a = -1$ case because we would arrive at $a = \frac{1}{a}$, which is another contradiction. Thus, $a < -1$."

 

[mfb]

Now suppose that $-1 < a < 0$. Since $a < 0 \to \frac{1}{a} < 0$, we know that $\frac{1}{a} < 0$. Therefore, $-\frac{1}{a} > 0$.

I would shorten that part to
Now suppose that $-1 < a < 0$. Then $a<0 \Rightarrow \frac{1}{a} < 0 \Rightarrow -\frac{1}{a} > 0$
but the longer version is not wrong.

Looks good. You used more things than stated in (2), however:
a<b and c>0 => ac<bc
a<b<c => a<c

For the case -1<a<0 there is a shortcut by using the first "relevant equation" with b=1: $0 < -a < 1 \Rightarrow \frac{1}{1} < \frac{1}{-a} \Rightarrow -a < \frac{1}{-a} \Rightarrow a > \frac{1}{a}$, contradiction.

The proof gets easier to read if you make separate parts for the different ranges of a you consider.

 

[jfierro]

Thanks! I have revised the proof as follows, and it does seem to have gotten clearer to read.
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Facts used (implicitly or otherwise):

• $0 < a < b \to \frac{1}{b} < \frac{1}{a}$
  
• $a < 0 \to -a > 0$
• $(a < b \wedge c > 0) \to ac < bc$
• $a < b < c \to a < c$
• $(-1)\frac{1}{-a} = \frac{1}{a}$
• $(-1)(-1) = 1$ (Ok, I'm going to stop here lest we get down to the field, distributive and ordering properties of the reals.)

---
From the contrapositive of $0 < a < b \to \frac{1}{b} < \frac{1}{a}$ we can conclude that $\frac{1}{a} \le \frac{1}{b} \to a \le 0 \vee b \le a$. But then since $b > a$ and $a \ne 0$, it follows that $a < 0$. This means that $a$ can fall under one of three cases:

1. $-1 < a < 0$
2. $a = -1$
3. $a < -1$

Suppose that $a$ is as in the first case. This implies that $0 < -a < 1$. From $0 < a < b \to \frac{1}{b} < \frac{1}{a}$ with $b = 1$ it follows that $1 < \frac{1}{-a}$. Since $-a < 1$, this would also imply that $-a < \frac{1}{-a}$ and thus that $a > \frac{1}{a}$, which is a contradiction.

Now suppose that $a = -1$. This would mean that $a = \frac{1}{a}$, which is again a contradiction.

Therefore, it must be the case that $a < -1$.