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

• jfierro
In summary: We can also discard the ## a = -1 ## case because we would arrive at ## a = \frac{1}{a} ##, which is another contradiction. Thus, ## 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 ##."

jfierro said:
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.
---
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 ##.

Last edited:

## 1. How can you prove that if a < 1/a < b < 1/b then a < -1?

The proof involves using the properties of inequalities and the fact that the product of two negative numbers is a positive number. First, we can multiply both sides of the inequality a < 1/a by -1 to get -a > -1. Similarly, we can multiply both sides of the inequality b < 1/b by -1 to get -b > -1. Since a < b, we can substitute -b for b in the inequality -a > -b. This gives us -a > -b > -1. Finally, we can divide both sides by -1 to get a < b < 1. Since we know that 1 is greater than any positive number, this means that a must be less than -1.

## 2. Are there any other ways to prove this statement?

Yes, there are other ways to prove this statement. One way is to use contradiction by assuming that a ≥ -1 and showing that this leads to a contradiction. Another way is to use mathematical induction to prove the statement for all values of a and b. However, the method described in the previous answer is a simple and straightforward way to prove the statement.

## 3. Can this statement be generalized to include other numbers besides -1?

Yes, this statement can be generalized to include any negative number. The proof would follow a similar logic, but instead of using -1, we would use the specific negative number in the inequality. For example, if we wanted to prove that if a < 1/a < b < 1/b then a < -2, we would multiply both sides by -2 and substitute -2 for -1 in the proof.

## 4. Is there a visual representation that can help understand this statement?

Yes, we can represent this statement on a number line. Since we know that a < 1/a < b < 1/b, we can plot these numbers on a number line with a and b being closer to 0 and 1/a and 1/b being closer to 1. Then, we can see that a must be less than -1 since it is to the left of all the other numbers on the number line.

## 5. Can this statement be applied to other mathematical concepts?

Yes, this statement can be applied to other mathematical concepts that involve inequalities and negative numbers. For example, it can be used in calculus to prove the convergence of a series, or in geometry to prove the properties of a triangle. It can also be used in economics and social sciences to analyze the relationship between variables with negative values.