Can You Solve This Proof Involving Inequalities and Negations?

[ForceBoy]

Homework Statement

The chapter of the book this exercise is found in is titled Proofs Involving Negations and Conditionals.

The problem is as follows:

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

Homework Equations

The contrapositive law will be used as well as inequalities

The Attempt at a Solution

[/B]
My scratch work is as follows:

$\text{RTP: } a < \frac{1}{a} < b < \frac{1}{b} \to a < -1$
because the conclusion has nothing to do with $b$,
$a < \frac{1}{a} \to a < -1$

The contrapositive is then

$a \geq -1 \to a \geq \frac{1}{a}$

I then break this up into the cases that $a > 0$ and $-1 \leq a < 0$
(I do not include the case that $a = 0$ because it was given that $a \neq 0$)

For the first case I work backwards to see how I would arrive at $a \geq \frac{1}{a}$ from $a \geq -1$

$a \geq \frac{1}{a} \\ a^2 \geq 1 \\ |a| \geq 1$
I choose the lower restriction:
$a \geq -1$

I now know how to proceed in my proof for the first case.

My problem comes up with the second case, $-1 \leq a < 0$

$a \geq \frac{1}{a} \\ a^2 \leq 1$The inequality is flipped because of multiplication by a negative number
But now $|a| \leq -1$

But this is impossible! I know I made a mistake but I don't know what it is. Could someone point it out to me?

 

[member 587159]

If you ignore that there is given something about $b$, the statement is false: Take $a= 1/2 >-1$. You must somehow be able to use the relation of $a$ and $b$, even if $b$ doesn't appear in the conclusion.

 

[ForceBoy]

All that is given about $b$ is that it is a real nonzero number and that it satisfies the given inequality. I don't see how this would be useful for finding a restriction on $a$. When writing the proof, I will mention $b$ but only in the first line stating my givens:

Suppose $a$ and $b$ are real nonzero numbers where...

I won't ignore $b$ but I won't use it. By this I mean I will acknowledge $b$ and the information on it but this information won't be used. Is there something wrong with this?

 

[member 587159]

All that is given about $b$ is that it is a real nonzero number and that it satisfies the given inequality. I don't see how this would be useful for finding a restriction on $a$. When writing the proof, I will mention $b$ but only in the first line stating my givens:

Suppose $a$ and $b$ are real nonzero numbers where...

I won't ignore $b$ but I won't use it. By this I mean I will acknowledge $b$ and the information on it but this information won't be used. Is there something wrong with this?

If you never use anything about $b$, you won't find the result. I provided a counterexample if you drop that assumption.

 

[ForceBoy]

I will see what I can do with $b$, then. Thank you.

 

[Ray Vickson]

Homework Statement

The chapter of the book this exercise is found in is titled Proofs Involving Negations and Conditionals.

The problem is as follows:

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

Homework Equations

The contrapositive law will be used as well as inequalities

The Attempt at a Solution

[/B]
My scratch work is as follows:

$\text{RTP: } a < \frac{1}{a} < b < \frac{1}{b} \to a < -1$
because the conclusion has nothing to do with $b$,
$a < \frac{1}{a} \to a < -1$

The contrapositive is then

$a \geq -1 \to a \geq \frac{1}{a}$

I then break this up into the cases that $a > 0$ and $-1 \leq a < 0$
(I do not include the case that $a = 0$ because it was given that $a \neq 0$)

For the first case I work backwards to see how I would arrive at $a \geq \frac{1}{a}$ from $a \geq -1$

$a \geq \frac{1}{a} \\ a^2 \geq 1 \\ |a| \geq 1$
I choose the lower restriction:
$a \geq -1$

I now know how to proceed in my proof for the first case.

My problem comes up with the second case, $-1 \leq a < 0$

$a \geq \frac{1}{a} \\ a^2 \leq 1$The inequality is flipped because of multiplication by a negative number
But now $|a| \leq -1$

But this is impossible! I know I made a mistake but don't know what it is. Could someone point it out to me?

Your first error is the statement $a < \frac{1}{a} \Rightarrow a < -1$. That is false. What we can say is:
$$
a < \frac 1 a \Rightarrow \begin{cases} a^2 < 1 & \text{if} \;\; a > 0\\
a^2 > 1 & \text{if} \;\; a < 0
\end{cases}
$$
So, if $a > 0$ it must lie between 0 and 1, while if $a < 0$ it must be less than -1.

 

[RPinPA]

The only thing you know about $b$ is that it exists and has those properties. What does that existence tell you?
We've established what $a < 1/a$ implies about the set $a$ belongs to.
Since $b < 1/b$, then $b$ belongs to the same set.

So is there something about the relationship between $a$ and $b$, something in those inequalities, that eliminates the possibility of $a > 0$? That won't allow $b$ to exist which satisfies all the conditions?

Try to for some counterexamples and see why they won't work. That often helps me understand why all of the premises are important in some proof.
For instance, let $a = 1/2$ and $b = 2/3$. Then $a < b$, $a < 1/a$, $b < 1/b$. But $1/b = 3/2 < 1/a = 2$ and also $1/a > b$ so not all the inequalities are satisfied.

 

[SammyS]

It looks like OP has been given plenty of good help for now.

Let's wait for a substantial reply from @ForceBoy before heaping more help on this thread.

 

[ForceBoy]

Thank you all for your help and time. I now understand the problem better and have a new mindset on how to tackle it. I will write an attempt soon.

 

[ForceBoy]

Suppose there are two nonzero real numbers, $a$ and $b$, such that $a < \frac{1}{a} < b < \frac{1}{b}$. Because $a < b$, $a$ cannot be positive while $b$ is negative. Then there are three cases to consider.
Case 1: Suppose $a > 0$ and $b > 0$. Then $a < \frac{1}{a}$ yields that $a^2 < 1$ so $|a| <1$ but $a > 0$ so $0 < a < 1$. Through a similar process we see that $0 < b <1$. We can then choose two arbitrary numbers, $a$ and $b$, in the range $(0, 1)$ such that $a<b$. Dividing both sides of the inequality by $ab$ we see that $\frac{1}{b} < \frac{1}{a}$ but $\frac{1}{a} <\frac{1}{b}$. Because the assumption that $a < 0$ and $b < 0$ has led to a contradiction, we know that this is false.
Case 2: Suppose $a < 0$ and $b > 0$. Then $a < \frac{1}{a}$ yields that $a^2 > 1$ so $|a| >1$ but $a < 0$ so $a < -1$. $b$ lies in the range $(0, -1)$. We can then choose some numbers, one, $a$ less than $-1$ and the other, $b$, greater than zero and less than one, such that $a < b$. Dividing both sides of the inequality by $ab$, we see that $\frac{1}{a} <\frac{1}{b}$ and the initial inequality is satisfied.
Case 3: Suppose $a < 0$ and $b < 0$. Then $a < \frac{1}{a}$ yields that $a^2 < 1$ so $|a| <1$ but $a > 0$ so $a< -1$. Through a similar process we see that $b< -1$. We can then choose two arbitrary numbers, $a$ and $b$, in the range $[-∞, 1)$ such that $a<b$. Dividing both sides of the inequality by $ab$ we see that $\frac{1}{b} < \frac{1}{a}$ but $\frac{1}{a} <\frac{1}{b}$. Because the assumption that $a < 0$ and $b < 0$ has lead to a contradiction, we know that this is false.
In the cases that the initial inequality held true, $a < -1$. Therefore if there exist two nonzero real numbers, $a$ and $b$, such that $a < \frac{1}{a} < b < \frac{1}{b}$ then $a<-1$.

 

[ForceBoy]

In line seven, I meant $b$ lies in the range $(0,1)$