- #1
ForceBoy
- 47
- 6
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?