Proving 0 < x < 1 -> x^2 < 1: A Discrete Math Textbook

[agro]

A Discrete Math textbook first proved that the statement:

0 < x < 1 -> x^2 < 1

is true (I have no problem following the proof).

It then went to prove the contrapositive:

x^2 >= 1 -> x <= 0 or x >= 1

Here's the proof:

Assume x^2 >= 1. (no problem here)
If x <= 0, we have the desired result, so assume x > 0. (what?!?)

The last part I wrote bedazzled me. What does the book mean when it says: "If x <= 0, we have the desired result"? For an implication to be true, the hypothesis and conclusion must be true. We already assumed that the hypothesis is true. Now for the conclusion to be either the left side of the OR statement is true or the right side is true. How can the book conclude that x <= 0 is true?

thanks a lot

 

[Stevo]

Firstly, $$0<|x|<1 \Longleftrightarrow 0<x^2<1$$

Secondly, $$x^2\geq1 \Longleftrightarrow |x|\geq1$$

 

[Hurkyl]

The last part I wrote bedazzled me. What does the book mean when it says: "If x <= 0, we have the desired result"?

Well, you're trying to prove that x <= 0 or x >= 1. If it so happens that x <= 0, then you don't have to do any work to prove that x <= 0 or x >= 1.


The book is doing a proof by cases: x <= 0 or x > 0.

 

[Gokul43201]

Another way to prove the converse is to let x = 1 + h, and then show that h < 0. This is pretty straightforward. Try it yourself.

 

[HallsofIvy]

That's a fairly standard technique so you might want to make certain you understand:

If we want to prove "A is true or B is true", it is sufficient to prove that "If A is NOT true, then B is true".