Proof by Contradiction - There exists a positive number $x$ such that $x - \frac{2}{x} <1$ and $x \leq 2$.

Question

I want to prove that there exists a positive number $x$ such that $x - \dfrac{2}{x} <1$ and $x \leq 2$.

Starting a proof by contradiction, I assumed that there is a positive number x such that x - (2/x) > 1 and x <= 2. I multiplied the first inequality by x and got x^2 - x - 2 > 0. I'm confused where to go from here and how to use the fact that x must also be less than or equal to 2.

Answer 1

Take ANY positive number such that:

$$x<2\tag{1}$$

This implies:

$$\frac2x>\frac{2}{2}$$

$$\frac2x>1$$

$$-\frac2x<-1\tag{2}$$

Add (1) and (2) and you get:

$$x-\frac2x<1$$

Actually all positive numbers less than 2 are solutions to your problem. The only exception is $x=2$.

Answer 2

You made a error in "I assumed that there is a positive number x such that x - (2/x) > 1 and x <= 2."

To use proof by contradiction, it should be "For all positive x, x - (2/x) > 1 or x <= 2."

Actually the original statement is wrong. There does NOT exist a positive number x such that x - (2/x) > 1 and x <= 2.