What Values of x Satisfy the Inequality \(\frac{x-1}{x+1} > 0\)?

[Saladsamurai]

Homework Statement

Find all x for which $\frac{x-1}{x+1}>0 \qquad(1)$

Homework Equations

(2) AB > 0 if A,B >0 OR A,B < 0

(3) 1/Z > 0 => Z > 0

The Attempt at a Solution

Since (1) holds if:

$(x-1) > 0 \text{ and } (x+1) > 0 \qquad x\ne -1$

then we must have x>1 AND x>-1

and since (1) also will hold if:

$(x-1) < 0 \text{ and } (x+1) < 0 \qquad x\ne -1$

then we must have x<1 AND x<-1

So that the solution is x on the interval $(-\infty,-1) \cup (1,\infty)$.

What is the proper way to write the solution using set builder notation?

 

[jbunniii]

You could write, for example:

$$\left\{x : \frac{x-1}{x+1} > 0\right\} = (-\infty, -1) \cup (1, \infty)$$

Or:

$$\frac{x-1}{x+1} > 0 \iff x \in (-\infty, -1) \cup (1, \infty)$$