Inequality $\left(\frac{2x-1}{2x+1}\right)^2>0$ The inequality
$$\left(\frac{2x-1}{2x+1}\right)^2>0$$
has the same solution set as 
a) $(2x-1)^2 > 0$
b) $2x-1 \neq 0$
c) $2x-1 > 0$
d) None of the above
Is this equivalent to$$\frac{2x-1}{2x+1}>0 \ ? $$
 A: The answer for your question is shortly "No", because $$\left( \frac { 2x-1 }{ 2x+1 }  \right) ^{ 2 }>0\quad \Rightarrow x\in R-\left\{ \pm \frac { 1 }{ 2 }  \right\} \\ \\ \frac { 2x-1 }{ 2x+1 } >0\quad \Rightarrow x\in \left( -\infty ;-\frac { 1 }{ 2 }  \right) \cup \left( \frac { 1 }{ 2 } ;+\infty  \right) $$
A: No, it's not equivalent to $\frac{2x-1}{2x+1}>0$. Instead, note that the square of any real number $z$ is non-negative, and is zero iff $z=0$; thus $\left(\frac{2x-1}{2x+1}\right)^2>0$ implies $\frac{2x-1}{2x+1}\ne0$, which is equivalent to $2x-1\ne0$. But $2x+1\ne0$ too because division by zero is not allowed, so the answer is (d).
A: The square of any expression is zero or positive, provided it is defined. Here the expression


*

*is zero when $2x-1=0$,

*is undefined when $2x+1=0$.
Hence, d) i.e. $|x|\ne\frac12$.
And no, the square of an expression being positive doesn't mean that the expression is positive.
A: They are not equivalent because the square allows for "negative" values of the expression that make the inequality true when squared, stripping the square, a common mistake, only finds the "positive" values of the expression. The values of x that make the inequality true are (2x-1)^2 > 0 as long as x does not equal -1/2, which causes division by zero. Parcly Taxel writes a perfectly concise and beautiful mathematical response to the problem, I'm just illuminating the reason why.
