$\sqrt{x^2-1} =\sqrt{x+1}\cdot\sqrt{x-1}.$ Problem: The equation $$\sqrt{x^2-1} =\sqrt{x+1}\cdot\sqrt{x-1},$$
holds true:
a) $\forall \ x\in\mathbb{R}$
b) $\forall \ x\in\{\mathbb{R}:|x|\geq1\}$
c) $\forall \ x\in\{\mathbb{R}:x\geq1\}$
d) None of the above.

Reasoning:
Using $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ I get that $$\sqrt{x^2-1}=\sqrt{(x+1)(x-1)}=\sqrt{(x+1)}\cdot\sqrt{(x-1)}.$$
This means that $$\sqrt{x^2-1}\Longleftrightarrow\sqrt{(x+1)}\cdot\sqrt{(x-1)}.$$ Thus, this should be true for all reals, for example $x=0$ one would obtain $$\begin{array}{lcl}
\text{LHS}  & = & \sqrt{0^2-1} = \sqrt{-1} = i.\\
\text{RHS}  & = & \sqrt{0+1}\cdot \sqrt{0-1} = \sqrt{1}\cdot\sqrt{-1}=1\cdot(-1)=i. \\
\end{array} $$
The answer is c). However, if you ask me, it's true that everything becomes undefined (complex, if you will) once we go under $x=1$ but the question asks for which real values of $x$ the equation holds. Clearly the equation holds for $x<1$ as well. 
If you disagree with me, please state why. Thanks in advance!
 A: Just so this has an answer: Algebraic identities (like legal contracts) come with stipulations that must be met if the conclusions are to be true.
In the real setting, the bare equation $\sqrt{ab} = \sqrt{a\vphantom{b}}\sqrt{b}$ is not an "identity" at all. The correct statement is
$$
\sqrt{ab} = \sqrt{a\vphantom{b}}\sqrt{b}\quad\text{for all real $a \geq 0$, $b \geq 0$.}
$$
The quantifier "for all real $a \geq 0$, $b \geq 0$" is not a decorative flourish, but an essential part of the identity.
Here, consequently, the stated algebraic relationship holds if and only if $x + 1 \geq 0$ and $x - 1 \geq 0$, i.e., if and only if $x \geq 1$.

The deeper point is, if as a student you find yourself memorizing fragments such as
$$
a^{2} - b^{2} = (a - b)(a + b)\quad\text{or}\quad
0 \leq x^{2},
$$
or even worse,
$$
y = mx + b\quad\text{or}\quad
a^{2} + b^{2} = c^{2}\quad\text{or}\quad
x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a},
$$
you need to shift the focus of your attention: The meaning of symbols matters.
