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.


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!

  • 3
    $\begingroup$ I disagree with you purely on the basis that $\sqrt{-1}$ is either $i$ or $-i$, and you have no idea of knowing whether the left-hand side and the right-hand side chooses the same one. $\endgroup$ – Arthur May 23 '17 at 11:13
  • 1
    $\begingroup$ Hint: How about if $x < 0$? (For example, $x = -2$.) $\endgroup$ – Arnie Bebita-Dris May 23 '17 at 11:13
  • 4
    $\begingroup$ In general, the formula $$\sqrt{ab} = \sqrt{a}\sqrt{b}$$ breaks down when $a < 0$ and $b < 0$. $\endgroup$ – Arnie Bebita-Dris May 23 '17 at 11:14
  • 1
    $\begingroup$ Let $x=-2$. Then $\sqrt{x^2-1}=\sqrt{3}$, but $\sqrt{x+1}\sqrt{x-1}=\sqrt{-1}\sqrt{-3}=-\sqrt{3}$. $\endgroup$ – Arastas May 23 '17 at 11:17
  • $\begingroup$ Damn! You guys are right...how could I look past such a trivial thing... I always get a pathetic "ofcourse"-moment once you guys point it out for me. Thank you very much! $\endgroup$ – Parseval May 23 '17 at 11:18

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.

  • $\begingroup$ Wouldn't "the correct statement" be $\sqrt{ab}=\sqrt{|a|}\sqrt{|b|}$? $\endgroup$ – Michael Hoppe May 23 '17 at 18:42
  • $\begingroup$ @MichaelHoppe: I'd prefer to think of your equation as a correct statement (for all real $a$ and $b$). The proximate point is, I omitted the absolute values following OP's reasoning, and because the motivating question doesn't have any. :) $\endgroup$ – Andrew D. Hwang May 23 '17 at 19:09
  • $\begingroup$ But that what it's all about: the equation is true iff $x^2-1\geq0$ and $x+1$ and $x-1$ are both non-negative. $\endgroup$ – Michael Hoppe May 23 '17 at 19:14
  • $\begingroup$ @MichaelHoppe: Yes, but since the arguments are non-negative, there's no need for explicit absolute value signs. (In other words, it's true that $\sqrt{|x^{2} - 1|} = \sqrt{|x + 1|} \sqrt{|x - 1|}$ for all real $x$, but that's not relevant to OP's question.) $\endgroup$ – Andrew D. Hwang May 23 '17 at 19:53

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