The Absolute Value can be defined in many ways, but these are the two most common :

1. As a Piecewise Function

$$ |x|= \begin{cases} -x&\text{if } x < 0\\ x&\text{if } x\geq 0 \end{cases} $$

2. As The Principle Square Root of a Square

$$|x| = \sqrt{x^2}$$

In the second definition that I've included here, what stops us from doing the following, and reaching a contradiction?

$$|x| = (x^{2})^{\frac{1}{2}} = x \ \ \ \ \ \ \ \ \ \ \ \text{Contradiction}$$

Likewise, if I have $f(x) = \ln(|x|)$, what is the reason why the following contradiction can't be reached :

$$f(x) = \ln(|x|)$$ $$\implies f(x)=\ln(\sqrt{x^2})$$ $$\implies f(x) = \ln[(x^2)^{\frac{1}{2}}]$$ $$\implies f(x) = \frac{1}{2}\ln(x^2)$$ $$\implies f(x) = \frac{1}{2} \cdot 2 \ \ln(x)$$ $$\implies f(x) = \ln(x) \ \ \ \ \ \ \ \ \ \ \ \text{Contradiction}$$

  • $\begingroup$ When you are using different definitions of a mathematical function it's not difficult to arrive at a contradiction. If you really think the square root does what you've written down in the third equation, then $|.|$ in the second equation is not correctly defined. Fortunately, it's the other way round. $\endgroup$ – Thomas May 15 '16 at 14:46
  • $\begingroup$ What stops us from doing $$\lvert x\rvert =(x^2)^{1/2} \text{ [whatever it means]} = x\quad ?$$ Well, assuming $a^{1/2}$ is defined as $\sqrt a$, the fact that $\sqrt{x^2}\ne x$ when $x<0$. $\endgroup$ – user228113 May 15 '16 at 15:01
  • $\begingroup$ A better definition could be $|x|=D$, where $D$ is the distance from $0$, allowing us to deal with complex numbers better without the contradictions you state. $\endgroup$ – Simply Beautiful Art Jun 1 '16 at 12:14

$\sqrt{x^2} = (x^2)^{\frac{1}{2}}=x$, only when $x>0$, otherwise it equals $-x$.

This is because for any $x>0$, $|x|=x$ and for any $x<0$, $ |x|=-x$

So, no contradiction is reached.

  • $\begingroup$ I didn't downvote, but would be better if you finish saying otherwise it equals $-x$. $\endgroup$ – Integral May 15 '16 at 15:05

Because $(x^2)^{1/2}$ is not in general equal to $x$. The squaring operation lost the information about the sign of $x$.

With the log example, it's because you need to pick a branch.

  • $\begingroup$ this comment has to do with the linked answer you provided. In it, it says $(a^{b})^{n} \neq a^{bn}$ in Complex Analysis (i.e. when working with complex variables), however are there any special cases in Real Analysis, where it does break down, I'm guessing this would be one of them? $\endgroup$ – Perturbative May 15 '16 at 18:17
  • $\begingroup$ Yes, this is one of them. When everything is positive, you're fine. When $a$ is negative, then you might have problems. $\endgroup$ – Patrick Stevens May 15 '16 at 19:35

What stops us from saying $|x|= (x^2)^{1/2}= x$ is that the second equality simply is NOT true! $a^{1/2}$ is defined as "the positive number $x$ such that $x^2= a$". In particular, if $x= -2$, then $x^2= (-2)^2= 4$ and then $((-2)^2)^{1/2}= 4^{1/2}= 2$, not $-2$.


First off, I want to thank everyone for their answers here, but I found this to be the most general answer after thinking about this question for a while.

$$(a^{n})^{\frac{1}{n}} = a \ \ ,\ \text{if} \ \ n \ \ \text{is odd}$$ $$(a^{n})^{\frac{1}{n}} = |a|\ \ ,\ \text{if} \ \ n \ \ \text{is even}$$

Stated a bit more formally :

$$(a^{2n+1})^{\frac{1}{2n+1}} = a \ \ ,\ \forall n\in\mathbb{Z}$$ $$(a^{2n})^{\frac{1}{2n}} = |a|\ \ ,\ \forall n\in\mathbb{Z}$$

Parts of this answer and the intuition developed behind it have come from Paul's Online Notes, so full credit must go to that site.


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