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I need help with algebra of exponentiation.

$$\begin{align*} \sqrt{x^2} &= (x^2)^{1/2} &\qquad &\text{(since }\sqrt[x]{y}=y^{1/x}\text{)}\\ &= x^{2(1/2)} &&\text{(since }(x^y)^z = x^{yz}\text{)}\\ &= x^{(1/2)2} &&\text{(since }xy=yx\text{)}\\ &= (x^{1/2})^2 &&\text{(since }(x^y)^z = x^{yz}\text{)}\\ &= \left(\sqrt{x}\right)^2 &&\text{(since }\sqrt[x]{y} = y^{1/x}\text{)} \end{align*}$$


$\sqrt{x^2}$ is a real number.

$(\sqrt{x})^2$ is a real number, if $x\geq 0$.

$\sqrt{x^2} = (\sqrt{x})^2$.

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A lengthy calculation! Can you try to frame an explicit question? It would help if you used more words. –  André Nicolas Jul 14 '11 at 22:11
3  
This is an utter mess. The first line makes absolutely no sense. The equal sign means "these two things are equal", it does not mean "and here's something completely different but somewhat related in my head to what came before". $\sqrt{x^2}$ is not equal to $\sqrt[x]{y}$. $(x^2)^{1/2}$ is not equal to $(x^y)^z$. $x^{2\times(1/2)}$ is not equal to $xy=yx$. In short, this is a "stream of consciousness" list of incorrect equalities. Rule 1: An equal sign goes only between two things that are equal. –  Arturo Magidin Jul 14 '11 at 22:14
    
@Arturo: This is really unfair. The original text made sense. (at least to me) –  t.b. Jul 14 '11 at 22:19
    
@Theo: The original text has the exact same incorrect equalities: the first line says "root 2(x^2)", the second line asserts this is equal to "root x y " and to "y^{1/x}". Curly brackets do not indicate comments, they indicate sets; so this makes it even more nonsensical. It's still a "stream of consciousness" list of incorrect equalities. –  Arturo Magidin Jul 14 '11 at 22:22
    
@Arturo: If you look closely enough, Fox is changing the numbers to free variables, writing out the formulas, and then changing the variables back to numbers. Definitely quirky and amateurish the way it's written, but the motivating idea is sound algebra in the end. I think we all agree OP needs to learn how to do it the standard way so that he or she is more efficient and better able to communicate his or her thoughts. EDIT: Theo has apparently edited the question to make apparent Fox's modus here. –  anon Jul 14 '11 at 22:25

2 Answers 2

With real numbers, the exponent rule $(x^y)^z = x^{yz}$ only holds for nonnegative $x$; it does not hold for arbitrary $x$. So you cannot go from line 1 to line 2 (in the edited version) except if you know $x\geq 0$. If you don't know whether $x\geq 0$ or not, then the equality $(x^y)^z = x^{yz}$ need not hold.

For example: $\sqrt{(-1)^2}$ is not equal to $(-1)^{(1/2)2}$.

In short,
$$\large(x^y)^z = x^{yz}$$ is not universally true; for integer exponents it holds for any base, but for nonintegral exponents it need not hold with nonpositive bases. In particular, for $y=\frac{1}{2}$, it can only hold if $x\geq 0$.

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The equality $$ (x^2 )^{1/2} = x^{2(1/2)} $$ holds only for $x \geq 0$. If $x < 0$, then $$ (x^2 )^{1/2} = |x| > 0, $$ but $$ x^{2(1/2)} = x < 0. $$

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