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When we derive some formula and have to do huge algebraic expansions that deal with raising powers we use exponent rules mindlessly and we never write down the $\pm$ symbol. Why is this right?

My reasoning is that we write $\pm$ when were looking for a set of solutions (e.g., $x^2=4\implies x=\pm2$) but not when we know the exact identity of $x$. If we know the exact identity of a symbol we can use exponent rules. For example, $2=\sqrt{2^2}$ seems more valid than $2=\sqrt{4}$ (since in the former case we're clearly undoing an operation on the number 2, which could've been a symbol like $a$), if that makes sense at all. I know its a stupid thing to say, but well, I'd like to see what you think.

One problem with the $\pm$ notation is that we won't be able to do things like $a\sqrt{4}=\sqrt{a^24}$ where tricks like these are done all the time.

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  • $\begingroup$ Edited the question, forgot to include the squared symbol... $\endgroup$ – DLV Aug 21 '15 at 22:12
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    $\begingroup$ Whatever you do, don't forget that $a=\pm b$ isn't an equality, $\pm b$ isn't an object. The symbol $a=\pm b$ abbreviates the disjunction $a=b\lor a=-b$. $\endgroup$ – Git Gud Aug 21 '15 at 22:14
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    $\begingroup$ The square-root symbol $\sqrt x$ means "the nonnegative square root." Thus, $\sqrt9=\sqrt{3^2}=\sqrt{(-3)^2}=3$. Now, this means that $\sqrt{x^2}=|x|$, which is annoying. $\endgroup$ – Akiva Weinberger Aug 21 '15 at 22:14
  • $\begingroup$ @GitGud (I should mention for OP's sake that $\lor$ means "or") $\endgroup$ – Akiva Weinberger Aug 21 '15 at 22:15
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    $\begingroup$ The book I use for this material has something called the "square root property", which says that the equation $X^2 = k$ has two solutions, $X = \pm \sqrt{k}$. So, the $\pm$ comes from solving equations; $\sqrt{k}$ always means the non-negative root. $\endgroup$ – pjs36 Aug 21 '15 at 22:15
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Actually, never. $\sqrt{b}$, by definition, is the nonnegative solution to the equation $x^2 = b$.

This means that, by definition, $\sqrt{a^2} = |a|$ for all real values of $a$.

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  • $\begingroup$ Would my message have been different if I have said raise to the power 1/2? Or do you also assume the convention of the positive solution? $\endgroup$ – DLV Aug 21 '15 at 22:17
  • $\begingroup$ Since $\sqrt{a^2} = |a|$ for all real values of $a$. Just wondering, how about $(-a)$ as a solution? $\endgroup$ – NoChance Aug 21 '15 at 22:19
  • $\begingroup$ @David Since the definition of $x^{\frac12}$ is that $x^{\frac12} = \sqrt{x}$, the message would not change. $\endgroup$ – 5xum Aug 21 '15 at 22:19
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    $\begingroup$ @EmmadKareem $-a$ is a solution to the equation $x^2=a^2$. In fact, the equation has two solutions (if $a\neq 0$). The solutions are $a$ and $-a$. $\endgroup$ – 5xum Aug 21 '15 at 22:20
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    $\begingroup$ @David, who said these rules are valid outside positive reals? en.wikipedia.org/wiki/… $\endgroup$ – Ennar Aug 22 '15 at 0:40
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When you write $x^2=4\implies x=\pm2$, what exactly justifies that implication?

You start with two equal expressions: $x^2=4$.

You are allowed to apply the same function to both sides of the equation, and still have equality. So $\sqrt{x^2}=\sqrt{4}$.

Now here's where many students have been taught poorly. They've been taught that the $\sqrt{}$ "cancels" the ${}^2$, leaving $x$ on the left. And they've been taught that $\sqrt{4}=\pm2$. Both of these are incorrect. What really should be the next line is $$|x|=2$$ because $\sqrt{4}$ is unambiguously $2$, and $\sqrt{x^2}$ is, for all real $x$, $|x|$.

So now that you know $|x|=2$, what must $x$ be? How many real numbers are there with absolute value $2$? Just $2$ and $-2$. So $x$ could be either one of these. And we are left with the sloppy summary $x=\pm2$.

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