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A very simple example is the following; suppose I wish to factor out $\pm$ from $\mp 7$, then by my logic $$\mp 7= \color{blue}{\pm \mp} (\mp 7)=\begin{cases}+(-(-7)) & =+7\\ -(+(+7)) & =-7\end{cases}=\pm 7\ne \mp 7\tag{*}$$

So something has gone wrong, as equality is not satisfied in $(*)$. The reason why I wrote $\color{blue}{\pm \mp}$ is because if I factor out a $\pm$ then to compensate I must include a $\mp$ sign (at least I thought).

Just like if I took a factor of $x$ out of $$1+x=x\left(\frac{1}{x}+1\right)$$ so that $$x\times \frac{1}{x}=1$$

But, this same logic does not seem to apply to $(*)$.

Clearly, I am missing something very simple, but right now I can't understand what I'm doing wrong. Could someone please explain how to factor out $\pm$ while maintaining equality?

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    $\begingroup$ "If I factor out a $\pm$ then to compensate I must include a $\mp$" ... Consider this: If you factor-out a $-$, then you must compensate with another $-$; likewise, if you factor-out a $+$, then you must compensate with another $+$. The signs match. So to compensate for factoring-out $\pm$, you use another $\pm$. $\endgroup$ – Blue Sep 29 '18 at 13:59
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    $\begingroup$ We simply have $$\mp 7= \mp1\cdot(+7)= \pm1\cdot(-7)$$ $\endgroup$ – gimusi Sep 29 '18 at 14:00
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    $\begingroup$ @BLAZE: your breakdown of the signs is correct, demonstrating that your initial equation is not. $\endgroup$ – abiessu Sep 29 '18 at 14:06
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    $\begingroup$ @BLAZE: You should have $\mp 7 = \pm\pm(\mp 7) $, since $\pm\pm = +$. $\endgroup$ – Blue Sep 29 '18 at 14:08
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    $\begingroup$ I am not sure that $\pm$ always has these connotations. If it is a signal that the sign is ambiguous and either can be chosen, then the expression in which it occurs is in fact two expressions - one for each choice of sign. If there are a number of ambiguous signs $\mp$ can be used to show that the top signs are to be taken consistently throughout, or alternatively the bottom signs. But in an expression like $\pm \sqrt {49}\pm \sqrt {49}$ there are four choices and three values. So how you deal with the symbol depends on he context in which it is used. $\endgroup$ – Mark Bennet Sep 29 '18 at 14:12
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Note that

$$\color{blue}{\pm1\cdot \mp1} =-1 \implies \color{blue}{\pm \mp} (\pm 7)=\mp7$$

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  • $\begingroup$ Thank you for your answer, please take another look at $(*)$ and can you tell me at what point I am making a mistake? $\endgroup$ – BLAZE Sep 29 '18 at 14:01
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    $\begingroup$ @BLAZE First step of the first line $$\mp 7\neq \color{blue}{\pm \mp} (\mp 7) $$ since as noticed $\color{blue}{\pm1\cdot \mp1} =-1$. $\endgroup$ – gimusi Sep 29 '18 at 14:02
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    $\begingroup$ @BLAZE: the thing to note in this answer is that $\pm\mp x=-x$, therefore what you have written in $(*)$ is $x=-x$. $\endgroup$ – abiessu Sep 29 '18 at 14:03
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    $\begingroup$ @gimusi So to summarize, factoring out a $\pm$ from $\mp 7$ just changes the sign of $7$, in other words; $\mp(7)= -\pm(7)$. Is this the correct? $\endgroup$ – BLAZE Sep 29 '18 at 14:22
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    $\begingroup$ @BLAZE Yes the last one is correct if we mean $$\mp7=-1 \cdot \pm1\cdot(+7)$$ $\endgroup$ – gimusi Sep 29 '18 at 14:26
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In a ring you have the following rules: $a\cdot 0 = 0 = 0\cdot a$, $-(-a)=a$, $(-a)b = a(-b)=-(ab)$, and $(-a)(-b)=ab$, where $-a$ is the additive inverse of $a$. From here, everything clarifies.

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