# How to correctly factor out a $\pm$ from a number that has a $\mp$ in front?

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?

• "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 $\pm$. – Blue Sep 29 '18 at 13:59
• We simply have $$\mp 7= \mp1\cdot(+7)= \pm1\cdot(-7)$$ – gimusi Sep 29 '18 at 14:00
• @BLAZE: your breakdown of the signs is correct, demonstrating that your initial equation is not. – abiessu Sep 29 '18 at 14:06
• @BLAZE: You should have $\mp 7 = \pm\pm(\mp 7)$, since $\pm\pm = +$. – Blue Sep 29 '18 at 14:08
• 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. – Mark Bennet Sep 29 '18 at 14:12

$$\color{blue}{\pm1\cdot \mp1} =-1 \implies \color{blue}{\pm \mp} (\pm 7)=\mp7$$
• Thank you for your answer, please take another look at $(*)$ and can you tell me at what point I am making a mistake? – BLAZE Sep 29 '18 at 14:01
• @BLAZE First step of the first line $$\mp 7\neq \color{blue}{\pm \mp} (\mp 7)$$ since as noticed $\color{blue}{\pm1\cdot \mp1} =-1$. – gimusi Sep 29 '18 at 14:02
• @BLAZE: the thing to note in this answer is that $\pm\mp x=-x$, therefore what you have written in $(*)$ is $x=-x$. – abiessu Sep 29 '18 at 14:03
• @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? – BLAZE Sep 29 '18 at 14:22
• @BLAZE Yes the last one is correct if we mean $$\mp7=-1 \cdot \pm1\cdot(+7)$$ – gimusi Sep 29 '18 at 14:26
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.