The numerator is structured like: $$(-b)\pm\sqrt{b^2- 4ac}.$$

Is it confusing or acceptable to distinguish between the following two things?

  1. An idiom; and
  2. What is or seems to be a compositionally transparent and faithful representation of meaning. Observe that if we replace expressions (a,b) with either "open_interval_(a,b)" or "ordered_pair_(a,b)", then we have taken a step towards disambiguation. However, that is technically not a matter of direct disambiguation of individual morphograph-like vocabulary items, unless we actually introduce four symbols: open_interval_left_bracket, open_interval_right_bracket, ordered_pair_left_bracket, ordered_pair_right_bracket.

Should "plus or minus" be introduced as an operation, or as an informal notation that is a memory aid, with set theory not being involved?

Suppose that we consider an operation on class variables in set theory: could we get something like an unordered pair of proper classes that isn't merely the empty set, or does this get into an area of taboo, like ancient taboos associated with zero, or more recent taboos associated with infinity or infinitesimals?


2 Answers 2


When solving a quadratic equation,

$$ax^2+ bx+ c = 0$$

we use shorthand for the two solutions, to include both $$x_1 = \frac {-b +\sqrt{b^2-4ac}}{2a}$$

and $$x_2 = \frac{-b - \sqrt{b^2 -4ac}}{2a}$$

Hence, the shorthand, $$x_i = \frac{-b\pm \sqrt{b^2-4ac}}{2a}.$$

I.e., the solutions to $ax^2+bx+c = 0$ are given by $$x\in \left\{ \frac {-b +\sqrt{b^2-4ac}}{2a},\frac{-b - \sqrt{b^2 -4ac}}{2a}\right\}.$$

So $\pm$ is shorthand, and is not itself an arithmetic operation. So set theory is not involved.

  • $\begingroup$ Note, that when $b^2 - 4ac = 0$, then we have $x_1=x_2 = \dfrac{-b}{2a}$, e.g. in the case of $x^2+4x +4 = 0$, in which case $x_1=x_2 = -2$. $\endgroup$
    – amWhy
    Commented Nov 26, 2019 at 21:39

Here's why you should take great care when considering $\pm$ as an operator. It's not unusual to see a sentence of the form

We deduce that $A=\pm B$ and hence that $C=D\pm E$.

This isn't simply saying that both ($A$ is either $B$ or $-B$) and (either $C=D-E$ or $C=D+E$). When two $\pm$'s appear in the same sentence it is implied that they are both to be read together, in this case as either ($A=B$ and $C=D+E$) or ($A=-B$ and $C=D-E$).

So it makes more sense here to read $\pm$ as shorthand.

  • 4
    $\begingroup$ This is why LaTeX has both $\pm$ and $\mp$, in case the sign flips. $\endgroup$
    – Aeryk
    Commented Nov 27, 2019 at 0:25
  • 2
    $\begingroup$ When I want to write hopefully unambiguous statements involving $\pm$ on both sides of an equality/inequality, I've sometimes put in parentheses afterwards phrases like "2 possibilities" and "4 possibilities", depending on what is intended. For example, $\pm |y| = 3 \pm |x|$ (2 possibilities) vs. $\pm |y| = 3 \pm |x|$ (4 possibilities). By the way, a nice precalculus-level student exercise is to graph "by hand" all 4 possibilities. $\endgroup$ Commented Nov 27, 2019 at 9:06
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    $\begingroup$ While taking a shower just now it occurred to me that I've also used this parenthetical device in writing certain algebraic expressions, although I can't find an explicit example online right now where I've done this. But here's an example in which this could be done: To rationalize a sum of three square roots, you make use of a product of factors of the form $(a \pm b \pm c)$ (4 factors), whose product $(a^2 - b^2)^2 - 2c^2(a^2 + b^2) + c^4$ only involves even powers of $a,$ $b,$ $c$ (see here for more about this particular topic). $\endgroup$ Commented Nov 27, 2019 at 10:10
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    $\begingroup$ I rather think TeX has both ± and ∓ because it's been a standard notation for hundreds of years. Girard used ± in 1626. Experimentation for opposing pairs of signs went for a long time. Newton used $\perp$ and $\top$. Da Cunha introduced ∓ in 1790. See Cajori, Math. Notations. $\endgroup$
    – user1815
    Commented Nov 28, 2019 at 16:31

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