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Q&A

# The meaning of $\pm$

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Consider the claim $|x| = \pm x$.

I would interpret it as stating that $|x| = x$ and $|x| = -x$, thereby implying that $x = 0$.

A user at Matheducators stackexchange interprets it as saying that $|x| = x$ or $|x| = -x$, which holds for all real numbers.

I would typically use the notation to index solutions to an equation or some numbers I am going through; the roots of a second order polynomial being an elementary case, but not the only one. I might also use it as a shorthand if I need to calculate something for two things of opposite sign and figured I could both calculations at once.

It is pretty clear that my intuitions are in conflict here; in one case I consider $\pm$ to mean and, while in other, or. Is this just a mistake or is there something deeper going on here?

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I think I would describe that as ‘the two claims $|x| = \pm x$’, not a single claim. It's comparable to saying ‘the two roots of $x^2 - 4$ are $x = \pm2$’ and not ‘the root of $x^2 - 4$ is $x = \pm2$’.

Whether those two claims are meant to be and-ed or or-ed would depend on context. In a vacuum, I don't think I would be able to unambiguously assign meaning to ‘the claim $|x| = \pm x$’.

More broadly, I think I've seen only two general conventions for $\pm$-containing equations, and they both seem fairly rigorous to me if slightly informal. The first is that an expression or equation containing one or more ‘$\pm$’ and optionally ‘$\mp$’ signs is always shorthand for two of the same: one in which each $\pm$ and $\mp$ is replaced by its upper component, and one in which the lower components are used instead. The second is that an expression or equation using $n$ ‘$\pm$’ signs is a shorthand for $2^n$ of the same in which every combination is represented. I generally expect an author to clarify which convention is in use if more than one ‘$\pm$’ is present, and as above I also expect these conventions to be used in a context in which multiple equations/expressions are expected.

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I think I can explain what's going on by interpreting things more formally, though it could be gotten at informally as well; it's just clearer to see formally.

One way to interpret "$y = \pm x$" (and other informal notations like $y = 1,2,3$) is as $y \in \{x, -x\}$ ($y \in \{1,2,3\}$). This is logically equivalent to $y = x \lor y = -x$. So that's one direct answer.

However, because $(A \lor B) \implies C$ is equivalent to $(A \implies C) \land (B \implies C)$ (and $\forall x \in S. P(x)$ is equivalent to $\forall x. x \in S \implies P(x)$), the text might have a more conjunctive rendition. To be clear, we'd still interpret "$y = \pm x$" as $y = x \lor y = -x$, but a statement like "$y^2 = 1$ for $y = \pm 1$" could mean $\forall y. y = \pm 1 \implies y^2 = 1$ and could be rendered as "$y^2 = 1$ for $y = 1$ and $y = -1$".

Of course, for roots of polynomials we often take something like "$y^2 = 1$ for $y = \pm 1$" to mean $y^2 = 1 \iff y\in\{1, -1\}$. In this case, if we take the left-to-right implication, i.e. using the quadratic equation to constrain the values of $y$, then clearly viewing $y=\pm 1$ as "$y = 1$ or $y = -1$" is natural. Going the other way, i.e. showing that these values of $y$ satisfy the quadratic equation leads to the situation of the previous paragraph.

I don't think I've ever seen anyone write "$y = \pm x$" to mean $y = x \land y = -x$. In most contexts, this immediately implies $y=0$ so you may as well write that. If the point is to prove $y=0$ by showing $y = x \land y = -x$, it would be clearer to write something like $x = y = -x$.

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