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As the title states, is $\pm1=1$ a valid statement since $\pm1$ means $1$ or $-1$?

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    $\begingroup$ The notation $\pm$ is used when there are multiple solutions as a shorthand. It is not meant to be used as part of an equality, but rather a definition. So this is invalid, yes. $\endgroup$ Commented Nov 14, 2021 at 23:24
  • $\begingroup$ $1$ and $\pm 1$ are not interchangeable. They cannot be equal. $\endgroup$ Commented Nov 15, 2021 at 14:43
  • $\begingroup$ We cannot infer from $x=\pm 1$ that $x=1$. $\endgroup$ Commented Nov 15, 2021 at 14:57

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You should always view any mathematical proposition involving $\pm$ as a shorthand for one which does not involve $\pm$. For example, I might say that $x^2 = 1$ if and only if $x = \pm 1$. Here, it is clear that when I write $x = \pm 1$, I mean that either $x = 1$ or $x = -1$.

It is not obvious what the statement $1 = \pm 1$ means. I’d say the most obvious interpretation is that either $1 = 1$ or $1 = -1$, so I’d say it’s a true statement. But it depends on context, and you should never use $\pm$ unless it’s clear how one could rewrite the statement to avoid using $\pm$.

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Technically, $\pm 1$ is always an abuse of notation. We just always know what we “really mean.” I would never write $1=\pm1$ because it would be unclear what I was trying to say.

Writing has the intent of communication, and we use $\pm$ to simplify what we write for the reader, even if it is an abuse of notation. Never abuse notation to confuse.

Most of the time, when we write an expression like the one from the quadratic formula: $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a},$$ we mean that both possible values on the right side are valid solutions, not that one of them is.

The non-abuse of notation would be:

If $b,c\in\mathbb R$ then for $x\in \mathbb C,$ $x^2+2bx+c=0$ iff $x=-b-\sqrt{b^2-c}$ or $x=-b+\sqrt{b^2-c}.$

This is longer and less clear - the reader has to scan the two values to see the pattern.

The repetition can also cause errors by the writer, as well.

Ultimately, we accept the abuse of notation because it is clearer.

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No. Maybe you are thinking that $\pm1=1$ means $-1 = 1 \lor 1 = 1$, but that is not the case. In fact, I would say that $\pm1=1$ isn't even a meaningful statement, since $\pm1$ doesn't refer to a number.

What mathematicians sometimes do say is something like $x = \pm1$, which means that $x = -1 \lor x = 1$, but from that you cannot conclude $x = 1$

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No. $\pm 1$ is the set $\{-1,1\}$. $1$ is the number $1$. These are not equal since not both are a set and not both are a number.

If we happen to be working in a context where it is meaningful to have a number be a one-element set (for instance, we are finding a solution set and right now there is only one solution), then $1$ can be the same as the set $\{1\}$, but then these two sets are not equal.

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    $\begingroup$ So why do we write $x^2=4\implies x=\pm2$ instead of $x^2=4\implies x\in\pm2$? $\endgroup$ Commented Nov 14, 2021 at 23:37
  • $\begingroup$ @BarryCipra : Because $x$ is the solution set, not an element of that set. Also, because in elementary mathematics we have several otherwise precise notations bunged together to make "simple to write" statements. $\endgroup$ Commented Nov 14, 2021 at 23:39
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    $\begingroup$ Yeah, this doesn’t explain why we can write $x=\pm1.$ we wouldn’t write $x=\{+1,-1\}.$ $\pm1$ doesn’t have a formal meaning. It is always an abuse of notation. But it is a convenient abuse of notation. $\endgroup$ Commented Nov 15, 2021 at 0:08
  • $\begingroup$ "we wouldn’t write $x=\{+1,−1\}$" [citation needed]. I have seen that written by a variety of instructors/lecturers in a variety of settings. Frequently, the words being said are "... the solution set is one and minus one ...". $\endgroup$ Commented Nov 15, 2021 at 13:25
  • $\begingroup$ @ThomasAndrews : In fact the most common way to write the use of the quadratic equation is to write $x = \text{some set}$. Examples: p. 140 of College Algebra. And writing a set without the braces is common; we have examples on this site (e.g., here: "...for every $j=1,2,3$.", having the meaning "for every $j \in \{1,2,3\}$.) $\endgroup$ Commented Nov 15, 2021 at 19:53

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