How to notate the universal quantifier when applied to an equation?

Question

So I have an equation with infinite regular solutions. Let's say this equation is $$\sin^{-1}(0)=\pi n$$ where $n$ is any integer. How do I express this formally using the universal quantifier?

Do I say

$\forall n\in\mathbb Z: \sin^{-1}(0)=\pi n$
$\sin^{-1}(0)=\forall n\in\mathbb Z:\pi n$
$\sin^{-1}(0)=\pi n, \forall n\in\mathbb Z$

Or is there a more aesthetic way to express this?

In other words, by convention, where am I supposed to put the universal quantifier when an equality sign is involved?

Btw just as clarification, when we do inverse trigonometry, the input and output domains have a bijective relationship, meaning that $$\sin^{-1}(0) = 0$$ If you want to consider other angles, you can instead write $$\sin(\pi n) = 0$$ — Prometheus, Nov 23 '21 at 02:14
I would write $\sin^{-1}(0)={n\pi\mid n\in \mathbb Z}$. Translated to English, "The inverse image of $0$ with respect to sine is the set of number of the form $n\pi$ as $n$ ranges over the integers." — Mike Earnest, Nov 23 '21 at 02:20
$\sin^{-1},$ as written, is a function, so it only has one value for any argument. — Thomas Andrews, Nov 23 '21 at 02:23
But you can write:$$\forall x\in\mathbb R\left(\sin(x)=0\iff\left(\exists n\in\mathbb Z(x=n\pi)\right)\right)$$ — Thomas Andrews, Nov 23 '21 at 02:27
@ThomasAndrews $f(x)=\sqrt{x}$ is a function, but it has two values for all but one argument. I have never heard of a definition for a function in mathematics that disallows multiple solutions, unless we are working under type theory or something like that. — Vlad, Nov 23 '21 at 03:30
@Prometheus Interesting note about the bijective nature of the inverse trig functions, I wasn't aware of that. Is there some kind of elusive algebraic reason that we don't assign countably infinite solutions to these functions? — Vlad, Nov 23 '21 at 03:32
A function is literally defined as something that takes any input to give a single output. If it gives out two outputs, it's not even a function anymore. — Prometheus, Nov 23 '21 at 03:35
@Prometheus The definition of function that I've always heard is an assignment from a domain to a codomain. I.e. an assignment of every element of a set X to any element in a set Y. I have never, ever, heard that an element of the domain couldn't be assigned to multiple elements of the codomain, since that's precisely what $n$th roots do. This is what I've always been taught. Why would it even be useful to restrict functions to only have one output to begin with? — Vlad, Nov 23 '21 at 03:51
What you have described is just a general mapping. A function is a special type of mapping. — Prometheus, Nov 23 '21 at 03:55
@Vlad Restricting functions to one output is useful because the notation $f(x)$ does not really work otherwise. In your example, $\sin^{-1}(0) = 0$ and $\sin^{-1}(0) = \pi$, so can I conclude that $0 = \pi$? In general, mathematicians really want equality to work like that. That is why, for real numbers, $\sqrt{a}$ only refers to the positive solution of $x^2 = a$. For complex numbers, there is no natural choice for a distinguished root anymore and people do talk about “multivalued functions” but they generally require much more care when working with them. [...] — Eike Schulte, Nov 23 '21 at 09:17
[...] In this context, there are even tools (like Riemann surfaces) that are specifically designed to allow you to avoid talking about multivalued functions (at the expense of introducing new objects). For general pre-images, no such tools exist, so we usually write them as sets as in Mike Earnest’s comment, i.e. $\sin^{-1}(0) = { n \pi | n \in \mathbb Z }$. Only if the pre-image is unique (which for $\sin$ it isn’t) do we sometimes use $f^{-1}(x) = y$ and even then it’s abuse of notation unless $f^{-1}$ denotes the inverse function (not multivalued function!) instead of the pre-image. — Eike Schulte, Nov 23 '21 at 09:25
@EikeSchulte Thank you for your input. I have just now discovered that there is a special name for the kind of function that I erroneously thought all functions were. A great example of such a function is the Lambert W. — Vlad, Nov 24 '21 at 05:09
@Vlad the problem is the use of $=$ with multivalued functions. For example, if you take the multivalued functions $\sqrt{\cdot}$ and you write $1=\sqrt{1}$ and $-1=\sqrt{1}$ you might conclude $-1=1.$ So, while sometimes we do use $=$ this way, it is an abuse of notation and technically wrong, and you should really say $-1\in\sqrt1.$ — Thomas Andrews, Nov 24 '21 at 05:40
@ThomasAndrews There I absolutely agree, and the only way to get out of such a paradox is by treating all numbers as sets of themselves, but this becomes cumbersome to notate explicitly for all equations. (For example $\sqrt{{1}}={1,-1}$.) Although, the example you chose is a bit unlucky since in this case there is a very natural and intuitive notation, i.e. $\sqrt{1}=\pm 1$. — Vlad, Nov 24 '21 at 05:54
But $a=\pm1$ is also an abuse of notation, and for the same reasons. @vlad — Thomas Andrews, Nov 24 '21 at 06:39

score 2 · Accepted Answer · answered Nov 23 '21 at 09:11

Using natural language (like you did in the first paragraph of your question) is a very common, and in many cases, the preferred way of writing a quantification.

But there are occasions where using symbols is preferable. Formal definitions of formulas usually dictate that the quantifier must come first, i.e. $$ \forall x \in X: f(x) = g(x). $$ If you work with formulas in logic or model theory, this is the only way.

Outside of this context, it can be more pleasant to put the formula up-front, and most people would find $$ f(x) = g(x) \qquad \forall x \in X $$ acceptable as well. It is generally a bad idea to mix quantifiers before and after the formula, though, because there is no convention in which order they apply (and $\forall$ and $\exists$ cannot be interchanged in general).

Your second option really makes no sense at all because it looks like $\sin^{-1}(0)$ is somehow equal to the “statement” $\forall n \in \mathbb Z : \pi n$ (which isn’t even a statement because $\pi n$ isn’t).

score 1 · Answer 2 · answered Nov 23 '21 at 09:52

1

$$\sin(x)=0\iff\exists\ n\in\mathbb Z:x=n\pi.$$

answered Nov 23 '21 at 09:52

ryang · Answer 3 · 2021-11-25T02:54:51.907

$$\sin^{-1}(0)=\pi n$$ where $n$ is any integer

If you accept that $\sin^{-1}(0)$ is just the single value $0$ (as pointed out in the comments), then technically $$\forall n{\in}\mathbb Z\:\:\sin^{-1}(0)=\pi n$$ is false, but $$\exists n{\in}\mathbb Z\:\:\sin^{-1}(0)=\pi n$$ is true.
Both these statements (which I think you, in trying to convey that every integer multiple of $\pi$ is a solution of $\sin\theta,$ actually meant) are false!
$$\forall\theta{\in}\mathbb R\,\Big(\quad\forall n{\in}\mathbb Z\quad \big(\sin\theta=0 \implies \theta=\pi n\big)\quad\Big)$$

$$\forall\theta{\in}\mathbb R\,\Big(\quad\sin\theta=0 \implies \big(\forall n{\in}\mathbb Z\quad\theta=\pi n\big)\quad\Big)$$
These statements are all true (the last one vacuously true): $$\forall\theta{\in}\mathbb R\,\Big(\quad\exists n{\in}\mathbb Z\quad \big(\sin\theta=0 \iff \theta=\pi n\big)\quad\Big)$$

$$\forall\theta{\in}\mathbb R\,\Big(\quad\sin\theta=0 \iff \big(\exists n{\in}\mathbb Z\quad\theta=\pi n\big)\quad\Big)$$

$$\forall\theta{\in}\mathbb R\,\Big(\quad\forall n{\in}\mathbb Z\quad \big(\theta=\pi n \implies \sin\theta=0\big)\quad\Big)$$

$$\forall\theta{\in}\mathbb R\,\Big(\quad\big(\forall n{\in}\mathbb Z\quad\theta=\pi n\big) \implies \sin\theta=0\quad\Big)$$

$\forall n\in\mathbb Z: \sin^{-1}(0)=\pi n$

The colon (“such that”) is, at best, unnecessary.

$\sin^{-1}(0)=\forall n\in\mathbb Z:\pi n$

Syntactically very wrong.

$\sin^{-1}(0)=\pi n, \forall n\in\mathbb Z$

Syntactically wrong because in symbolic logic, quantifiers are always in front of the formula. Even in natural language (e.g., English), when there are multiple quantifiers, placing one or all of them behind the formula is called hanging the quantifier; to prevent scope ambiguities, avoid this practice.

How to notate the universal quantifier when applied to an equation?

3 Answers3