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

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

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?

Answer 1

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).

Answer 2

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

Answer 3

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

1. 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.

2. 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!
   $$\forall\theta{\in}\mathbb R\,\Big(\quad\forall n{\in}\mathbb Z\quad \big(\sin\theta=0 \implies \theta=\pi n\big)\quad\Big)$$

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

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

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

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

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

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

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

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

Syntactically very wrong.

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

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.