How do you prove a statement in the form of "for all integers $x$, there is some integer $y$, such that $3 \mid x + y$". I believe the statement is true. I know you start by assuming $x$ is an integer, and you pick a $y$. Let's say $y = 3$. And next you need to prove that $3 \mid x + y$ in order to prove the statement is true.
I think the next step is to say $3k = x + y$ where $k$ is an integer. I'm just not sure where to go from here. My initial thought was that $3k = x + y$ where $k$ is an integer proves that $3$ divides $x + y$, but then what is the point of picking a value for $y$? And if I'm wrong here, how can I prove that $3 \mid x + y$ based on my assumption that $x$ is an integer and $y = 3$?
 A: Let $y$ be dependent on $x$. $y$ can't be independent of $x$, this can be seen by trying a few values of $x$.
Given $x\in \mathbb{Z}$, let $y=3-x \in \mathbb{Z}$, then we have $x+y=3$. This would answer your question.
However, I would encourage you to practice more:

*

*Try to think of a different choice of $y$ as an exercise.

*Also try to think given an $x$, find all the possible choice of $y$.

A: Logically, the statement that you’ve given as an example means that if you pick any integer $x$ - you can always find some another integer $y$ such that their sum $x + y$ is divisible by $3$.
Than, I guess you interpret it in a slightly different way: you say that suppose $x$ is (some) integer and we pick a $y$.
The point of the statement though is different: it says that we fix $x$ first - it’s arbitrary in the sense that we can give it any value before fixing, but after that - $x$ is defined for remaining part of the statement - and it says that we can pick some $y$ for that $x$, such that given condition is satisfied.
As far as practice, I think any beginner calculus book( What are the recommended textbooks for introductory calculus? )will give you some practice of understanding such things, since you have to understand quantifiers($\forall$, $\exists$, etc) in order to figure out limits, on which plenty of basic concepts are based. But you can read a book on geometry or algebra too.
A: If $x \in \mathbb{Z}$, then choose $y=2x \in \mathbb{Z}$. This gives $x+y=x+2x=3x$.
Since $3 | 3x$ we conclude $3|x+y$
A: So we know $x$ is an integer.  And we want $3|x+y$.  So we want $x+y = 3k$ for some $y$ and $k$
$y$ is the one we want to find.  $x$ is the one we have that we are committed to and can't change.  And $k$ is what we can manipulate any way we want to find the necessary $y$.
So we want $y = 3k-x$ which.... is all we need.  For any $x$ just let $k = 1$ and $y = 3 - x$.
That's it $x+y = x+(3-x) = 3$ and $3|x+y$.
That will work for every $x$.
