# 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 $$y = 3$$?

• It's better to think about the statement as, "For each integer $x$ there is some integer $y$ such that $3 \mid (x+y)$." This (correctly) emphasizes that your choice of $y$ depends on what value of $x$ you're starting with. As you've written it, you can confuse yourself into thinking there's supposed to be some $y$ that works no matter what $x$ you choose, but that's not what's meant and it's not true. Oct 2 '20 at 3:42
• "I know you start by assuming x is an integer, and you pick a y" NO! The question isn't "for every $x$ then any $y$ will be $3|x+y$. The question is for every $x$ than there is some $y$. You can't just pick it. you must find it. Now $x$ is what you are given. $y$ is the unknown you must solve for. You know $x + y = 3k$ for some $k$. And to solve for $y$ we must have $y = 3k - x$ for some $k$. So we can pick the $k$ (but not the $y$) to get $y = 3-x$ for example. ... for any $x$ then there exists $y = 3-x$ so that $3|x+y$ (as $x+y=x+(3-y)=3$ and $3|3$. Done. Oct 2 '20 at 4:20
• "And if I'm wrong here, how can I prove that 3∣x+y based on my assumption that x is an integer and y=3?" You can't if it's false. ANd if $x = 2$ and $y=3$ then $3\not \mid x+y$ so that statement isn't true at all. I think you are confuse "for some $y$" with for any $y$" Oct 2 '20 at 4:23

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 $$y$$.

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.

If $$x \in \mathbb{Z}$$, then choose $$y=2x \in \mathbb{Z}$$. This gives $$x+y=x+2x=3x$$.

Since $$3 | 3x$$ we conclude $$3|x+y$$

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 $$x$$.