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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 $y = 3$?

metamorphy
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    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. – Robert Shore Oct 02 '20 at 03:42
  • "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. – fleablood Oct 02 '20 at 04:20
  • "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* $y$" – fleablood Oct 02 '20 at 04:23

4 Answers4

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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 $y$.
J. W. Tanner
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Siong Thye Goh
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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.

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

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

GraduateStudent
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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 $x$.

fleablood
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