0

How to rewrite this problem?

I have a problem. "Let $x,y$ $\in$ $5\mathbb{Z}$. prove that $x+y$ $\in$ $5\mathbb{Z}$ ."

I did try to write this problem in the form of "something is an element of (set builder notation)" like this.

prove that $x+y$ $\in$ {$5\mathbb{Z} | $x,y$ \in 5\mathbb{Z}$}

Is it true?

Mathskid
  • 87

2 Answers2

1

Suppose that $x\in 5\Bbb Z$ and that $y\in 5\Bbb Z$.

This means by definition that there exists some integer $k\in \Bbb Z$ such that $x = 5k$ and some $\ell \in \Bbb Z$ such that $y = 5\ell$.

Then $x+y$ can be written as... which is... and so...

Then $x+y$ can be written as $x+y=5k + 5\ell = 5(k+\ell)$ which is $5$ times an integer and so $x+y$ satisfies the definition of being an element of $5\Bbb Z$.

As for "rewriting the problem", the original $x,y\in 5\Bbb Z$ implies $x+y\in 5\Bbb Z$ is clear enough. You could also have opted to write this as $5\Bbb Z + 5\Bbb Z \subseteq 5\Bbb Z$ or similar where here we take "$A+B$" to mean $\{a+b~:~a\in A,b\in B\}$

JMoravitz
  • 79,518
1

You can define $E=\{(x,y)\in 5\mathbb Z\times 5\mathbb Z\mid x+y\in5\mathbb Z\}$

The question resumes to show that $E$ is the whole space, i.e. " does $E=5\mathbb Z\times 5\mathbb Z$ ? " since this happen to be always true.

Conversely with $E^\complement=\{(x,y)\in 5\mathbb Z\times 5\mathbb Z\mid x+y\notin5\mathbb Z\}$ this is equivalent to $E^\complement=\varnothing$.

An alternative writing could be:

  • let $x\in 5\mathbb Z+5\mathbb Z$, show that $x\in 5\mathbb Z$
  • or show $5\mathbb Z+5\mathbb Z\subset 5\mathbb Z$ in pure set notation

And you can rewrite the set (in the set builder notation) $5\mathbb Z+5\mathbb Z=\{a+b\mid a\in 5\mathbb Z,\ b\in 5\mathbb Z\}$

zwim
  • 28,563