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I'm trying to show that this relation is an equivalence relation. $$xRy \longleftrightarrow 7|(2x+5y) \text{ for all } x,y\in Z$$ I need to show that $R$ is reflexive. If I take every $x$ and $y$ are the same I can see that it's fine, how to write it in formal way?

For the next conditions I need some advice.

  1. $(a,b) \in R \Rightarrow (b,a) \in R$, i.e., Symmetry
  2. $(a,b) \in R , (a,c) \in R \Rightarrow (a,c)\in R$, i.e. Transitivity

Thanks!

MarnixKlooster ReinstateMonica
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Ofir Attia
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  • I think you mean $R$ instead of $A$. There are parentheses missing in the definifion of $R$. – Git Gud Jul 31 '13 at 16:14
  • What is the universe, i.e. the set $\Omega$ such that $R \subseteq \Omega\times\Omega$? Is it $\mathbb{Z}$ or maybe $\mathbb{R}$ or yet something else? – dtldarek Jul 31 '13 at 16:21
  • Is your definition $xRy \longleftrightarrow (,,,7|(2x+5y) \rightarrow x,y\in Z,)$ or simply $xRy \longleftrightarrow 7|(2x+5y)$? I don't understand the '$\rightarrow x,y\in Z$' part. I'm assuming $R\subseteq \mathbb{Z}\times\mathbb{Z}$. – Alraxite Jul 31 '13 at 17:02
  • @Alraxite I've edited the question to replace the three different uses of $\rightarrow$ as I thought was intended. Apologies if I misinterpreted. – MarnixKlooster ReinstateMonica Jul 31 '13 at 19:18

2 Answers2

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For reflexivity, you need to show that $xRx$ for any $x\in\mathbb{Z}$. By your definition, $xRx$ if and only if $7$ divides $2x+5x=7x$; this is always true!

For the others, it really is just a matter of pushing through the definitions. For instance: for transitivity, suppose $xRy$ and $yRz$. We wish to show that $xRz$. By assumption, we must have $2x+5y=7k$ for some $k\in\mathbb{Z}$, and $2y+5z=7h$ for some $h\in\mathbb{Z}$. Then $$ 2x+5z=(7k-5y)+(7h-2y)=7k+7h-7y=7(k+h-y). $$ Since $k+h-y\in\mathbb{Z}$, you've shown that $7$ divides $2x+5z$, and hence $xRz$.

Try finishing up with symmetry similarly.

Nick Peterson
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Symmetry:

$$(x,y)\in R\iff 7\mid(2x+5y)$$

Now, we have that

$$2y+5x=7(x+y)-(2x+5y) \implies 7\mid (2y+5x)\implies (y,x)\in R$$

DonAntonio
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  • Can you explain a little more? I didnt understand it. – Ofir Attia Jul 31 '13 at 16:26
  • How I can write the equivalence classes ( It's the right expression? ) $[x]_{R} = {x \rightarrow 2|x}$? its right? – Ofir Attia Jul 31 '13 at 16:44
  • The proof above shows symmetry: $;(x,y)\in R\iff (y,x)\in R;$... – DonAntonio Jul 31 '13 at 17:59
  • how did you get to $7(x+y)-(2x+5y) \implies 7\mid (2y+5x)$? there is something with Symmetric difference? I dont understand it. – Ofir Attia Aug 02 '13 at 11:16
  • Look at the answer!:$$2y+5x=7(x+y)-(2x+5y)$$ so $,2y+5x;$ is the difference of two integers each divisible by seven so it itself is divisible by seven... – DonAntonio Aug 02 '13 at 11:27
  • I was not able to find a solution alone,In addition take x and y and multiply by 7 and difference by the given its not obvious( its like a trick.. for me). thank you. – Ofir Attia Aug 02 '13 at 11:31
  • Oh, yes: of course it is a trick. There's nothing "obvious" or "immediate" here. You need to get used to try tricks like this in these cases. – DonAntonio Aug 02 '13 at 11:32