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if you proving someting like if a relation is symetric. say R is a relation on Z and $xRy$ if x+y is a multiple of 3. Then you want to prove symetry and say for some $x,y$ in Z $xRy$ if

$x+y=3k$ k some integer

This implies $y+x=3k$

$yRx$ so $R$ is symetric.

So my question is when i said "some x,y" are we know treating x and y as constants proving what we need to prove and then does this automatically prove its true for all $x,y$ in $Z$? Thanks

Carlos Bacca
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  • I'm not sure I understand your question, but if you wanted to prove that such a relation was symmetric, you would need to show that it held for all $x,y$. You couldn't make them specific elements in $\mathbb{Z}$. – Kevin Long Oct 20 '18 at 20:03
  • So is this proof wrong? – Carlos Bacca Oct 20 '18 at 20:22
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    The proof is correct. I think what you meant before is that you could substitute $x$ and $y$ for any pairs of elements of $\mathbb{Z}$ where $xRy$ and it would hold. In that case, that's the right idea; the proof works because if you set, for example, $x=2$ and $y=4$, the argument would work for those choices of $x$ and $y$, as well as any other such pair. – Kevin Long Oct 20 '18 at 20:25
  • oh i see thanks – Carlos Bacca Oct 20 '18 at 20:26

1 Answers1

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You want to prove that

$$(\forall (x,y)\in\Bbb Z^2) \;\; xRy \implies yRx$$

You say :

Let $(x,y)$ be an arbitrary element of $\Bbb Z^2$ such that $xRy.$

$$xRy \implies (\exists k\in\Bbb Z)\;\; :\; x+y=3k$$

$$\implies (\exists k\in\Bbb Z) \; :\; y+x=3k$$ $$\implies yRx.$$ thus $R$ is symetric.

hamam_Abdallah
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  • Thanks, why does showing it for an arbitrary element prove it for all elements? I kind of understand it does but wouldn't be able to explain it. – Carlos Bacca Oct 20 '18 at 20:13