0
$\begingroup$

I'm studying for my final exam of discrete mathematics, is an exercise in particular concerning equivalence relations do not know how to start:

$$ \text{Let } A = \left\{{3, 5, 6, 8, 9, 11, 13}\right\}\text{ and } R \subseteq A\times A: xRy\Longleftrightarrow{ x \equiv y}$$

How I can prove the symmetry, reflexivity and transitivity?

  • $(1)$ symmetry ($xRx$ for any $x$),

  • $(2)$ reflexivity ($xRy$ implies $yRx$), and

  • $(3)$ transitivity ($xRy$ and $yRz$ implies $xRz$)

I know clearly that the properties must be satisfied by other exercises I've done, but this one in specific, I do not know how to prove mathematically

  • 0
    what is the $x$ triple-bar $y$?2012-06-07
  • 0
    @ncmathsadist exactly equal or equivalent2012-06-07
  • 1
    These then follow because equality is an equivalence relation on any set.2012-06-07
  • 0
    Use \LaTeX to post, not an image. You have problems here and I can't edit the image.2012-06-07
  • 0
    What's A2? From your question, I presume it's the cartesian product of _A_ with itself, so what's _A_?2012-06-07
  • 0
    @RickDecker yes is the cartesian product of A x A2012-06-07
  • 0
    @ncmathsadist I don´t know how to post in \LaTeX2012-06-07
  • 0
    @Melkhiah66 That just leaves the question of what _A_ is.2012-06-07
  • 0
    @RickDecker Sorry...my bad...question edited2012-06-07

2 Answers 2

1

This should follow from the fact that equality is an equivalence relation under any set.

$x=x$, for all $x$

$x=y \to y=x$, for all $x$ and $y$

if $x=y$ and $y=z$, then $x=z$, for any $x$, $y$, and $z$

  • 0
    Sure, those are the properties of an equivalence relation, but how I prove it?2012-06-07
  • 1
    If you prove that it satisfies these three properties, then you prove it's an equivalence relation. Proving that these are true for equality will depend on whether you are taking equality as a defined relation or if it is a primitive.2012-06-07
0

Now that we have the clarification from my comment below, your proof should be straightforward. Equality on any set is an equivalence relation.

  • 0
    Yes, That's Right2012-06-07
  • 4
    This should be a comment, or CW.2012-06-07
  • 2
    So the answer should be (1) Reflexive: any _a_ in _A_ is equal to itself, (2) Symmetric: If _a_ and _b_ are in _A_ and _a R b_, Then _a_ = _b_ so _b_ = _a_ so _b R a_. I'll leave it up to you to show transitivity.2012-06-07
  • 4
    This post has been flagged by a user as "not an answer." Would you care to expand, sadist?2012-06-07
  • 0
    I think he is saying $A=\{3,5,6,8,9,11,13\}$ and $R=\{(x,y)\in A^2|x=y\}$.2012-06-07