# Proof of equivalence relation

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

• what is the $x$ triple-bar $y$? Jun 7, 2012 at 1:07
• @ncmathsadist exactly equal or equivalent Jun 7, 2012 at 1:09
• These then follow because equality is an equivalence relation on any set. Jun 7, 2012 at 1:15
• Use \LaTeX to post, not an image. You have problems here and I can't edit the image. Jun 7, 2012 at 1:16
• What's A2? From your question, I presume it's the cartesian product of A with itself, so what's A? Jun 7, 2012 at 1:17

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 $z$

• Sure, those are the properties of an equivalence relation, but how I prove it? Jun 7, 2012 at 1:46
• 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.
– A.S
Jun 7, 2012 at 2:25

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

• Yes, That's Right Jun 7, 2012 at 1:36
• This should be a comment, or CW.
– Pedro
Jun 7, 2012 at 1:37
• 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. Jun 7, 2012 at 1:42
• This post has been flagged by a user as "not an answer." Would you care to expand, sadist?
– anon
Jun 7, 2012 at 3:27
• I think he is saying $A=\{3,5,6,8,9,11,13\}$ and $R=\{(x,y)\in A^2|x=y\}$. Jun 7, 2012 at 12:20