1

I'm really confused about the concept of Reflexive and Symmetric relations. for an example from my textbook:

Let A = {b, c, d, e} and the relation on 'A' be defined as R = {(b,b),(b,c),(c,b),(c,c),(d,d),(b,d),(d,b),(c,d),(d,c)}

The book claims that the above relation is not reflexive, because by its definition, xRx must be true for all x ∈ A, however the pair (e,e) is not in R, so R is not reflexive.

But! The book also claimed that the relation above IS symmetric because R contains: bRc and cRb, bRd and dRb, dRc and cRd....but I can use the same argument that it made for the reflexive definition "for ALL of x,y ∈ A" and in this case, the relation R does not have (bRe and eRb), (cRe and eRc), (dRe and eRd) and etc... in the set R...

This is what I'm really confused about; why is R not reflexive because it doesn't have the element "e" in R, and on the other hand R is symmetric while R don't contain "e" at all? I mean the definition of symmetric is xRy -> yRx for ALL x,y ∈ A.

Thanks

Sam Kay
  • 117
  • A relation is symmetric if $aRb \iff bRa$ Whenever two items are related then the relation goes "both ways" if $a$ is related to $b$ then $b$ is related to $a$. Elements in the set that are unrelated don't matter. A relation is reflexive if every element is related to itself. – Doug M Nov 04 '17 at 01:16
  • Thank you all for the awesome answers! I was contemplating on this problem for sometime now. But I got it now! Thanks so much. – Sam Kay Nov 04 '17 at 02:26

3 Answers3

2

The key is in the definitions. Let's look at them more carefully.

A relation $R$ on $A$ is reflexive when for all $x\in A$ we have $xRx$. As you pointed out, $e\in A$ but we don't have $eRe$, and therefore by definition this is not a reflexive relation.

The definition for symmetric relations is a bit different. A relation $R$ on $A$ is symmetric when for all $x,y\in A$, $xRy$ implies $yRx$. In other words: if you give me $x$ and $y$, AND we know that $xRy$, then we must also have $yRx$.

So, the fact that we don't have $dRe$ isn't a problem. IF we knew that $eRd$, THEN symmetry would require $dRe$; but, because we don't have either, the symmetry relation is satisfied.

Another way of stating the definition of symmetry is that for all $x,y\in A$, we must either have (a) both $xRy$ and $yRx$, or (b) neither $xRy$ nor $yRx$.

Nick Peterson
  • 32,430
1

The definition of symmetric if, for all $x,y \in A$

$$xRy \iff yRx$$

So a symmetric relation must contain both $(x,y)$ and $(y,x)$, or neither. If it has only one out of the pair then it is not symmetric.

Adam Francey
  • 574
0
  • Reflexive: $~~~~\forall x\in A : (x,x)\in R $
  • Symmetric: $~\forall x\in A~\forall y\in A:\big((x,y)\in R\to(y,x)\in R\big)$

Reflexivity requires that for any $x$ which are in $A$, we shall definitely find the pair $(x,x)$ in $R$.

Symmetry requires that for all pairs $(x,y)$, if they are in $R$, then so is their mirror pair, $(y,x)$.   It is a conditional.   $(y,x)$ is only required to be in $R$ if $(x,y)$ is so; and vice versa.

Perhaps it will help to look at the negations.

  • Not Reflexive: $~~~\exists x\in A : (x,x)\notin R$
  • Not symmetric: $~\exists x\in A~\exists y\in A:\big((x,y)\in R\wedge (y,x)\notin R\big)$

$R$ will be not reflexive if there is some $(x,x)$ pair that is not included in $R$.

$R$ will be not symmetric if there is some $(x,y)$ pair that is in $R$ without its mirror is also being in $R$.   (or is not in $R$ while the mirror is.)

Graham Kemp
  • 129,094
  • OMG Facepalm, I forgot that symmetric is a conditional statement! lol, thank you sooo much for the detailed explanation, especially when you put the statement in its negation form! Thank you :) – Sam Kay Nov 04 '17 at 02:25