# B Why are these relations reflexive/symmetric/transitive?

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1. Nov 4, 2016

### Buffu

The definition of these relations as given in my textbook are :

(1):- Reflexive :- A relation $R : A \to A$ is called reflexive if $(a, a) \in R, \color{red}{\forall} a \in A$
(2):- Symmetric :- A relation $R : A \to A$ is called symmetric if $(a_1, a_2) \in R \implies (a_2, a_1) \in R, \color{red}{\forall} a_1, a_2 \in A$
(3):- Transitive :- A relation $R : A \to A$ is called transitive if $(a_1, a_2) \in R \land (a_2, a_3) \in R \implies (a_1, a_3) \in R, \color{red}{\forall} a_1, a_2,a_3 \in A$

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Here are examples in my textbook that i guess are violating 'for all' part of the definition.

$\large{\cdot}$ Let $R$ be a relation in set $\{ 1, 2, 3 \}$. $R = \{(1,2), (2,1)\}$. It is given that this relation is symmetric.

My confusion :- Why is this symmetric ? It clearly violates the 'for all' part of the definition $(2)$. It does not contain $(1, 3), (3,1), (2,3), (3,2)$.

$\large{\cdot}$ A relation $R$ on set $\{ 1,2,3,4\}$ given by $R = \{(1,2), (2,2), (1,1), (4,4), (1,3), (3,3), (3,2)\}$ is both reflexive and transitive.

Same as one, Why is this thing transitive, it does not have $(1,4), (4, 3), (1,3) \cdots$.

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After a bit of thinking i reached the conclusion that definition for symmetric means that, If $R$ is symmetric then $(a,b) \in R \implies (b,a)\in R$ and it does not have to be that every doublet in $A \times A$ have to be in $R$.
I reached the similar conclusions for reflexive and transitive definitions.

But then I saw this example :-

$R =\{(1,1)\}$ is not reflexive for a set $A = \{1,2\}$.And the reason given is that; $R$ is not reflexive because it does not contain $(2,2)$.

So why can't i argue on the basis of the example that previous two examples of mine are not symmetric and transitive, respectively ?

All three definitions have $\forall$ in them but only for reflexive it really mean "for all" for, the rest it simply mean "for". Why is this so ?
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I think i am so confuse that some parts of this thread are not understandable by the readers.

So if the thread is too confusing to understand, I will glad if some one can explain why examples $(1),(2)$ are symmetric and transitive respectively when they clearly violates the "for all" part of the definition.

2. Nov 4, 2016

### Staff: Mentor

No, it doesn't, because as $(1,3) \notin R$ so it isn't required that $(3,1) \in R$.
The condition says: If $(a_1,a_2) \in R$ then $(a_2,a_1) \in R$. There is no statement about the case $(a_1,a_2) \notin R$.
Same reason here. No statement is made for elements which are not related. These properties describe a relation as a subset of $A \times A$. It does not describe unrelated pairs, i.e. $(A \times A) - R$.

3. Nov 4, 2016

### Buffu

Not different from what i thought.

So why isn't $R = \{ (1,1) \}$ reflexive ? $R$ is relation on set $A = \{1,2\}$ (my third example).

Last edited: Nov 4, 2016
4. Nov 4, 2016

### Staff: Mentor

Here you have a for all statement on $A$ whereas in the case of symmetry or transitivity it is a for all statement on $R$.

5. Nov 4, 2016

### Buffu

But my textbook has for all on A for all three of them. Maybe it is wrong ?

6. Nov 4, 2016

### Staff: Mentor

Not really.
It states: For all pairs in $A \times A$ holds the implication [Each related pair has to have its symmetric counterpart.]
This can as well be formulated as: $\forall_{(a,b)\in R} \; (b,a) \in R$ and similar with transitivity.

If one defines reflexivity in the same way as
$R$ is reflexive $\Longleftrightarrow \; \forall_{(a,b)\in R} \; (a,a) \in R \, \wedge \, (b,b) \in R$
then this will complicate things in applications such as $R = \text{ "equivalence" }$ or $R = \text{ "equality" }$ for $1=1$ might suddenly not hold.

But imagine all pairs $(a,b) \in A \times A$ where also in $R$. What sense would it make to define $R$ at all?

7. Nov 4, 2016

### Buffu

What is $R = \text{ "equality" }$ ?

8. Nov 4, 2016

### Staff: Mentor

A relation. Preferably one which is reflexive, i.e. every element of a set $A$ equals itself, not only a few.
But not every element is equal to all others. But if $a=b$ then also $b=a$ holds, which doesn't mean, there cannot be a $c$ with $a\neq c$.

9. Nov 4, 2016

### Buffu

I am not sure if i understood properly (No doubt I am a moron). Is it same as equivalence relation (google search for "equality relation" brings links related to "Equivalence relation ") ? Can you give a example ?

How so ?

10. Nov 4, 2016

### Staff: Mentor

Well, I consider "$=$" to be a special equivalence relation, e.g. $1=\frac{2}{2}$ is equal but not the same. However, I know that not everybody shares this opinion. Some insist to call it the same. In my opinion a pie is different from two halves. Only equivalent in mass, volume and taste. Nevertheless, you may also call it the same. I won't.

An important example of an equivalence relation is ($6$ here is only an example, could be any integer):

Two integers are called equivalent, if their remainder in a division by six is the same. (The remainders be taken from zero to five.)

Here one gets, e.g. $7 \equiv -5$ but obviously $7 \neq -5$. As you've probably read, any relation is called an equivalence relation, if it satisfies the three conditions above.

If you define $R=\{(2,2)\}$ on $A=\{1,2\}$ and call it reflexive, then $R="="$ wouldn't be reflexive for $(1,1) \notin R$. But we want "$=$" to be reflexive and $1=1$ to hold. (However, if it confuses you, then better forget it. Not the luckiest example I've ever chosen.)

11. Nov 4, 2016

### Buffu

I am still a bit confused but i guess i am too illiterate to understand more.
I think I understood enough to proceed with the exercises in the book, maybe someday I might understand the rest.

12. Nov 4, 2016

### Stephen Tashi

I agree that this is confusing terminology. The technical difficulty is interpretation the meaning of " A $\rightarrow$ A". The usual way to define a relation would be to define the statement "R is a relation on the set S" to mean that R is a subset of the cartesian product S $\times$ S. So R need not contain all possible ordered pairs (x,y) that exist in S$\times$S.

As far as I can tell, the notation "R:A$\rightarrow$ A" doesn't imply that "R is a relation on the set A". Instead, we apparently have the situation where "R is a relation on the set S" and A is some subset of S. The signficance of putting an "A" on both sides of the "$\rightarrow$" isn't clear to me.

The examples you gave say things like "R is a relation on {1,2,3,4}" and those examples make sense from the definition of "R is a relation on S".

13. Nov 4, 2016

### Buffu

That is the creativity of my mind. In all the definition, it is actually "Relation R on A" not $R: A\to A$. I thought it won't make any difference so I used it in hurry.

14. Nov 4, 2016

### Stephen Tashi

In that case the, problems you have reconciling the definitions of "symmetric" and "transitive" with examples are defects in you own thinking. The problem you have in reconciling the definition of "reflexive" with examples is a defect in your textbook - and the world at large.

An important feature of the definitions of "symmetric" and "transitive" is that they contain an "if" clause. For example, there is a distinction between saying

$\forall a_1 \in A$ and $\forall a_2 \in A$, if $(a_1,a_2) \in R$ then $(a_2,a_1)\in R$

versus

$\forall a_1 \in A$ and $\forall a_2 \in A$, $(a_1,a_2) \in R$ and $(a_2,a_1) \in R$

In the world at large, the definition of "reflexive relation on A" is given in various way. We can formulate at least two different concepts.

Concept 1: Require that for each $a \in A$, $(a,a) \in R$. By this definition, R {(1,1),(2,2)} is not a reflexive relation on the set {1,2,3} because R omits (3,3).

Concept 2: Only require that if an element $a \in A$ appears in some ordered pair in R then $(a,a) \in R$.
This can be stated using the "existential quantifier" $\exists$.
$\forall a \in A,$, if $\exists x \in A$ such that $(a,x) \in R$ or $(x,a) \in R$ then $(a,a) \in R$.

A book that intends for {(1,1),(2,2)} to be a reflexive relation "on the set" A = {1,2,3} should use Concept 2 - or some similar concept.

15. Nov 4, 2016

### Buffu

Why won't two different concepts for same thing a problem ?
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I have one more example that i can't understand,
$R = \{(1,2),(2,1), (1,1) \}$ is transitive on $A= \{1,2\}$.
Because $(1,2),(2,1) \in R \implies (1,1) \in R$ is sufficient for $R$ to be transitive on $A$.

I can say, by similar logic that $R= \{(1,1),(2,2), \cdots (n,n)\}$ is symmetric on $A={1,2, \cdots , n. n \in mathbb{N}}$ ?
Because $(a, b) \in R \implies (b,a) \in R$ where $a = b$.

16. Nov 4, 2016

### Staff: Mentor

It is not transitive, as $(2,1) \in R$ and $(1,2) \in R$ but $(2,2) \not \in R$.

Yes, that is symmetric. It is also reflexive and transitive.

17. Nov 4, 2016

### Stephen Tashi

Yes, it will be a bad problem if the same book uses both concepts. There are some mathematical concepts that different books define differently. Different concepts used in different books is not a bad problem. That kind of inconsistency is familiar to students of mathematics.

It is not clear what you mean. . I don't know how to interpret the sentence structure "Because...implies .. is sufficient for...".

R isn't transitive. R contains (2,1) and (1,2). In order for it to be transitive, it should contain (2,2).

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Your statement would be clear if your sentence had the form: "<statement 1> because <statement 2>". I would interpret that structure to mean: "<statement 2> implies <statement 1>". However you are using the structure: "Because <statement 2>. I don't know what that means.

In that example, R is symmetric. It is permissible in the definition of symmetric to have the situation a = b. The definition of symmetric does not forbid examples where the only instances of (a,b) $\in$ R are those where a=b.

18. Nov 5, 2016

### Buffu

I apologies for confusion.

I did not create that example, it is given in the textbook.
$R = \{(1,2),(2,1),(2,2) \}$ is transitive on set $A = \{1,2\}$.
The reason given is that because $R$ satisfies the condition $(1,2) \in R$ and $(2,1) \in R \implies (1,1) \in R$.
I think the reasoning is correct.
In the definition of transitivity, we can take $a_1 = a_3 = 1, a_2 = 2$.

I can't understand why it should not be transitive ? Am i missing something ?

19. Nov 5, 2016

### Staff: Mentor

If (1,2) and (2,1) are both in R, then transitivity with $a_1 = a_3 = 1$ and $a_2=2$ requires (1,1) to be in the set, and transitivity with $a_1 = a_3 = 2$ and $a_2=1$ requires (2,2) to be in the set. You need both, because transitivity is "for all ...".

20. Nov 5, 2016

### Buffu

Can it be that the book is wrong ?
I think they missed (2,2).

Thanks for you time.