# A question on the properties of a relation

Suppose we have some relations on the set [1,2,3,4].

1. $$R_{1}=\{(2,2),(2,3),(2,4),(3,2),(3,3),(3,4)\}$$

2. $$R_{2}=\{(1,1),(1,2),(2,1),(2,2),(3,3),(4,4)\}$$

3. $$R_{3}=\{(2,4),(4,2)\}$$

4. $$R_{4}=\{(1,1),(2,2),(3,3),(4,4)\}$$

5. $$R_{5}=\{(1,2),(2,3),(3,4)\}$$

6. $$R_{6}=\{(1,3),(1,4),(2,3),(2,4),(3,1),(3,4)\}$$

I need to determine if these relations are reflexive/symmetric/anti-symmetric and transitive.

Can someone please give me a detailed explanation on what we need to check to determine these properties. I'm very confused. I'm getting my answers wrong.

I will be very grateful, thanks a lot.

• Check if the relations $R_1$ etc satisfy the properties defining symmetry etc. Symmetric means : $\forall x,y [xRy \leftrightarrow yRx]$. – Mauro ALLEGRANZA May 25 at 12:00
• Consider e.g. $R_1$; we have $(2,4) \in R_1$. What about $(4,2)$ ? – Mauro ALLEGRANZA May 25 at 12:01
• Yes, R1 is not symmetric. I'm getting more confused on anti symmetric and transitive. – Sonny Jordan May 25 at 12:45

All you have to do is checking the definitions.

For example, let’s work $$R_1$$ out.

• Can $$R_1$$ be reflexive? Recall the definition:

A relation $$R$$ over a set $$A$$ is said to be reflexive iff for each $$a \in A$$ one has $$(a,a) \in R$$.

The answer is no, because $$(1,1) \notin R_1$$.

• Can $$R_1$$ be antireflexive? Recall the definition:

A relation $$R$$ over a set $$A$$ is antireflexive iff for each $$a \in A$$ one has $$(a,a) \notin R$$.

The answer is no, because $$(2,2) \in R_1$$.

• Can $$R_1$$ be symmetric? Recall the definition:

A relation $$R$$ over a set $$A$$ is symmetric iff for each $$a \in A$$ one has $$(a,b) \in R \Rightarrow (b,a) \in R$$.

The answer is no, because $$(2,4) \in R_1$$ but $$(4,2) \notin R_1$$.

• Can $$R_1$$ be antisymmetric? Recall the definition:

A relation $$R$$ over a set $$A$$ is antisymmetric iff for each $$a, b\in A$$ one has $$(a,b) \in R \land (b,a) \in R \Rightarrow a=b$$.

The answer is no, because $$(2,3) \in R_1$$ and also $$(3,2) \in R_1$$, but $$2\neq 3$$.

• Can $$R_1$$ be asymmetric? Recall the definition:

A relation $$R$$ over a set $$A$$ is asymmetric iff for each $$a,b\in A$$ one has $$(a,b) \in R \Rightarrow (b,a) \notin R$$.

The answer is no, because $$(2,3) \in R_1$$ and $$(3,2) \in R_1$$.

• Can $$R_1$$ be transitive? Recall the definition:

A relation $$R$$ over a set $$A$$ is transitive iff for each $$a,b,c \in A$$ one has $$(a,b) , (b,c) \in R \Rightarrow (a,c) \in R$$.

The answer is yes, because a direct inspection shows that each couple of possible consecutive connections $$(a,b) , (b,c) \in R_1$$ is closed by an arc $$(a,c) \in R_1$$. In fact:

• the path $$(2,2)$$ & $$(2,2)$$ is closed by $$(2,2)$$;
• the path $$(2,2)$$ & $$(2,3)$$ is closed by $$(2,3)$$;
• the path $$(2,2)$$ & $$(2,4)$$ is closed by $$(2,4)$$;
• the path $$(2,2)$$ & $$(2,3)$$ is closed by $$(2,3)$$;
• the path $$(2,3)$$ & $$(3,2)$$ is closed by $$(2,2)$$;
• the path $$(2,3)$$ & $$(3,3)$$ is closed by $$(2,3)$$;
• the path $$(2,3)$$ & $$(3,4)$$ is closed by $$(2,4)$$;
• the path $$(3,2)$$ & $$(2,2)$$ is closed by $$(3,2)$$;
• the path $$(3,2)$$ & $$(2,3)$$ is closed by $$(3,3)$$;
• the path $$(3,2)$$ & $$(2,4)$$ is closed by $$(3,4)$$;
• the path $$(3,3)$$ & $$(3,2)$$ is closed by $$(3,2)$$;
• the path $$(3,3)$$ & $$(3,3)$$ is closed by $$(3,3)$$;
• the path $$(3,3)$$ & $$(3,4)$$ is closed by $$(3,4)$$.

Remaining relations can be studied in a similar fashion.

• Did we conclude that this is transitive by using logic? Such as p implies q, but since p is false, q being false or true will mean our proposition is always true. Sorry if I am incorrect, I am confused in the relations chapter. – Sonny Jordan May 25 at 13:22
• @SonnyJordan : No, it’s not logic. See the edited post. ;-) – Pacciu May 25 at 14:03
• Why didn't we check (2,4) or (3,4)? – Sonny Jordan May 25 at 14:07
• Because there are no pairs in $R_1$ with $4$ in the first coordinate, i.e. there are no connections which are consecutive to $(2,4)$ and $(3,4)$. – Pacciu May 25 at 14:15
• Have you ever tried to draw the oriented graph of $R_1$? – Pacciu May 25 at 14:17

A relation $$R$$ on set $$A$$ is said to be reflexive if $$\forall a \in A ~ (a,a) \in R$$.

A relation $$R$$ on set $$A$$ is said to be symmetric if $$(a,b) \in R \mathrm{~then~} (b,a) \in R$$. $$\phi$$ is also symmetric.

A relation $$R$$ on set $$A$$ is said to be anti-symmetric if $$(a,b) \in R \mathrm{~and~} (b,a) \in R \mathrm{~then~} a = b$$. No symmetric pair must be present.

A relation $$R$$ on set $$A$$ is said to be Transitive if $$(a,b) \in R \mathrm{~and~} (b,c) \in R \mathrm{~then~} (a,c) \in R$$.

$$R_1$$ does not have $$(4,2)$$ hence it is not symmetric although it is reflexive, transitive. However $$R_1$$ is not antisymmetric (as we have symmetric pairs).