Trying to Understand the Meaning of Transitivity in Relation to a Particular Problem

Question

I was trying to understand transitive relation and so I was solving a problem. The question is : $R_1 = \{(a,b)| a =b \text{ or }a = -b\} , R_2 = \{(a,b)| a =b \}, R_3 = \{(a,b)| a =b+1\}$, which one is transitive and why?

As far as I know transitive relation is, if $a>b$ and if $b > c$, then $a> c$. I am assuming that the given a and b are real numbers but I am not sure where I will get C so that I can show that which one is transitive.

I saw many youtube tutorials and read my book but I am very confused with this math. I am new in this topic. It will be really helpful if someone can please explain how can I solve this problem.

Thank you very much.

Answer 1

Relations are just collections of ordered pairs. For example, a relation on the set of real numbers $\mathbb{R}$ would be $(x,y)$ pairs. Formally, a relation on a set $S$ is simply some subset of the cartesian product $R \subseteq S \times S$. We say a relation $R$ is transitive if whenever the pairs $(a,b), (b,c) \in R$ then $(a,c) \in R$ as well. In the first relation, if we know $(a,b), (b,c) \in R_1$, then we know either $a = b$ or $a = -b$. Similarly, either $b = c$ or $b = -c$. Clearly this implies either $a = c$ or $a = -c$ as we know the magnitudes of a,b,c must be the same. Thus $(a,c) \in R_1$ and the relation is transitive.

Answer 2

$R_2$ is the identity relation. Or simply, the equality $=$ is transitive. That is, if $(a,b)$ and $(b,c)$ are the both in $R_2$, then $a=b$ and $b=c$. Thus $a=c$, i.e. $(a,c)\in R_2$.

Answer 3

Try replacing, for example, your $a \gt b$ with $\{a,b\} \in R_1$, $b \gt c$ with $\{b,c\} \in R_1$ and $a \gt c$ with $\{a,c\} \in R_1$. Then replace $R_1$ with $R_2$ and $R_3$ to see for which of these the first $2$ statements (e.g., if $\{a,b\} \in R_1$ and $\{b,c\} \in R_1$) means the third one must hold as well (e.g., $\{a,c\} \in R_1$).

If you do this, you should find that $R_1$ and $R_2$ are transitive, while $R_3$ is not (because $a = b + 1$ and $b = c + 1$ means $a = c + 2$, not $a = c + 1$). Can you finish the rest yourself?