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Given $S = \{1,2,3,4,5\}$

Relation $R=\{(x,y)|x-y=0\}\subset S\mathbb x S$

Create the set $M = S\mathbb x S$

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

Applying the relation $R=\{(x,y)|x-y=0\}\subset M$

Gives the $OUTPUT = \{(1,1),(2,2),(3,3),(4,4),(5,5)\}$

I understand this relation is reflexive. And I need help in understanding, how this relation is also transitive?

For a relation to te transitive $x=y$ and $y=a$ then $x=a$. How can I apply the transitive rule in this case?

Asaf Karagila
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Abbas
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  • Your relation $R$ (on the set $S$) is equality, so yes, it is transitive, there is nothing more to write than the definition in your last line. – Anne Bauval Sep 12 '22 at 10:51

1 Answers1

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In order to show transitivity of the relation $R$ we have to show that for all $x,y,z\in S$ \begin{align*} (x,y)\in R\quad\mathrm{and}\quad (y,z)\in R \quad\mathrm{implies }\quad (x,z)\in R\tag{1} \end{align*}

Let $x,y,z\in S$. We obtain \begin{align*} \begin{matrix} (x,y)\in R\quad\Rightarrow\quad x=y\\ (y,z)\in R\quad\Rightarrow\quad y=z\\ \end{matrix}\ \Bigr\} \quad\Rightarrow\quad x=z \quad\Rightarrow\quad (x,z)\in R \end{align*} and (1) follows.

Markus Scheuer
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