I encountered a question in doing my exercise about relations: {1; 2; 3; 4; 5}.
R = {(1; 1);(1; 2);(1; 4);(3; 3);(4; 2);(4; 4);(5; 3);(5; 5)}
I need to answer whether the relation is transitive.
My guess is YES, because for every ordered pair (x,y)∈ R and ordered pair (y,z)∈ R, it implies the ordered pair (x,z)∈ R:
(1,1) (1,2) --> (1,2) (1,1) (1,4) --> (1,4) (4,4) (4,2) --> (4,2) (5,3) (3,3) --> (5,3)
But my classmate says NO. He suggests one counter-example:
If (x,y) = (1,2) ∈ R and (y,z) = (2,4) ∉ R, then (x,z) = (1,4) ∈ R
Which is correct? Thank you for your advice.
The relation in question is transitive, but you have not provided a proof according to the rules.
In order to test for transitivity we only have to test pairs of entries $(x,y)\in R$ with $x\ne y$. In your case these are the entries $$(1,2),\quad(1,4),\quad (4,2),\quad(5,3)\ .$$ Here only $4$ appears both as first and as second coordinate of some entry. It follows that $(1,4)$ and $(4,2)$ is the only pair of entries we have to test, and as $(1,2)\in R$ we are done.