In my previous assignment, I was given two sets, A and B, both containing the elements 1, 2, 3, and 4. Additionally, I was provided with a relation C, which consisted of pairs of elements: (1, 1), (3, 4), (2, 2), and (3, 3).
I argued that relation C was transitive based on the definition of transitivity. I believed that by considering (3, 3) as (a, b) and (3, 4) as (b, c), I could conclude that (a, c) would be (3, 4), which already existed in the set.
To my surprise, my teacher disagreed and claimed that relation C was not transitive, reflexive, or symmetric. Now, I am wondering if I can contest my grade because I strongly believe that my reasoning is correct. I even sought assistance by posting a question on the Mathematics Stack Exchange website months ago, which can be found at the provided link: Transitive relation of non function
You have not given your reasoning, however, it is transitive. Suppose $xRy$ and $y Rz$. (Note that this implies that $(x,y),(y,z) \in C$.)
We must have $x=y$ for the only element of $C$ that has different components is $(3,4)$ and there is no element of $C$ with $4$ as the first element.
In particular we must have $(x,z) \in C$ and so it is transitive.
It is not reflexive as $(4,4) \notin C$ and it is not symmetric as $(4,3) \notin C$.