# Discrete Math Relations on the set {1, 2, 3}

I was just to make sure that I am doing these correctly. Here is what I have:

1. Describe a binary relation on 1,2,3 that is reflexive and symmetric, but not transitive:
And I have: {(1,1), (2,2), (3,3)} it is obviously reflexive and I figured this would be true that it is symmetric as well.

2. Binary relation on 1,2,3 that is reflexive and transitive, but neither symmetric or antisymmetric:
My answer: {(1,1), (2,2), (3,3), (1,2), (2,3), (2,1)}

3. Binary relation on 1,2,3 that is antisymmetric and transitive, but not reflexive:

Help is very much appreciated. Thank you

• Your first relation is transitive. Your second relation is not transitive because it contains $(1, 2)$ and $(2, 3)$ but it does not contain $(1, 3)$. Your last relation is fine. Commented Mar 14, 2015 at 17:53

$1$. The function you gave is symmetric, transitive and reflexive.

For a function that is symmetric, reflexive but not transitive take $\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}$

$2$. the function you gave is not transitive since we have $(1,2)$ and $(2,3)$ but not $(1,3)$.

For an example that works take $(1,1),(2,2),(3,3),(1,2),(2,3),(1,3),(1,3)$. So just add $(1,3)$ to make it transitive and it works.

$3$. your example is good. You can also take $\emptyset$.

• How is {(1,2),(2,1),(2,3),(3,2)} reflexive if it does not contain (1,1), (2,2), and (3,3)? Commented Mar 14, 2015 at 18:00
• Oh yeah my bad. Commented Mar 14, 2015 at 18:02
• Would you be able to explain why my first relation is transitive? I don't understand how it could be? Thanks a lot Commented Mar 14, 2015 at 18:04
• Transitivity mean that if you have $(a,b)$ and $(b,c)$ then you have $(a,c)$. In this case this is true because there is only one relation including $b$, and that is $(b,b)$. SO you just have to prove if you have $(b,b)$ and $(b,b)$ then you have $(b,b)$. Which holds. Commented Mar 14, 2015 at 18:05
• Ohhh haha wow okay, yeah didn't think of that. Thank you very much! Commented Mar 14, 2015 at 18:06