# binary relation that is both symmetric and irreflexive

I was asked to find a binary relation on $A$ that is symmetrical and irreflexive, and which is also a function from $A$ to $A$ where $A=\{1,2,3,4\}$

so i don't know if its correct but i came up with these $\{<1,2>, <1,3>, <1,4>, <2,1>, <2,3>, <2,4>, <3,1>, <3,2>, <3,4>, <4,1>, <4,2>, <4,3>\}$

• This is not a function. For a binary relation $R$ to be a function, you cannot have both $<a,b>$ and $<a,c>$ in $R$ Commented Oct 8, 2017 at 2:29
• if its not a function then does it mean that it can't be either surjective or injective?
– j77
Commented Oct 8, 2017 at 15:39
• No, it doesn't mean that. A binary relation can be both surjective and injective without being a function. E.g in this context $\{<1,1>,<1,2>,<1,3>,<1,4> \}$ is injective and surjective but not a function. Commented Oct 8, 2017 at 17:02
• So in this case the binary relation of the answer that i came up is injective
– j77
Commented Oct 8, 2017 at 19:59
• No, because you have multiple elements mapping to the same element. For a binary relation $R$ to be injective, you cannot have both $<b,a>$ and $<c,a>$ in $R$. Commented Oct 8, 2017 at 21:49

For a relation to be a function on a $4$-element set, it needs to have exactly $4$ ordered pairs in it, one to map each value somewhere.