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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 $\{<1,2>, <1,3>, <1,4>, <2,1>, <2,3>, <2,4>, <3,1>, <3,2>, <3,4>, <4,1>, <4,2>, <4,3>\}$

G Tony Jacobs
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j77
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  • This is not a function. For a binary relation $R$ to be a function, you cannot have both $<a,b>$ and $<a,c>$ in $R$ – Bram28 Oct 08 '17 at 02:29
  • if its not a function then does it mean that it can't be either surjective or injective? – j77 Oct 08 '17 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. – Bram28 Oct 08 '17 at 17:02
  • So in this case the binary relation of the answer that i came up is injective – j77 Oct 08 '17 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 $R$. – Bram28 Oct 08 '17 at 21:49

2 Answers2

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hint:

rephrase as find a function which has no fixed points and whose square is the identity.

Your example is not a function since it relates the same element to multiple other elements.

Mark Joshi
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This relation appears to be symmetric and irreflexive, but it's not a function, because the same domain values appear multiple times with different range values, i.e., it does not pass the "vertical line test".

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.

G Tony Jacobs
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