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Determine which of the reflexive, symmetric, and transitive properties are satisfied by the given relation R defined on set S...

S= {1,2,3}
R= {(1,1), (1,3), (2,2), (2,3), (3,1), (3,2), (3,3)}
I've concluded that it is not reflexive due to that there is no (1,2) or (2,1) in R. I'm not too confident to say its not symmetric... Basically would anyone be able to properly simplify an explanation on how to find whether its reflexive, transitive, and/ or symmetric?

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  • $\begingroup$ I guess it is reflexive ... if $x \sim y$ then $y \sim x$. $\endgroup$ – Donald Splutterwit Jan 29 '17 at 21:21
  • $\begingroup$ If it's reflexive, $(1,1)$, $(2,2)$ and $(3,3)$ must be in it (we must have xRx for all x). If it is symmetric then if $(a,b)$ is in R, $(b,a)$ is in R too (xRy iff yRx). So if you found $(1,3)\in R$ but $(3,1)\notin R$ then it would not be symmetric. If no such example exists, it is. $\endgroup$ – spaceisdarkgreen Jan 29 '17 at 21:21
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Reflexivity means that for all $x \in S$, we have that $(x,x) \in R$. Looking at your example, we see that $(1,1), (2,2)$ and $(3,3)$ are in $R$, so this relation is reflexive.

Symmetry of a relation means that if $(x,y) \in R$, then $(y,x) \in R$. This is the case: we have that $(1,3) \in R$ and so is $(3,1)$. Analogous for $(2,3)$. Note that neither $(1,2)$ nor $(2,1)$ is in $R$, but this is not a problem, since symmetry of a relation states that $\textbf{if}$ $(x,y) \in R$ $\textbf{then}$ $(y,x) \in R$. So if $(x,y)$ is not in $R$, we don't need to worry about $(y,x)$ being in $R$.

Transitivity of a relation means that if $(x,y \in R$ and $(y,z) \in R$, then so is $(x,z)$. This is not the case, can you find a counter example?

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Reflexive means you have $(x,x) \in R$ for any $x \in S$ ... which you do .. so it is reflexive.

Symmetric means you have $(x,y) \in R$ whenever $(y,x) \in R$ .. which again is the case. So it is symmetric.

Transitive means that you have $(x,z) \in R$ whenever $(x,y) \in R$ and $(y,z) \in R$ .. which is not the case: you have $(2,3)$ and $(3,1)$ ... but you don't have $(2,1)$ So it is not transitive.

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  • $\begingroup$ well this just goes to show how well i savvy this subject.... $\endgroup$ – Madara Jan 29 '17 at 21:21
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Draw the picture with points marked by the elements of $S$ (here just 1,2,3) and arrows between points such that $(a,b)\in R$ iff there is an arrow from $a$ to $b$. Reflexive means there is an arrow from each point to itself, symmetric that if there is arrow from one point to another, then there is a reverse arrow. Transitive means that if you can go from one point to another by some arrows then they are connected with the arrow directly.

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