If R1 is reflective and not transitive, R2 is transitive but not symmetric and R3 is symmetric but not reflexive. We need to find an example of a set S and the three relations R1 R2 R3.
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what i did was find out the definitions of the following terms reflexive, transitive and symmetric. but i cannot find an example of a set that has all this 3 relations – Shavneel Prasad Sep 05 '13 at 22:13
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http://math.stackexchange.com/questions/482620/relations-examples-and-counterexamples – njguliyev Sep 05 '13 at 22:14
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1Yep, this looks like a duplicate to me. At the time I saw it I couldn't flag it as such because @njguliyev's link didn't have an upvoted answer, but it had a good answer, so I upvoted it :) – Ben Millwood Sep 05 '13 at 22:31
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$R_1$ use the $\delta_{1}$ relation "made" reflective ($=1 \vee =$)
$R_2$ use the implication relation ($\Rightarrow$) in any fashion;
$R_3$ use the inequality relation ($\neq$) in any fashion.
$S$ could be $\{1,2,3\}$ for example.
For the specific $S$ the relations could be $$\begin{align*}R_1 = & \{(1,1), (1,2), (1,3), (2,1), (3,1) , (2,2), (3,3)\} & = \{1\} \times S \cup S \times \{1\} \cup {\rm id}\\ R_2 = & \{(1,2), (2,3), (1,3)\} \\ R_3 = & \{(1,2), (1,3), (2,3), (2,1), (3,1), (3,2)\} \end{align*}$$
AlexR
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but we still did not learn about the use of Kronecker delta relation. – Shavneel Prasad Sep 05 '13 at 22:17
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@BenMillwood not with two arguments I think... as there are only $$a\sim b, c \sim b \Leftrightarrow a=b=c \Rightarrow a\sim c$$ Maybe that's not the exact name... – AlexR Sep 05 '13 at 22:18
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@ShavneelPrasad I think it's called "identical relation" anyways. $$a\sim b :\Leftrightarrow a==b$$ "is exactly the same element as" – AlexR Sep 05 '13 at 22:20
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@BenMillwood I didn't mean $R: \chi_R(b) = \delta_{ab}$ but rather $(a,b) \in R :\Leftrightarrow \delta_{ab} = 1 \Leftrightarrow a = b$ – AlexR Sep 05 '13 at 22:21
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@BenMillwood Alright sorry for the mistake. My new example should work though, as $2\not\sim 3$ but $2\sim 1\sim 3$ – AlexR Sep 05 '13 at 22:24
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@AlexR: okay, better, but your $R_2$ is not transitive, and your $R_3$ is not symmetric, and there's a perfectly good answer to this question here already anyway – Ben Millwood Sep 05 '13 at 22:29
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@BenMillwood Why is $R_2$ not transitive? It's $1\Rightarrow 2\Rightarrow 3$ converted to set notation... – AlexR Sep 05 '13 at 22:31
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Sorry, deleted my comment because I realised I should have just fixed it myself. – Ben Millwood Sep 05 '13 at 22:34
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Suppose $S$ is the set of all integers and $n$ be some fixed integer. Take the relation on $S$, that $x$ is related to $y$ if $y$ when divided by $n$ leaves the same remainder as $x$ leaves when divided by $n$. i.e. $$x \sim y :\Leftrightarrow x \equiv y \quad ({\rm Mod}\ n)$$
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This gives an equivalence relation, which is none of the things asked for. – Ben Millwood Sep 05 '13 at 22:30