I am trying to answer the following question:

R, S, T: 0...10 ⟺ 0...10

R= {m,n:N∥m≠n∧m+n=8}

S= {r:R∥r.1 < r.2}


Give the reflexive, symmetric and transitive closures for R, S and T

For R I have: R = {(0,8),(1,7),(2,6),(3,5),(5,3),(6,2),(1,7),(8,0)}

Therefore Reflexive Closure: R∪{(0,0),(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8),(9,9),(10,10)}

Symmetric Closure: {(0,8),(1,7),(2,6),(3,5),(5,3),(6,2),(7,1),(8,0)}

Transitive Closure: R∪{(0,0),(1,1),(2,2),(3,3),(5,5),(6,6),(7,7),(8,8)}

Now, I am trying to establish the same for S and T which are based on R

For S, I have:

S = {(0,8),(1,7),(2,6),(3,5)}

Reflexive Closure: SU{(0,0),(1.1),(2.2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8),(9,9),(10,10)}


Symmetric Closure: {(0,8),(1,7),(2,6),(3,5),(5,3),(6,2),(7,1),(8,0)}

Transitive Closure: {(0,8),(1,7),(2,6),(3,5)}

Can you please let me know if this is correct so far. Also, what do you think the outcome for T would be?

  • $\begingroup$ What do you mean with "R, S, T: 0...10 ⟺ 0...10"? $\endgroup$ – md2perpe Jun 17 '17 at 10:29
  • $\begingroup$ Its a three pronged question. i.e.: R:0..10⟺0..10 R={m,n:N∥m≠n∧m+n=8} S:0..10⟺0..10 S= {r:R∥r.1 < r.2} T:0..10⟺0..10 T=R;S $\endgroup$ – user450432 Jun 17 '17 at 10:30
  • $\begingroup$ I don't understand the definiton of T ... Also, to show your work, cN you pleas just tell us what you have for $R$, $S$, and $T$ before prociding any kind of closure? $\endgroup$ – Bram28 Jun 17 '17 at 10:31
  • $\begingroup$ I've never seen the notation "0...10 ⟺ 0...10", but I suppose that it means "relation between $\{ 0, \ldots, 10 \}$ and $\{ 0, \ldots, 10 \}$". $\endgroup$ – md2perpe Jun 17 '17 at 10:35
  • $\begingroup$ Is $T$ the composition of $R$ and $S$, i.e. $xTy$ if there exists $z$ such that $xRz$ and $zSy$? $\endgroup$ – md2perpe Jun 17 '17 at 10:36

Any clsure (whether reflexive, symmetric, or transitive) will have to contain the relation itself ... You are only indicating the pairs that are being added.

Also, I think $S$ is defined as those pars from $R$ where the first number of the pair is smaller han the second. You must have interpreted it differently, otherwise you would not be adding elements like $(1,2)$ or $(3,6)$ to the symmetric closure of $S$

$T$ should be the composition of $R$ and $T$, so this should contain all $(a,c)$ such that there is some $b$ such that $(a,b) \in R$ and $(b,c) \in S$. For example, since $(8,0) \in R$, and $(0,8) \in S$, $(8,8) \in T$.

  • $\begingroup$ Thanks for clarifying, I definitely interpreted incorrectly. In other words you're saying it should be: {(0,8),(1.7),(2,6),(3,5)} $\endgroup$ – user450432 Jun 17 '17 at 10:59
  • $\begingroup$ @KingGeorge Exactly! I see you updated your post, and your $R$ and $S$ are correct. Also, your closures are all correct as well, except for the reflexive closure of $S$, which should be $S$ together with $(0,0),(1,1)$, etc. Note that the symmetric closure of $R$ is $R$ itself, since $R$ is symmetric already. let me see if I can help you out with $T$... $\endgroup$ – Bram28 Jun 17 '17 at 18:23
  • $\begingroup$ @KingGeorge OK, that weird semicolon-like symbol is indeed for relational composition ... See the Wikipedia article on Composition of Relations. in my Answer I explain what this means, and give an example of what you should be putting into $T$ ... Can you now figure out $T$? $\endgroup$ – Bram28 Jun 17 '17 at 18:32
  • $\begingroup$ Thanks for this. you are right, I have now updated my reflexive closure for S. I think I repeated the elements of S by accident $\endgroup$ – user450432 Jun 18 '17 at 22:31
  • $\begingroup$ @KingGeorge OK, that looks good, though I am not sure what that set is that is listed between the reflexive closure of $S$ and the symmetric closure of $S$. Note that the symmetric closure of $S$ is $R$. Have you made any progress on $T$? $\endgroup$ – Bram28 Jun 18 '17 at 22:34

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