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Working with functions, more specifically function composition. In my case, for starters,

We have an island with the places haven, dale, sun, and ness.

We have a smaller island with the places east, bent and big.

And a tiny island with the place smallest.

We have relations between different places represented like Relation ⊆ place × place

Place ={Little, east, bent, big, ness, haven, sun, dale}

The relation is represented like (this is how the places are connected/related by roads etc)

Relation ={(haven, dale),(haven, sun),(haven, ness),(dale, sun),(east, bent),(east, big)}

The islands are connected by ferry- represented like

Ferry ⊆ place × place

Specifically:

Ferry ={(ness, smallest),(smallest, haven),(haven, ness)}

Question: what is the result/way to solve

  1. Relation ∘ Ferry

  2. Ferry ∘ Relation

  3. Ferry−1∘ Relation

I have read up on composition of relations online and in books, but I don't understand it and I am stuck on how to solve these tasks/what the correct answer is. If someone could show me the correct answer in a simple way that would be highly appreciated.

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  • $\begingroup$ "odd" is not on your map? $\endgroup$ Commented May 10, 2018 at 13:51
  • $\begingroup$ @mol3574710n0fN074710n edited my typo, it was supposed to say ness not odd! $\endgroup$ Commented May 10, 2018 at 14:18

1 Answer 1

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Guide.

For finding Relation$\circ$Ferry start looking for an ordered pair $(a,b)$ in Ferry such that for some $c$ we have $(b,c)$ in Relation. That gives three questions associated with the three pairs in Ferry:

  • Do we have a pair $(\text{smallest},\dots)$ in Relation? No!

  • Do we have a pair $(\text{haven},\dots)$ in Relation? Yes! We can substitute $\text{dale}$, $\text{sun}$ and $\text{ness}$ for $\dots$ and the conclusion is that $(\text{smallest},\text{dale})$, $(\text{smallest},\text{sun})$ and $(\text{smallest},\text{ness})$ are elements of Relation$\circ$Ferry.

  • Do we have a pair $(\text{ness},\dots)$ in Relation? No!

Now we found: Relation$\circ$Ferry$=\{(\text{smallest},\text{dale}),(\text{smallest},\text{sun}),(\text{smallest},\text{ness})\}$

edit:

For finding Ferry$\circ$Relation start looking for an ordered pair $(a,b)$ in Relation such that for some $c$ we have $(b,c)$ in Ferry. That gives six questions associated with the six pairs in Relation:

  • Do we have a pair $(\text{dale},\dots)$ in Ferrry? No!
  • Do we have a pair $(\text{sun},\dots)$ in Ferry? No!
  • Do we have a pair $(\text{ness},\dots)$ in Ferry? Yes! We can substitute $\text{smallest}$ for $\dots$ and the conclusion is that $(\text{haven},\text{smallest})$ is an elements of Ferry$\circ$Relation.
  • Do we have a pair $(\text{sun},\dots)$ in Ferry? No!
  • Do we have a pair $(\text{bent},\dots)$ in Ferry? No!
  • Do we have a pair $(\text{big},\dots)$ in Ferry? No!

Now we found: Ferry$\circ$Relation$=\{(\text{haven},\text{smallest})\}$

For finding Ferry$^{-1}\circ$Relation do exactly the same thing but now not for Ferry but for Ferry$^{-1}$ which is the relation $\{(\text{smallest},\text{ness}),(\text{haven},\text{smallest}),(\text{ness},\text{haven})\}$

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  • $\begingroup$ thank you so much! but how do i go ahead with the other two? im sorry but i have autism and sometimes i simply dont understand it until someone explains the spesific task 100% like you just did. $\endgroup$ Commented May 10, 2018 at 15:45
  • $\begingroup$ I have extended in an edit. $\endgroup$ Commented May 10, 2018 at 17:36

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