3

How many reflexive relation are there on a set with n elements ?

Git Gud
  • 31,706
Salman
  • 41
  • 1
    In here where I study maths, it's defined that a relation is a subset of Decartes product $A\times A$. So a reflexive relation is a subset contains all $(a,a)$. There are remaining $2^{n^2-n}$ ways to construct a subset – anonymous67 Nov 01 '14 at 13:10
  • for symmetric and transitive and antisymmetric ? – Salman Nov 01 '14 at 13:14
  • @Salman If you keep pumping for more answers to other questions, I'll flag. If you have more questions, post as a new question. – amWhy Nov 01 '14 at 13:17
  • Symmetric means if there is $(a,b)$ in the subset, there should also be $(b,a)$ (if $a\ne b$). Then you should see how many choice you have. ($2^n$ for $(a,a)$ and something divide $2$ for $(a,b)$) Transitive is a little bit more tricky I think. But you should work by yourself and see what happen. Then you should ask by pointing out where you are stuck. – anonymous67 Nov 01 '14 at 13:19

1 Answers1

11

Let $A$ be this set. Write $A\times A$ in a grid. If the square $(j,k)$ is marked, that means that $jRk$.

Since the relation is reflexive you must mark the main diagonal. They are $n$ squares. Now the question is: 'How many ways are there to mark or not the other squares of the grid?'. There are $n^2-n$ squares, and there are $2$ possibilities for each. Then, there are $$2^{n^2-n}$$ possibilities.

ajotatxe
  • 66,849
  • for symmetric and transitive and antisymmetric ? – Salman Nov 01 '14 at 13:13
  • 3
    @Salman Don't get greedy, pumping for more answers to other questions. Your question was answered, and very well, at that. If you have another question, post it as a new question. – amWhy Nov 01 '14 at 13:15