2
$\begingroup$

let S = {1,2,3,4}

Explain why each of the below are not equivalence relation.

{ (1,1), (1,2), (2,1), (2,2), (3,3) }

{ (1,1), (1,2), (2,3), (1,3), (2,2), (3,3), (4,4) }

{ (1,1), (2,2), (3,3), (4,4), (2,3), (3,2), (2,4), (4,2)}

I am having difficulty trying to understand the 3 condition

  1. Reflective
  2. Symmetric
  3. Transitive

Would appreciate if anyone would to provide me with the guidance.

Thanks

$\endgroup$
  • 1
    $\begingroup$ so do you mean that for 2nd set, (1,2) & (2,1) is not within the set and hence not symmetric? $\endgroup$ – Tapwater May 11 '14 at 8:50
1
$\begingroup$
  1. The first one is surely not an equivalence relation as it is not reflective: $(4,4)$ does not belong, this is $4$~$4$ is false, this is $x$~$x$ for all $x$ is false.

  2. The second one is exactly as you wrote in your comment: $(1,2)$ belongs to the relation, $(2,1)$ does not, hence symmetric property fails.

  3. Third: we have the pairs $(3,2)$ and $(2,4)$, but not $(3,4)$. Hence is not true that
    $x$~$y\ \&\ y$~$z\ \Rightarrow\ x$~$z\quad \forall\,x,y,z$ (not transitive).

All cool about it? ;)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.