Question: Let A = {1,2,3,4,} and R = { ( x ,y ) E A x A : x + 1 = y}.

(i) Represent R in numerical form as a set of ordered pairs.
(ii) Show that R is not an equivalence relation of
(iii) Determine the transitive closure of R

My question is, what do I do with R, the x+1 confuses me. I know if r was = {1, 2, 3, 4, 5} The ordered pairs would be {(1, 1), (1, 2), (1,3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5)}
Am I right?

Question

Question: Let A = {1,2,3,4,} and R = { ( x ,y ) E A x A : x + 1 = y}.

(i) Represent R in numerical form as a set of ordered pairs.
(ii) Show that R is not an equivalence relation of 
(iii) Determine the transitive closure of R

My question is, what do I do with R, the x+1 confuses me. I  know if r was = {1, 2, 3, 4, 5} The ordered pairs would be {(1, 1), (1, 2), (1,3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5)}
Am I right?

watchmath · Answer

Not quite
Here is one member of $R$: $(3,4$). Why this is a member of $R$? Because for $x=3,y=4$ we have $x+1=y$. Make sense?

anonymous · Answer

Oh I think I get it. So would it be R= {(1,2), (2,3), (3, 4), (4, 5)} ?

watchmath · Answer

Perfect!! :D. You deserve a medal!!

anonymous · Answer

Haha thanks so much =D I have an exam tomorrow and you helped me so much thank you!