anonymous
  • anonymous
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?
Mathematics
chestercat
  • chestercat
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watchmath
  • watchmath
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
  • anonymous
Oh I think I get it. So would it be R= {(1,2), (2,3), (3, 4), (4, 5)} ?
watchmath
  • watchmath
Perfect!! :D. You deserve a medal!!

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

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