Proving p is an equivalence relation on AxB

  • Thread starter Thread starter Natasha1
  • Start date Start date
  • Tags Tags
    Relations Sets
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
2 replies · 2K views
Natasha1
Messages
494
Reaction score
9
Let AxB be the set of ordered pairs (a,b) where a and b belong to the set of natural numbers N.

A relation p: AxB -> AxB is defined by: (a,b)p(c,d) <-> a+d = b+c

(i) Is (2,6) related to (4,8)? Give three ordered pairs which are related to (2,6)

ANSWER:

Yes (2,6)p(4,8) as 2+8 = 6+4 = 10

(2,6)p(2,6) = 2+6 = 6+2 = 8
(2,6)p(6,10) = 2+10 = 6+6 = 12
(2,6)p(12,16) = 2+16 = 6+12 = 18

OR should it be written like this as the question stipulates ordered pairs related to (2,6) i.e. p(2,6)

(2,6)p(2,6) = 2+6 = 6+2 = 8
(6,10)p(2,6) = 6+6 = 2+10 = 12
(12,16)p(2,6) = 6+12 = 2+16 = 18

ANY SUGGESTIONS?

(ii) Is (2,6) related to (a, a+4)? Justify your answer

ANSWER

A = 2, B= 6, C= a and d = a+4

as (a,b)p(c,d) <-> a+d = b+c

Then a+d = 2+a+4 = 6+a and b+c = 6+a

Therefore a+d = b+c and (a,b)p(c,d)
Hence (2,6) is related to (a, a+4)

CORRECT, ANY IMPROVEMENT?


(iii) Prove (i.e. using general letters) that p is an equivalence relation on the set AxB

An equivalence relation on a set X is a binary relation on X that is reflexive, symmetric and transitive.

a) This set AxB is reflexive as any ordered pair is related to itself i.e. (a,b)p(a,b) <-> a+b = b+a
i.e. (2,6)p(2,6) = 2+6 = 6+2 = 8

b) This set AxB is symmetric as if we take any 2 ordered pairs (a,b), (c,d) then (a,b)p(c,d), hence a+d = b+c. And then (c,d)p(a,b), hence c+b = d+a.

i.e. (2,6)p(10, 14) we get 2+14 = 6+10 = 16
and (10,14)p(2,6) we get 6+10 = 2+14 = 16

c)This set AxB is transitive as if we take 3 ordered pairs (a,b), (c,d) and (e,f) we have (a,b)p(c,d), hence a+d = b+c and (c,d)p(e,f), then c+f = d+e, which gives (a,b)p(e,f) hence a+f = b+e

i.e. (2,6), (10,14) and (16,20)

(2,6)p(10,14) we get 2+14 = 6+10 = 16
and (10,14)p(16,20) we get 10+20 = 14+16 = 30

Then (2,6)p(16,20) we get 2+20 = 6+16 = 22

ANY SUGGESTIONS FOR IMPROVEMENT?
 

Discussion

Physics news on Phys.org
Natasha1 said:
c)This set AxB is transitive as if we take 3 ordered pairs (a,b), (c,d) and (e,f) we have (a,b)p(c,d), hence a+d = b+c and (c,d)p(e,f), then c+f = d+e, which gives (a,b)p(e,f) hence a+f = b+e
There doesn't seem to be a logical implication here. How do the previous lines imply the last phrase, that (a,b)p(e,f) ?
 
(i) and (ii) look good. Incidentally, ApB is read as "A is related to B". (at least when p doesn't need to be stated explicitly)


For (iii), I think you're close, but not quite there conceptually.

First off, you made a notational mistake: your goal is to prove that p is reflexive, symmetric, and transitive.

Secondly, why are you saying AxB instead of NxN?

Finally, you're slightly off in how to prove this. For example, in part (b), you never proved p is symmetric. All you've said is:

(a,b) p (c,d) implies a+d=b+c
(c,d) p (a,b) implies c+b=d+a

(Actually, you also said that (a,b)p(c,d) for any two ordered pairs.)

In particular, you've not said that (a,b)p(c,d) if and only if (c,d)p(a,b), which is what you're trying to prove.