Let AxB be the set of ordered pairs (a,b) where a and b belong to the set of natural numbers N.(adsbygoogle = window.adsbygoogle || []).push({});

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 reflexiveas 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 symmetricas 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 transitiveas 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?

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# Homework Help: Sets and Relations (just needs checking please)

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