DISCRETE MATH Verify whether the given binary relation R on the set A is an equi

Question

DISCRETE MATH

Verify whether the given binary relation R on the set A is an equivalence relation: Let A = {1, 2, 3, 4, 5} and let R be the binary relation whose elements are the pairs (1, 1), (1, 3), (1, 4), (2, 2), (2, 5), (3, 1), (3, 3), (3, 4), (4, 1), (4, 3), (4, 4), (5, 2), (5, 5), and only these. Let A be the set Z of integer numbers and let R be the relation given by: R = {(n, m): [[n elementof Z logicaland m elementof Z] logicaland [there exists k elementof Z: n - m = 8k]]}. (Notice that R may be equivalently stated by setting: R = {(n, m) elementof Z times Z: [there exists k elementof Z: n - m = 8k]}.) Let A be the set Q of rational numbers and let R be the relation defined by: R = {(p/q, s/r) elementof Q times Q: [pr = qs]}. Remind that every rational number can be expressed as a function of integer numbers p/q, such that q notequalto 0. Let A be the set N natural numbers and let R be the relation given by: R = {(n, m) elementof N times N: (m = 2^n]}. Let A = {} and let R be the relation given by: R = {}. Let A be the class of all the possible sets, and let R be the binary relation given by: R = {(S, Q): [[S elementof A logicaland Q elementof A] logicaland [there exists f: S rightarrow Q bijective function]]}. If you found an affirmative answer in 1.(a), then find quotient set A/R.

Explanation / Answer

For equivalence relation, we must have a relation reflexive, symmetric and transitive all.

So for reflexive : aRa or (a,a) should be in R

and in given relation it is as (1,1),(2,2),(3,3), (4,4) and (5,5) are in R, so it is reflexive.

Further for symmetry if aRb then bRa

And pairs (1,3),(3,1),(2,5),(5,2) etc, show that symmetry is there.

Further for transitive, we have if aRb and bRc, then aR c

But here we do not find any such relation of three pairs that follow above rule, so it is not transitive and hence it is not equivalence.

This is the answer of part (a)

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Again as n-m =8k

so clearly n-n = 0 = 8(0) , so it is reflexive.

Further if n-m = 8k, then m-n = 8k ( or a multiple of 8 also where k is an integer)

So clearly based on above equation, it is symmetric also.

And lastly if n-m = 8k and m-p = 8d, ( where d is any integer) then

n-(8d+p) = 8k

or n-p = 8k+8d =8(k+d) = 8 ( an integer)

So clearly transitivity also hold here.

Thus R is equivalence here.

This is the answer of part (b).

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