Part II: Relations 1. For each of the following relations, determine where the r

Question

Part II: Relations 1. For each of the following relations, determine where the relation is: . Reflexive Anti-reflexive Anti-symmetric · Transitive Justify you answers, a. R is a relation on the set {1,2,3,4} such that R = b. R is a relation on the set of all people such that (a, b) E R if and only if a and b have a common grandparent c. R is a relation on the power set of a set A such that (x,y) E R if and only if x c y. d. R is a relation on Z such that (x,y) E R if and only if x y e. R is a relation of Z+ such that (x,y) e R if and only if y is divisible by x. Hint: And integer y is divisible by an integer x with x integer k such that y = xk. 0 if and only if there exists an

Explanation / Answer

a.)

Given the set S= {1,2,3,4}

and the relation R= {(1,1),(1,3),(2,2),(3,1),(3,3),(4,4)}

Reflexive:

A relation is reflexive on the set S, if a S then (a,a) R or say aRa (means a is related to itself by the relation R)

here (1,1) R, (2,2) R, (3,3) R, (4,4) R so the Relation R is reflexive on the set S

As it is reflexive so it will not be the anti-reflexive

A relation is anti-symmetric on the set S,only if aRb then b~Ra (i.e. bis not related to a by relation R) should be true, Above relation is not anti-symmetric as 1R3 and but also 3R1.

A relation is transitive on the set S, only when aRb and bRc implies aRc means (a,b) R and (b,c) R then

(a,c) R should also be true.

Here the relation is transitive as (1,3) R and (3,1) R and also (1,1) R.

b) a R a for this relation will be true because obviously the same person will have common grandparent.

So it is Reflexive, so not anti-reflexive. It will be symmetric too, as if grandparent of a is the grandparent of b then vice versa also will be true. So, it is not anti-symmetric.

It will be transitive too as, if grandparent of a is the grandparent of b and grandparent of b is the grandparent of c then grandparent of a is also the grandparent of c. So it is transitive.

c)Sorry I couldn't solve this question

d)This relation will be anti-reflexive as a number always will equal to itself so (a,a) R, so it is not reflexive.

Let ab then it is guaranteed that ba, so the relation is symmetric, it is not anti-symmetric.

It is not transitive as let ab and bc but it doesn't imply that ac, but a can be equal to c, e.g. 13 and 31 but 1=1, so It is not transitive.

e) Given the relation R: x is divisible by y (obviously y can not be zero) or we can say that (x,y) R if x%y=0 (where % is the modulus operator which gives remainder when x is divided by y)

The relation will be reflexive as x%x=0 always true, so (x,x)R and the relation R is reflexive it is not anti-reflexive.

It will not be symmetric as if x%y=0 does not implies that y%x=0 e.g. 25%5=0 but 5%25=5 so it is anti-symmetric.

Let (x,y)R means x=k1*y where k1 is a constant and let (y,z)R means y=k2*z where k2 is another constant, substituting the y in first equation by second equation we get, x=k1k2*z or (x,z)R, so the given relation is transitive.

//If you have any doubt regarding the answer please ask in the comment section

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