Question

Consider the following relation R over the integers (Z).R={(a,b)∈Z²:∃ₖ∈ a=kb}.

a) Prove or disprove whether R is reflexive
b) Prove or disprove whether R is symmetric
c) Prove or disprove whether R is anti-symmetric
d) Prove or disprove whether R is transitive

Answer 1

The relation R is reflexive, not symmetric, anti-symmetric, and transitive.

The correct option is A.

Step-by-step explanation:

Properties of the Relation R

Let's examine each property in turn:

a) Reflexivity

A relation R is reflexive if every element is related to itself. In this case, for any integer a, we can choose k=1, which gives us a=1*a. Therefore, (a,a) is in R for all a in Z, proving that R is reflexive.

b) Symmetry

R is symmetric if whenever (a,b) is in R, then (b,a) is also in R. However, if a=kb for some k in Z, b does not necessarily equal ka for any k in Z. For example, if a=2 and b=1, then a=2*b, but b is not equal to any integer times a. Thus, R is not symmetric.

c) Anti-symmetry

R is anti-symmetric if whenever both (a,b) and (b,a) are in R, then a must equal b. Since we have k such that a=kb and if we assume b=la for some l in Z, then a=kla. If (a,b) and (b,a) are in R only when a=b, then (k,l)=(1,1), and thus R can be considered anti-symmetric.

d) Transitivity

R is transitive if whenever (a,b) and (b,c) is in R, then (a,c) must also be in R. If a=kb and b=lc, then a=k(lc)=(kl)c, which means (a,c) is in R. Therefore, R is transitive.

The correct option is A.