Solution to For each of the following relations, determine whether they are reflexive, symmetric, anti- symmetric, and/or … - Sikademy
Author Image

Archangel Macsika

For each of the following relations, determine whether they are reflexive, symmetric, anti- symmetric, and/or transitive, and give a brief justification for each property. a) R ⊆ Z × Z where xRy iff x = y 2 b) The empty relation: R ⊆ A × A, where A is a non-empty set and R = ∅.

The Answer to the Question
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

For each of the following relations, let us determine whether they are reflexive, symmetric, anti- symmetric, and/or transitive.


a) R ⊆ \Z × \Z where xRy iff x = y^ 2

Since 2^2=4\ne 2, we conclude that the pair (2,2)\notin R, and hence R is not a reflexive relation.


Since 4=2^2, we conclude (4,2)\in R. On the other hand, 2\ne 4^2, and hence (2,4)\notin R. It follows that the relation is not symmetric.


If (x,y)\in R and (y,x)\in R, then x=y^2 and y=x^2. Therefore, x=x^4, and hence x=0 or x=1. If x=0, then y=0. If x=1, then y=1. It follows that If (x,y)\in R and (y,x)\in R,

then x=y=0 or x=y=1, and hence the relation is antisymmetric.


Taking into account that (16,4)\in R and (4,2)\in R, but (16,2)\notin R, we conclude that the relation R is not transitive.


b) The empty relation: R ⊆ A × A , where A is a non-empty set and R = ∅.

Let a\in A. Since (a,a)\notin\emptyset= R, we conclude that the relation R is not reflexive.


Since the statement [(x,y)\in\emptyset= R] is false, the implication [(x,y)\in R]\to[(y,x)\in R] is true, and hence this relation is symmetric.


Taking into account that the statement [(x,y)\in R\text{ and }(y,x)\in R] is false, the implication [(x,y)\in R\text{ and }(y,x)\in R]\to [x=y] is true, and hence this relation is antisymmetric.


Since the statement [(x,y)\in R\text{ and }(y,z)\in R] is false, the implication [(x,y)\in R\text{ and }(y,z)\in R]\to [(x,z)\in R] is true, and hence this relation is transitive.


Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-5-stid-8-sqid-1277-qpid-1015