edited by
212 views
1 votes
1 votes

$$
R=\left\{(x, y) \in \mathbb{N}^2: \exists n \in \mathbb{N}, x^n=y\right\}
$$
is a binary relation on the set of natural numbers $\mathbb{N}$.
Determine which of the following properties R satisfies:

  1. Reflexive
  2. Symmetric
  3. Anti-symmetric
  4. Transitive

For each property, either justify that the property always holds or show by a counterexample that the property does not hold.

edited by

1 Answer

0 votes
0 votes
A. Reflexive:

 

A binary relation R on a set S is reflexive if every element of S is related to itself. In other words, for all x ∈ S, (x, x) ∈ R.

 

In the given relation R, we need to check if every natural number x is related to itself, i.e., (x, x) ∈ R.

 

Let's consider an example:

For x = 2, we have (2^1, 2) = (2, 2) ∈ R since 2^1 = 2.

 

Therefore, for any natural number x, we can always find a power n (in this case, n = 1) such that x^n = x, which means (x, x) ∈ R. Hence, R is reflexive.

 

B. Symmetric:

 

A binary relation R on a set S is symmetric if for every (x, y) ∈ R, (y, x) ∈ R.

 

Let's consider an example to check symmetry:

For x = 2 and y = 4, we have (2^2, 4) = (4, 4) ∈ R since 2^2 = 4.

 

However, (4^1, 2) = (4, 2) ∉ R since 4^1 ≠ 2. Thus, (2, 4) ∈ R, but (4, 2) ∉ R.

 

Therefore, the relation R is not symmetric since there exists at least one pair (x, y) ∈ R where (y, x) ∉ R.

 

C. Anti-symmetric:

 

A binary relation R on a set S is anti-symmetric if for any distinct elements x and y in S, if (x, y) ∈ R and (y, x) ∈ R, then x = y.

 

To check anti-symmetry, let's consider an example:

For x = 2 and y = 4, we have (2^2, 4) = (4, 4) ∈ R since 2^2 = 4.

 

However, (4^1, 2) = (4, 2) ∉ R since 4^1 ≠ 2.

 

Since (2, 4) ∈ R and (4, 2) ∉ R, and x ≠ y (x = 2 and y = 4), we can see that the relation R is not anti-symmetric.

 

D. Transitive:

 

A binary relation R on a set S is transitive if for any elements x, y, and z in S, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R.

 

Let's consider an example to check transitivity:

For x = 2, y = 4, and z = 16, we have (2^2, 4) = (4, 4) ∈ R since 2^2 = 4.

Also, (4^2, 16) = (16, 16) ∈ R since 4^2 = 16.

 

Thus, (2, 4) ∈ R and (4, 16) ∈ R, and we can see that (2, 16) ∈ R since 2^4 = 16.

 

Therefore, the relation R is transitive.

 

Summary:

 

A. Reflexive: R is reflexive.

B. Symmetric: R is not symmetric.

C. Anti-symmetric: R is not anti-symmetric.

D. Transitive: R is transitive.

Related questions

1 votes
1 votes
1 answer
2