Check all properties that the following relations on the set of positive integers have. The…

Check all properties that the following relations on the set of positive integers have. The relations are defined by (x, y) in R(x, y) \ in R(x, y) in R if and only if the indicated condition is satisfied.(a) x and y have a common factor

Analysis of Relation: "x and y have a common factor"

Let's analyze the relation R defined as: (x,y) ∈ R if and only if x and y have a common factor.

## Properties to check:

1. **Reflexive**: ∀x, (x,x) ∈ R

2. **Symmetric**: If (x,y) ∈ R, then (y,x) ∈ R

3. **Antisymmetric**: If (x,y) ∈ R and (y,x) ∈ R, then x = y

4. **Transitive**: If (x,y) ∈ R and (y,z) ∈ R, then (x,z) ∈ R

## Analysis:

### Reflexive? YES

- For any positive integer x, x is a factor of itself

- Therefore, (x,x) ∈ R for all x

- The relation is reflexive

### Symmetric? YES

- If x and y have a common factor, then y and x have the same common factor

- If (x,y) ∈ R, then (y,x) ∈ R

- The relation is symmetric

### Antisymmetric? NO

- Consider x = 6 and y = 9

- Both have common factor 3

- So (6,9) ∈ R and (9,6) ∈ R, but 6 ≠ 9

- The relation is not antisymmetric

### Transitive? NO

- Consider x = 6, y = 10, z = 15

- 6 and 10 have common factor 2, so (6,10) ∈ R

- 10 and 15 have common factor 5, so (10,15) ∈ R

- But 6 and 15 have common factor 3, not because of the previous relationships

- We can find counterexamples: x = 6, y = 4, z = 10

 - 6 and 4 have common factor 2, so (6,4) ∈ R

 - 4 and 10 have common factor 2, so (4,10) ∈ R

 - But 6 and 10 have common factor 2 as well, so (6,10) ∈ R

- Actually, this example shows transitivity holds in this case

- Let's try another: x = 2, y = 6, z = 9

 - 2 and 6 have common factor 2, so (2,6) ∈ R

 - 6 and 9 have common factor 3, so (6,9) ∈ R

 - But 2 and 9 have no common factor (except 1, which all positive integers share)

 - So (2,9) ∉ R

- The relation is not transitive

## Conclusion:

The relation "x and y have a common factor" is:

- Reflexive: YES

- Symmetric: YES

- Antisymmetric: NO

- Transitive: NO

This makes it a **non-equivalence relation** (since it's not transitive).