Question

Let \( R \) be a relation defined on a set \( \mathbb{N} \) of natural numbers such that \( R = \{(x, y) : xy \text{ is a square of a natural number}, x, y \in \mathbb{N} \} \). Determine if the relation \( R \) is an equivalence relation.

Hint:

Quick Tip: For checking symmetry, remember that the relation \( xy = k^2 \) implies \( yx = k^2 \), satisfying the symmetry condition. For transitivity, check if multiplying the two conditions leads to a square. In this case, the absence of a guaranteed square for transitivity means the relation fails to be transitive.

Answer 1

To determine if the relation \( R \) is an equivalence relation, we need to check if it satisfies the three properties of equivalence relations: reflexivity, symmetry, and transitivity.
1. Reflexivity:
A relation is reflexive if for every element \( x \in \mathbb{N} \), \( (x, x) \in R \).
For \( (x, x) \) to be in \( R \), we need \( x \cdot x = x^2 \) to be a square of a natural number. Clearly, \( x^2 \) is always a square for any \( x \in \mathbb{N} \). Hence, the relation \( R \) is reflexive.
2. Symmetry:
A relation is symmetric if for every pair \( (x, y) \in R \), \( (y, x) \) must also be in \( R \).
Let \( (x, y) \in R \), so \( x \cdot y = k^2 \) for some \( k \in \mathbb{N} \). Since multiplication is commutative, we have \( y \cdot x = x \cdot y = k^2 \), which is a square. Therefore, \( (y, x) \in R \). Hence, the relation \( R \) is symmetric.
3. Transitivity:
A relation is transitive if whenever \( (x, y) \in R \) and \( (y, z) \in R \), we have \( (x, z) \in R \).
Suppose \( (x, y) \in R \) and \( (y, z) \in R \). This means: \[ x \cdot y = k_1^2 \quad \text{and} \quad y \cdot z = k_2^2 \] Multiplying these two equations, we get: \[ (x \cdot y) \cdot (y \cdot z) = k_1^2 \cdot k_2^2 \] \[ x \cdot y^2 \cdot z = (k_1 k_2)^2 \] For \( x \cdot z \) to be a square, we need \( y^2 \) to be a perfect square, which is not necessarily true. Hence, the relation is not transitive.
Conclusion: Since the relation \( R \) is reflexive and symmetric but not transitive, it is not an equivalence relation.