Question

Is \(R\) an equivalence relation on the set \(J\) of all integers, and, if so, what are the \(R\) -classes, if (a) \(R=\\{(x, y) \mid x-y\) is divisible by a fixed \(n\\}\) (b) \(R=\\{(x, y) \mid x-y\) is \(o d d\\}\) (c) \(R=\\{(x, y) \mid x-y\) is a prime \(\\}\). \((x, y, n\) denote integers.)

Short answer

(a) Yes, congruence classes mod \(n\); (b) No; (c) No.

Step-by-step solution

Understanding equivalence relation

An equivalence relation on a set must satisfy three properties: reflexivity, symmetry, and transitivity. We will check these properties for each given relation \(R\).

Analyzing part (a)

*Relation*: \(R = \{(x, y) \mid x-y \text{ is divisible by a fixed } n\}\). 1. **Reflexivity**: For any integer \(x\), \(x-x = 0\), which is divisible by \(n\). Thus, \((x, x) \in R\). Reflexivity holds.2. **Symmetry**: If \((x, y) \in R\), then \(x-y\) is divisible by \(n\). Hence, \(y-x\) is also divisible by \(n\) (as it is just \(-(x-y)\)), which means \((y, x) \in R\). Symmetry holds.3. **Transitivity**: If \((x, y) \in R\) and \((y, z) \in R\), then \(x-y\) and \(y-z\) are divisible by \(n\). Thus, \(x-z = (x-y) + (y-z)\) is also divisible by \(n\). Hence, \((x, z) \in R\). Transitivity holds. Since all three properties hold, \(R\) is an equivalence relation for part (a). The \(R\)-classes are the sets of integers that differ by multiples of \(n\), or equivalently, the congruence classes mod \(n\).

Analyzing part (b)

*Relation*: \(R = \{(x, y) \mid x-y \text{ is odd}\}\). 1. **Reflexivity**: For any integer \(x\), \(x-x = 0\), which is not odd. So \((x, x) otin R\). Reflexivity fails.Since reflexivity does not hold, \(R\) is not an equivalence relation for part (b).

Analyzing part (c)

*Relation*: \(R = \{(x, y) \mid x-y \text{ is a prime number}\}\).1. **Reflexivity**: For any integer \(x\), \(x-x = 0\), which is not a prime number. Hence, \((x, x) otin R\). Reflexivity fails.Without reflexivity, there is no need to check the other properties. \(R\) is not an equivalence relation for part (c).

Key concepts

Reflexivity

To understand reflexivity, think about a property where each element relates to itself. This is a critical property in determining whether a relation is an equivalence relation.

For a relation to be reflexive, every element in the set must pair with itself. Mathematically, for a set \(A\), a relation \(R\) is reflexive if \((a, a) \in R\) for every \(a \in A\).

• In part (a), the relation \(R = \{(x, y) \mid x-y \text{ is divisible by a fixed } n\}\) is reflexive because for any integer \(x\), the expression \(x - x = 0\) is divisible by any \(n\).
• In part (b), the relation \(R = \{(x, y) \mid x-y \text{ is odd}\}\) isn't reflexive. Any integer minus itself is zero, which is not odd.
• Similarly, in part (c) with \(R = \{(x, y) \mid x-y \text{ is a prime number}\}\), we find that the relative difference \(x-x = 0\) is not a prime number.

Without reflexivity, a relation can't be considered an equivalence relation.

Symmetry

Symmetry in a relation implies that if one element is related to another, then the second is related back to the first. Imagine opening a door for someone, and then they open it for you the next time.

Formally, a relation \(R\) is symmetric if for any elements \(a\) and \(b\) in the set \(A\), whenever \((a, b) \in R\) then \((b, a) \in R\) as well.

• In part (a), the relation holds this property. If \((x, y) \in R\) meaning \(x-y\) is divisible by \(n\), then \(y-x\), which is \(-(x-y)\), is also divisible by \(n\). This confirms symmetry.
• In parts (b) and (c), symmetry is not tested since reflexivity already failed, disqualifying them from being equivalence relations.

Symmetry ensures a mutual relationship between any two elements, playing a key role in equivalence relations.

Transitivity

Transitivity is about creating a bridge between linked elements. If an initial element is related to a second, and this second is related to a third, transitivity requires that the initial is connected to the third.

In formal terms, a relation \(R\) over a set \(A\) is transitive if for any elements \(a, b, c \in A\), whenever \((a, b) \in R\) and \((b, c) \in R\), then \((a, c) \in R\) should also be in \(R\).

• For part (a), this holds true. If \(x-y\) and \(y-z\) are divisible by a fixed \(n\), then their sum \(x-z = (x-y) + (y-z)\) must also be divisible by \(n\), ensuring that \((x, z) \in R\).
• Just like before, parts (b) and (c) are skipped from further testing for equivalence due to failing reflexivity.

Transitivity helps maintain a consistent connection between sequences of elements, making it crucial for establishing equivalence relations.