Question

Define a relation \(R\) on \(\mathbb{Z}\) as \(x R y\) if and only if \(4 \mid(x+3 y) .\) Prove \(R\) is an equivalence relation. Describe its equivalence classes.

Short answer

The relation \(R\) is an equivalence relation. The equivalence classes of \(R\) are formed by grouping integers such that within each class \(4|(m+3n)\) holds true.

Step-by-step solution

Demonstrate Reflexivity

To prove reflexivity, we need to show that for every integer \(x \in \mathbb{Z}\), we have \(x R x\). The given relation indicates \(x R x\) if and only if \(4|(x+3x) = 4x\). By definition, every integer is a multiple of 4x for any integer \(x\), so \(x R x\) holds true. Thus, the relation is reflexive.

Prove Symmetry

To demonstrate symmetry, we need to prove that for any two integers \(x, y \in \mathbb{Z}\), if \(x R y\) then \(y R x\). From the given relation, if \(x R y\) then \(4|(x + 3y)\). This means there exists an integer \(k\), such that \(x + 3y = 4k\). Rearranging the equation, we get \(x = 4k - 3y\). This shows that 4 divides \(y + 3x\) , hence \(y R x\) holds true. Therefore, the relation \(R\) is symmetric.

Verify Transitivity

To prove that this relation is transitive, we must prove that for all integers \(x, y, z \in \mathbb{Z}\), if \(x R y\) and \(y R z\), then \(x R z\) holds. Assuming \(x R y\) and \(y R z\), we have \(4|(x + 3y)\) and \(4|(y + 3z)\). By adding these together, we have \(4|(x + 3y + y + 3z) \). This simplifies to \(4|(x + 3z)\), which is the definition for \(x R z\). Therefore, the relation \(R\) is transitive.

Describe Equivalence Classes

For integer \(n \in \mathbb{Z}\), the equivalence class of \(n\) under \(R\) is defined as \([n] = \{m \in \mathbb{Z} | m R n\}\). From the relation definition, \(m R n\) holds if and only if \(4|(m+3n)\). Thus, the equivalence class of \(n\) under \(R\) is the set \([n] = \{m \in \mathbb{Z} | 4|(m+3n)\}\).

Key concepts

Reflexivity

Understanding the concept of reflexivity is crucial when dealing with equivalence relations in mathematics. In simpler terms, reflexivity implies that every element is related to itself. For example, when examining our set of integers \(\mathbb{Z}\), to show that a relation \(R\) is reflexive, we have to demonstrate that each integer \(x\) is related to itself under \(R\). In the provided exercise, this was proven by showing \(x R x\) holds true for all \(x\), as \(4\) divides \(4x\), which is always the case for any integer \(x\).

Think of reflexivity as looking in the mirror. Just as you always see your reflection, in a reflexive relation, every element 'sees' itself within that relation.

Symmetry

Symmetry in an equivalence relation is akin to a two-way mirror effect: if you can see your friend through the mirror, they can see you too. To prove symmetry in our integer set, if one integer \(x\) relates to another \(y\) under \(R\), then \(y\) should also relate to \(x\) in the same way. This was shown in the solution by rearranging \(x + 3y = 4k\) to exhibit that \(4\) also divides \(y + 3x\), which confirms the symmetry of relation \(R\).

Always remember that symmetry in mathematics doesn't involve physical flips or reflections, but rather the reversibility of a relationship between elements.

Transitivity

To grasp the concept of transitivity in an equivalence relation, think of it as a 'domino effect.' If the first integer \(x\) is related to the second \(y\), and \(y\) is in turn related to a third integer \(z\), then transitivity ensures that the first integer \(x\) is also related to the third \(z\). The solution to our exercise demonstrates this by adding the two conditions \(x R y\) and \(y R z\) to eventually show that \(4\) divides \(x +3z\), hence confirming \(x R z\).

It's like a chain link: if the first link is attached to the second, and the second to the third, inevitably, the first and third link are connected.

Equivalence Classes

Equivalence classes are a fascinating feature of an equivalence relation and can be thought of as boxes categorizing elements that share a specific relationship. Each class contains all the elements that are related to each other under the given relation. In the context of our integer set and the relation \(R\), the equivalence class of an integer \(n\) includes all the integers \(m\) such that \(m R n\). The exercise defines this class as the set \( [n] = \(m \in \mathbb{Z} | 4|(m+3n)\) \) which are all the integers related to \(n\) by \(R\).

Imagine organizing books by genre; each equivalence class is like a bookshelf holding all books of the same genre together. The relation \(R\) determines how the 'books' (integers) are grouped into 'shelves' (equivalence classes).