Let's solve and prove each part of this question.

Part (a): Show that $R$ is an equivalence relation on $\mathbb{Z}$.

To show that $R$ is an equivalence relation on $\mathbb{Z}$, we need to verify that $R$ satisfies the properties of an equivalence relation: reflexivity, symmetry, and transitivity.

Reflexivity

To show that $R$ is reflexive, we need to show that for any integer $a \in \mathbb{Z}$, $a \, R \, a$.

Since $a - a = 0$ and $0$ is an even number, it follows that $a \, R \, a$ for all $a \in \mathbb{Z}$. Thus, $R$ is reflexive.

Symmetry

To show that $R$ is symmetric, we need to show that if $a \, R \, b$, then $b \, R \, a$ for any integers $a$ and $b$.

If $a \, R \, b$, then $a - b$ is an even number. By definition of even numbers, $a - b = 2k$ for some integer $k$. Then: $b - a = -(a - b) = -2k = 2(-k),$ which is also an even number (since the negative of an even number is still even). Therefore, $b \, R \, a$, proving that $R$ is symmetric.

Transitivity

To show that $R$ is transitive, we need to show that if $a \, R \, b$ and $b \, R \, c$, then $a \, R \, c$ for any integers $a$, $b$, and $c$.

If $a \, R \, b$, then $a - b$ is even, so $a - b = 2m$ for some integer $m$. Similarly, if $b \, R \, c$, then $b - c = 2n$ for some integer $n$. Now consider $a - c$: $a - c = (a - b) + (b - c) = 2m + 2n = 2(m + n).$ Since $2(m + n)$ is an even number, $a - c$ is also even. Therefore, $a \, R \, c$, which proves that $R$ is transitive.

Since $R$ is reflexive, symmetric, and transitive, we conclude that $R$ is an equivalence relation on $\mathbb{Z}$.

Part (b): Compute the equivalence classes of $R$.

The relation $R$ on $\mathbb{Z}$ is defined by $a \, R \, b$ if and only if $a - b$ is even. This means that $a$ and $b$ are in the same equivalence class if their difference is even, or in other words, if they are either both even or both odd.

Thus, there are two equivalence classes:

1. The set of all even integers: $[0] = \{ \dots, -4, -2, 0, 2, 4, \dots \}$.
2. The set of all odd integers: $[1] = \{ \dots, -3, -1, 1, 3, 5, \dots \}$.

We can express the equivalence classes more formally as:

• $[0] = \{ x \in \mathbb{Z} \mid x \equiv 0 \pmod{2} \}$, which represents the set of even integers.
• $[1] = \{ x \in \mathbb{Z} \mid x \equiv 1 \pmod{2} \}$, which represents the set of odd integers.

Thus, the equivalence classes of $R$ are $[0]$ and $[1]$, which correspond to the sets of even and odd integers, respectively.

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Would you like more details on any of the proof steps, or do you have questions? Here are related questions that build on this explanation:

1. How would the equivalence relation change if we defined $R$ such that $a - b$ is divisible by a number other than 2?
2. What is an example of a relation that is symmetric and reflexive but not transitive?
3. How can equivalence classes be represented for other modular relations (e.g., modulo 3)?
4. How do we determine the number of equivalence classes for a general modular relation on $\mathbb{Z}$?
5. How would the properties of an equivalence relation change in a non-integer set?

Tip: When showing that a relation is an equivalence relation, always go through reflexivity, symmetry, and transitivity in that order to ensure clarity and completeness.