Certainly! Let's go through the proof using $x$ and $y$ instead of $a$ and $b$.

Problem Restatement:

Consider the set $\mathbb{Z}$ of integers and an integer $m > 1$. We say that $x$ is congruent to $y$ modulo $m$, written $x \equiv y \ (\text{mod} \ m)$, if $m$ divides $x - y$. Show that this defines an equivalence relation on $\mathbb{Z}$.

Solution:

To show that the relation $x \equiv y \ (\text{mod} \ m)$ is an equivalence relation, we need to verify the three properties of equivalence relations: reflexivity, symmetry, and transitivity.

1. Reflexivity:

   • For any integer $x \in \mathbb{Z}$, we need to show that $x \equiv x \ (\text{mod} \ m)$.
   • $x \equiv x \ (\text{mod} \ m)$ means that $m$ divides $x - x$.
   • Since $x - x = 0$, and $m$ divides 0, this property holds true for any $x$.
   • Therefore, the relation is reflexive.

2. Symmetry:

   • We need to show that if $x \equiv y \ (\text{mod} \ m)$, then $y \equiv x \ (\text{mod} \ m)$.
   • If $x \equiv y \ (\text{mod} \ m)$, it means that $m$ divides $x - y$, i.e., $x - y = km$ for some integer $k$.
   • This can be rewritten as $y - x = -km$, and since $-k$ is also an integer, $m$ divides $y - x$.
   • Therefore, $y \equiv x \ (\text{mod} \ m)$, so the relation is symmetric.

3. Transitivity:

   • We need to show that if $x \equiv y \ (\text{mod} \ m)$ and $y \equiv z \ (\text{mod} \ m)$, then $x \equiv z \ (\text{mod} \ m)$.
   • If $x \equiv y \ (\text{mod} \ m)$, then $m$ divides $x - y$, i.e., $x - y = km$ for some integer $k$.
   • If $y \equiv z \ (\text{mod} \ m)$, then $m$ also divides $y - z$, i.e., $y - z = lm$ for some integer $l$.
   • Adding these two equations gives $(x - y) + (y - z) = km + lm = (k + l)m$, which implies $x - z = (k + l)m$.
   • Since $k + l$ is an integer, $m$ divides $x - z$, and thus $x \equiv z \ (\text{mod} \ m)$.
   • Therefore, the relation is transitive.

Since the relation $x \equiv y \ (\text{mod} \ m)$ satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation on the set $\mathbb{Z}$.

Summary:

The relation $x \equiv y \ (\text{mod} \ m)$ is an equivalence relation on the set of integers $\mathbb{Z}$ because it satisfies the three necessary properties: reflexivity, symmetry, and transitivity.

Would you like more details or have any questions?

Related Questions:

1. What are some real-life applications of equivalence relations?
2. How does modular arithmetic play a role in computer science?
3. Can you provide an example of an equivalence class under this relation?
4. How does the concept of congruence modulo $m$ relate to cyclic groups?
5. How is the Chinese Remainder Theorem related to modular arithmetic?

Tip:

When working with equivalence relations, always verify the three properties (reflexivity, symmetry, transitivity) as they are key to ensuring the relation behaves as expected in various mathematical contexts.