Q.Show that each of the relation R in the set $A = \{x \in Z : 0 \le x \le 12\}$, given by

Concept: Relation Properties — To prove equivalence, we check reflexivity, symmetry, and transitivity.

(i) $R = \{(a, b) : |a - b| \text{ is a multiple of } 4\}$ on $A = \{0,1,2,\dots,12\}$.

• Reflexive: For any $a \in A$, $|a-a| = 0$, which is a multiple of $4$. So $(a,a) \in R$.
• Symmetric: If $|a-b|$ is a multiple of $4$, then $|b-a| = |a-b|$ is also a multiple of $4$. So $(b,a) \in R$.
• Transitive: If $|a-b| = 4k$ and $|b-c| = 4m$, then $|a-c| \le |a-b| + |b-c| = 4(k+m)$. Also, $a-c = (a-b)+(b-c)$; the difference is a sum of multiples of $4$, hence a multiple of $4$. So $|a-c|$ is a multiple of $4$, giving $(a,c) \in R$. …

Why this works — the core idea

An equivalence relation is a way of saying "these things are the same in some specific sense." Three properties must hold:

• Reflexive: every element is related to itself.
• Symmetric: if $a$ is related to $b$, then $b$ is related to $a$.
• Transitive: if $a$ is related to $b$ and $b$ to $c$, then $a$ is related to $c$.

Once we verify these, the relation carves the set into disjoint equivalence classes — groups of elements that are all mutually related. The question then asks: which elements live in the same class as $1$?

The set here is $A = \{0,1,2,\dots,12\}$, a finite chunk of integers.

Case (i): $R = \{(a,b) : |a-b| \text{ is a multiple of } 4\}$

1. Reflexivity

For any $a \in A$, $|a-a| = 0$. Is $0$ a multiple of $4$? Yes — $0 = 4 \times 0$. So $(a,a) \in R$ for every $a$. Reflexive.

2. Symmetry

If $(a,b) \in R$, then $|a-b|$ is a multiple of $4$. But $|b-a| = |a-b|$, so it's the same number. Hence $(b,a) \in R$. Symmetric.

3. Transitivity

Suppose $(a,b) \in R$ and $(b,c) \in R$. Then $|a-b| = 4k$ and $|b-c| = 4m$ for some integers $k,m$.

We need to show $|a-c|$ is also a multiple of $4$. The triangle inequality gives:

But that only gives an upper bound — we need exact divisibility. A better approach: note that $|a-b|$ being a multiple of $4$ means $a \equiv b \pmod{4}$ (they leave the same remainder when divided by $4$). Similarly, $b \equiv c \pmod{4}$. By transitivity of congruence, $a \equiv c \pmod{4}$, so $|a-c|$ is a multiple of $4$. Hence $(a,c) \in R$. Transitive.

Since all three properties hold, $R$ is an equivalence relation.

4. Elements related to $1$

We want all $b \in A$ such that $|1-b|$ is a multiple of $4$. That means $1-b \equiv 0 \pmod{4}$, i.e. $b \equiv 1 \pmod{4}$.

Now list numbers in $A = \{0,1,\dots,12\}$ that are congruent to $1$ modulo $4$:

$1,\; 5,\; 9$ …

Method: Equivalence Relations by Congruence, and the Class of an Element

Use this when a relation is "$|a-b|$ is a multiple of $k$" or "$a=b$", to prove it is an equivalence relation and list the elements related to a given one.

Steps

Step 1: Translate to congruence

"$|a-b|$ is a multiple of $4$" means $a\equiv b \pmod 4$. Equality "$a=b$" is the finest such relation.

Step 2: Verify the three properties

Reflexive: $a-a=0$ is a multiple of $4$. Symmetric: $|a-b|=|b-a|$. Transitive: $a\equiv b$ and $b\equiv c$ give $a\equiv c$. …

Common Mistakes

Mistake 1: Including numbers outside the given set in a class

Why it's wrong: $13\equiv 1\pmod 4$, but $13\notin\{0,\dots,12\}$, so it is not in the class. Correct approach: intersect the congruence condition with the actual base set.

Mistake 2: Missing that $0$ is a multiple of $4$

Why it's wrong: reflexivity needs $|a-a|=0$ to count as a multiple of $4$, which it does. Correct approach: treat $0$ as a valid multiple. …