Relation \( ad = bc \) on \( \mathbb{Z} \times \mathbb{Z}_0 \)

📺 Video Explanation

📝 Question

Let \( \mathbb{Z} \) be the set of integers and \( \mathbb{Z}_0 \) the set of non-zero integers. Define relation \( R \) on \( \mathbb{Z} \times \mathbb{Z}_0 \) as:

\[ (a,b) R (c,d) \iff ad = bc \]

Show that \( R \) is an equivalence relation.

---

✅ Solution

🔹 Step 1: Reflexive

For reflexive, we need: \[ (a,b) R (a,b) \]

\[ ab = ba \]

This is always true (commutative property of multiplication).

✔ Therefore, the relation is Reflexive. :contentReference[oaicite:0]{index=0}

---

🔹 Step 2: Symmetric

Assume: \[ (a,b) R (c,d) \Rightarrow ad = bc \]

Then: \[ cb = da \]

So, \[ (c,d) R (a,b) \]

✔ Therefore, the relation is Symmetric. :contentReference[oaicite:1]{index=1}

---

🔹 Step 3: Transitive

Assume: \[ (a,b) R (c,d),\ (c,d) R (e,f) \]

\[ ad = bc,\quad cf = de \]

Multiply appropriately: \[ adf = bcf \]

Using \( cf = de \): \[ bcf = bde \]

So, \[ adf = bde \]

Since \( d \ne 0 \), divide both sides by \( d \): \[ af = be \]

Thus, \[ (a,b) R (e,f) \]

✔ Therefore, the relation is Transitive. :contentReference[oaicite:2]{index=2}

---

🎯 Final Conclusion

✔ Reflexive: Yes
✔ Symmetric: Yes
✔ Transitive: Yes

\[ \therefore R \text{ is an equivalence relation on } \mathbb{Z} \times \mathbb{Z}_0 \]

---

🚀 Exam Insight

• This represents equality of fractions: \( \frac{a}{b} = \frac{c}{d} \)
• Each equivalence class corresponds to a rational number
• Very important concept for construction of rational numbers