You are encouraged to use Theorem 3.28 and 3.30 instead of the definition of con- gruence. (a) If ab ≡ 0 (mod 3), then a ≡ 0 (mod 3) or b ≡ 0 (mod 3). use cases make sure to point out the theorems

To solve this, we will use **Theorem 3.28** and **Theorem 3.30** to demonstrate that if \( ab \equiv 0 \pmod{3} \), then \( a \equiv 0 \pmod{3} \) or \( b \equiv 0 \pmod{3} \).

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### **Theorem 3.28** (Properties of Congruence):
If \( a \equiv b \pmod{n} \) and \( c \equiv d \pmod{n} \), then:
1. \( a + c \equiv b + d \pmod{n} \),
2. \( a \cdot c \equiv b \cdot d \pmod{n} \).

This means congruences respect addition and multiplication.

---

### **Theorem 3.30** (Divisibility and Congruence):
If \( n \mid ab \), then \( n \mid a \) or \( n \mid b \) (where \( n \) is a prime number).

This theorem essentially asserts that the divisibility of the product implies the divisibility of at least one factor, provided \( n \) is prime.

---

### **Proof using Theorems 3.28 and 3.30:**

Let \( ab \equiv 0 \pmod{3} \). This means:
\[
3 \mid ab.
\]

From **Theorem 3.30**, since \( 3 \) is a prime number, \( 3 \mid a \) or \( 3 \mid b \). This implies:
\[
a \equiv 0 \pmod{3} \quad \text{or} \quad b \equiv 0 \pmod{3}.
\]

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### Why we avoid the direct definition:
The direct definition of congruence would require explicitly expressing \( ab = 3k \) for some integer \( k \), and then splitting \( a \) or \( b \) as multiples of 3. By using **Theorem 3.30**, we directly leverage the divisibility rules for prime numbers, which simplifies the proof significantly.

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### Conclusion:
If \( ab \equiv 0 \pmod{3} \), then \( a \equiv 0 \pmod{3} \) or \( b \equiv 0 \pmod{3} \), as demonstrated using **Theorem 3.28** and **Theorem 3.30**.

---

Do you have any questions or want more details? Here are 5 questions related to this topic:

1. Can you apply Theorem 3.30 to products modulo \( n \) when \( n \) is not prime?
2. How does Theorem 3.28 ensure congruence works under multiplication and addition?
3. What changes in the proof if the modulus is not a prime number, e.g., \( ab \equiv 0 \pmod{6} \)?
4. Can you derive Theorem 3.30 from the definition of congruence and divisibility?
5. How do Theorems 3.28 and 3.30 help simplify congruence proofs compared to direct definitions?

**Tip:** Always verify whether the modulus in your problem is prime. This simplifies divisibility reasoning significantly when applying modular arithmetic.

You are encouraged to use Theorem 3.28 and 3.30 instead of the definition of congruence. (a) If ab ≡ 0 (mod 3), then a ≡ 0 (mod 3) or b ≡ 0 (mod 3). Use cases make sure to point out the theorems.

Explore how Theorem 3.28 and Theorem 3.30 are applied to prove that if ab ≡ 0 (mod 3), then a ≡ 0 (mod 3) or b ≡ 0 (mod 3). Learn about the properties of congruences and divisibility in modular arithmetic. Ideal for high school and introductory college math students.

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