Math Problem Statement

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

Solution

To solve this, we will use Theorem 3.28 and Theorem 3.30 to demonstrate that if ab0(mod3)ab \equiv 0 \pmod{3}, then a0(mod3)a \equiv 0 \pmod{3} or b0(mod3)b \equiv 0 \pmod{3}.


Theorem 3.28 (Properties of Congruence):

If ab(modn)a \equiv b \pmod{n} and cd(modn)c \equiv d \pmod{n}, then:

  1. a+cb+d(modn)a + c \equiv b + d \pmod{n},
  2. acbd(modn)a \cdot c \equiv b \cdot d \pmod{n}.

This means congruences respect addition and multiplication.


Theorem 3.30 (Divisibility and Congruence):

If nabn \mid ab, then nan \mid a or nbn \mid b (where nn is a prime number).

This theorem essentially asserts that the divisibility of the product implies the divisibility of at least one factor, provided nn is prime.


Proof using Theorems 3.28 and 3.30:

Let ab0(mod3)ab \equiv 0 \pmod{3}. This means: 3ab.3 \mid ab.

From Theorem 3.30, since 33 is a prime number, 3a3 \mid a or 3b3 \mid b. This implies: a0(mod3)orb0(mod3).a \equiv 0 \pmod{3} \quad \text{or} \quad b \equiv 0 \pmod{3}.


Why we avoid the direct definition:

The direct definition of congruence would require explicitly expressing ab=3kab = 3k for some integer kk, and then splitting aa or bb as multiples of 3. By using Theorem 3.30, we directly leverage the divisibility rules for prime numbers, which simplifies the proof significantly.


Conclusion:

If ab0(mod3)ab \equiv 0 \pmod{3}, then a0(mod3)a \equiv 0 \pmod{3} or b0(mod3)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 nn when nn 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., ab0(mod6)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.

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Math Problem Analysis

Mathematical Concepts

Number Theory
Modular Arithmetic

Formulas

If ab ≡ 0 (mod n), then a ≡ 0 (mod n) or b ≡ 0 (mod n), where n is prime.

Theorems

Theorem 3.28 (Properties of Congruence)
Theorem 3.30 (Divisibility and Congruence)

Suitable Grade Level

Grades 11-12 or introductory college-level math