Math Problem Statement

Let m ∈ Z. Prove that the relation a ≡ b (mod m) is an equivalence relation.

Solution

To prove that the relation ab(modm)a \equiv b \pmod{m} is an equivalence relation, we must show that the relation satisfies three properties of an equivalence relation:

  1. Reflexivity: For any integer aa, we need to show that aa(modm)a \equiv a \pmod{m}.

  2. Symmetry: For any integers aa and bb, if ab(modm)a \equiv b \pmod{m}, we need to show that ba(modm)b \equiv a \pmod{m}.

  3. Transitivity: For any integers aa, bb, and cc, if ab(modm)a \equiv b \pmod{m} and bc(modm)b \equiv c \pmod{m}, we need to show that ac(modm)a \equiv c \pmod{m}.

1. Reflexivity

We need to show that aa(modm)a \equiv a \pmod{m}. By definition of congruence modulo mm, ab(modm)a \equiv b \pmod{m} means that mm divides the difference aba - b, i.e., ab=kma - b = km for some integer kk.

For aa(modm)a \equiv a \pmod{m}, the difference aa=0a - a = 0, which is clearly divisible by mm. Therefore, aa(modm)a \equiv a \pmod{m}, satisfying reflexivity.

2. Symmetry

We need to show that if ab(modm)a \equiv b \pmod{m}, then ba(modm)b \equiv a \pmod{m}.

By the definition of congruence, if ab(modm)a \equiv b \pmod{m}, then m(ab)m \mid (a - b). This means there exists an integer kk such that ab=kma - b = km.

Now, consider bab - a. We have: ba=(ab)=kmb - a = -(a - b) = -km Since km-km is also divisible by mm, it follows that m(ba)m \mid (b - a). Thus, ba(modm)b \equiv a \pmod{m}, proving symmetry.

3. Transitivity

We need to show that if ab(modm)a \equiv b \pmod{m} and bc(modm)b \equiv c \pmod{m}, then ac(modm)a \equiv c \pmod{m}.

From the assumption ab(modm)a \equiv b \pmod{m}, we know that m(ab)m \mid (a - b), so there exists an integer k1k_1 such that: ab=k1ma - b = k_1 m

Similarly, from bc(modm)b \equiv c \pmod{m}, we know that m(bc)m \mid (b - c), so there exists an integer k2k_2 such that: bc=k2mb - c = k_2 m

Now, consider the difference aca - c: ac=(ab)+(bc)=k1m+k2m=(k1+k2)ma - c = (a - b) + (b - c) = k_1 m + k_2 m = (k_1 + k_2) m Since aca - c is a multiple of mm, we conclude that m(ac)m \mid (a - c), so ac(modm)a \equiv c \pmod{m}. This proves transitivity.

Conclusion

Since we have shown that the relation ab(modm)a \equiv b \pmod{m} satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation.


Do you need further clarification or examples for any part of this proof?

Here are some related questions you might find useful:

  1. How do you define congruence modulo mm for integers?
  2. What are some examples of equivalence relations in other contexts (e.g., set theory)?
  3. Can you prove that congruence modulo mm is an equivalence relation in any other number system, such as real numbers?
  4. What is the relationship between the equivalence classes of aa and bb under modulo mm?
  5. How does the proof of transitivity differ if we are working with other types of relations, such as partial orders?

Tip: When working with equivalence relations, it's helpful to always verify the three properties (reflexivity, symmetry, and transitivity) in sequence for clarity and correctness.

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Math Problem Analysis

Mathematical Concepts

Number Theory
Modular Arithmetic
Equivalence Relations

Formulas

a ≡ b (mod m) means m | (a - b)

Theorems

Reflexive Property
Symmetric Property
Transitive Property of Equivalence Relations

Suitable Grade Level

Grades 10-12