Question

(a) For \(k, n_{1}, n_{2}, m_{1}, m_{2} \in \mathbb{N},\) show that if $$ n_{1} \equiv n_{2}(\bmod k) $$ and $$ m_{1}=m_{2}(\bmod k) $$ then $$ n_{1}+m_{1} \equiv n_{2}+m_{2}(\bmod k) $$ and $$ n_{1} \cdot m_{1} \equiv n_{2} \cdot m_{2}(\bmod k) $$ (b) Part (a) says that if we take two equivalence classes \([m]\) and \([n]\), then we can unambiguously define \([m]+[n]\) and \([m] \cdot[n]\). Pick any \(m_{1} \in[m]\) and any \(n_{1} \in[n],\) and define $$ [m]+[n]=\left[m_{1}+n_{1}\right] $$ and $$ [m] \cdot[n] \equiv\left[m_{1} \cdot n_{1}\right] $$ The definition is unambiguous since it doesn't matter which \(m_{1}\) and \(n_{1}\) we pick. Find the addition and multiplication tables for the equivalence classes of \(\equiv(\bmod 4)\) and \(\equiv(\bmod 5) .\) (Hint: For both \(\equiv(\bmod 4)\) and \(\equiv(\bmod 5),\) your answer should include $$ [0]+[0] \equiv[0],[0]+[1] \equiv[1],[0] \cdot[0] \equiv\\{0] $$ and $$ [1] \cdot[1] \equiv[1] $$ but, for \(\equiv(\bmod 4)\). $$ [2]+[2] \equiv[0] $$ whereas, that will be false for \(\equiv(\bmod 5) .\)

Short answer

The properties are proved by the definitions of congruences. Addition and multiplication tables for \(\pmod{4}\) and \(\pmod{5}\) can be constructed by applying these operations to each equivalence class.

Step-by-step solution

Understanding Modular Arithmetic

The problem involves proving expressions related to modular arithmetic. Recall that if two numbers are congruent modulo a number, they leave the same remainder when divided by that number. If \( n_1 \equiv n_2 \pmod{k} \), it means \( k \mid (n_1 - n_2) \), and similarly for \( m_1 \equiv m_2 \pmod{k} \).

Prove Addition Property

Let's prove that \( n_1 + m_1 \equiv n_2 + m_2 \pmod{k} \). By definition of congruence, \( n_1 - n_2 = ak \) and \( m_1 - m_2 = bk \) for some integers \( a \) and \( b \). Thus, \( (n_1 + m_1) - (n_2 + m_2) = ak + bk = (a + b)k \), which means \( k \mid (n_1 + m_1 - n_2 - m_2) \), and hence \( n_1 + m_1 \equiv n_2 + m_2 \pmod{k} \).

Prove Multiplication Property

Prove that \( n_1 \cdot m_1 \equiv n_2 \cdot m_2 \pmod{k} \). Using the congruences \( n_1 = n_2 + ak \) and \( m_1 = m_2 + bk \), multiplying these gives \( n_1 \cdot m_1 = (n_2 + ak)(m_2 + bk) = n_2m_2 + n_2bk + m_2ak + abk^2 \). The terms \( n_2bk \), \( m_2ak \), and \( abk^2 \) are divisible by \( k \), showing that \( n_1 \cdot m_1 - n_2 \cdot m_2 \) is divisible by \( k \). Thus, \( n_1 \cdot m_1 \equiv n_2 \cdot m_2 \pmod{k} \).

Define Addition and Multiplication for Equivalence Classes

The problem states that the results are similar for different representatives of classes \([n]\) and \([m]\). In modular arithmetic, any representative \( n_1 \) from \([n]\) and \( m_1 \) from \([m]\) are equivalent for operations under the same modulus.

Create Addition Table Modulo 4

For \( \equiv \pmod{4} \), calculate: - \([0] + [0] = [0]\), - \([0] + [1] = [1]\), - \([1] + [1] = [2]\), - \([2] + [2] = [0]\), - and so on for all combinations up to \([3].\) Utilize addition rules modulo 4.

Create Multiplication Table Modulo 4

For \( \equiv \pmod{4} \), calculate: - \([0] \cdot [0] = [0]\), - \([0] \cdot [1] = [0]\), - \([1] \cdot [1] = [1]\), - and similar computations for all equivalence class combinations, utilizing multiplication rules modulo 4.

Create Addition Table Modulo 5

For \( \equiv \pmod{5} \), calculate: - \([0] + [0] = [0]\), - \([0] + [1] = [1]\), - \([2] + [2] = [4]\), - and similar calculations for all combinations of equivalence classes using addition modulo 5.

Create Multiplication Table Modulo 5

For \( \equiv \pmod{5} \), calculate: - \([0] \cdot [0] = [0]\), - \([1] \cdot [1] = [1]\), - and similar computations for other class combinations, applying multiplication modulo 5.

Key concepts

Equivalence Classes

Equivalence classes are a fundamental concept in modular arithmetic, grouping elements with the same remainder when divided by a modulus. For example, when we work with numbers modulo 4, the equivalence classes are \([0], [1], [2], \text{ and } [3]\). Each of these classes represents a set of numbers sharing a common remainder.
To visualize this, think of dividing any integer \(x\) by 4. The remainder can be 0, 1, 2, or 3, which places \(x\) into one of these four equivalence classes. The number 5, for instance, belongs to \([1]\) because 5 divided by 4 leaves a remainder of 1. This simplification allows us to work with classes instead of individual numbers, simplifying calculations and revealing patterns.
Importantly, when performing operations like addition or multiplication, the outcome remains consistent across any representative chosen from the equivalence class. This makes working with these classes powerful and flexible.

Congruence Relations

Congruence relations describe when two numbers are equivalent under a specific modulus. When we say \( n_1 \equiv n_2 \pmod{k} \), it means \( n_1 \) and \( n_2 \) give the same remainder when divided by \( k \). In other words, \( k \mid (n_1 - n_2) \). This concept connects directly to the idea of equivalence classes by categorizing numbers based on their congruence.
Understanding these relations is crucial because they provide the foundation for many proofs and calculations. For example, if \( n_1 \equiv n_2 \pmod{k} \) and \( m_1 \equiv m_2 \pmod{k} \), then it follows that both their sums \( n_1 + m_1 \equiv n_2 + m_2 \pmod{k} \) and products \( n_1 \cdot m_1 \equiv n_2 \cdot m_2 \pmod{k} \) carry this congruence property.
This makes congruences a robust tool in mathematics, helping simplify otherwise complex operations and allowing mathematicians to work within a more manageable set of problem constraints.

Modular Addition and Multiplication

Modular arithmetic involves performing addition and multiplication operations within a restrictive range based on a modulus. For instance, with \( \equiv \pmod{4} \), you only have four possible results: 0, 1, 2, and 3. When two numbers are added or multiplied in this system, their results are brought back to one of these equivalence classes by reducing them modulo 4.
Let's take addition: \([2] + [3] \equiv [5] \equiv [1] \pmod{4} \). Similarly, for multiplication: \([2] \cdot [3] \equiv [6] \equiv [2] \pmod{4} \). Modulo reduction ensures that results remain within the expected range.
These operations are crucial because they preserve the mathematical structure and simplify computations, allowing mathematicians to construct concise and robust results even in more complex problems.