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Chapter 10: Problem 14

a. Let \(R\) be the relation of congruence modulo 3 . Which of the following equivalence classes are equal? \([7],[-4],[-6],[17],[4],[27],[19]\) b. Let \(R\) be the relation of congruence modulo 7. Which of the following equivalence classes are equal? $$ [35],[3],[-7],[12],[0],[-2],[17] $$

Short Answer

Expert verified
The equivalent classes under modulo 3 are: [7], [4], [19] are equal; [-4], [17] are equal; [-6], [27] are equal. Under modulo 7, the equal classes are: [35], [-7], [0] are equal; [3], [17] are equal; [12], [-2] are equal.

Step by step solution

01

Part a: Analyze equivalence classes under modulo 3.

To determine the equivalence classes that are equal under modulo 3, we need to find the remainder when each element is divided by 3. The equivalence classes and their remainders modulo 3 are: 1. [7] - \( 7 \equiv 1 \pmod{3} \) 2. [-4] - \(- 4 \equiv 2 \pmod{3} \) 3. [-6] - \(- 6 \equiv 0 \pmod{3} \) 4. [17] - \( 17 \equiv 2 \pmod{3} \) 5. [4] - \( 4 \equiv 1 \pmod{3} \) 6. [27] - \( 27 \equiv 0 \pmod{3} \) 7. [19] - \( 19 \equiv 1 \pmod{3} \)
02

Part a: Group equivalence classes with the same remainders modulo 3.

Now that we have the remainders, we can see which equivalence classes are equal under modulo 3: 1. [7], [4], and [19] are equal because they all have a remainder of 1 modulo 3. 2. [-4] and [17] are equal because they both have a remainder of 2 modulo 3. 3. [-6] and [27] are equal because they both have a remainder of 0 modulo 3.
03

Part b: Analyze equivalence classes under modulo 7.

Now we need to determine the equivalence classes that are equal under modulo 7. To do this, we will find the remainders when each element is divided by 7. The equivalence classes and their remainders modulo 7 are: 1. [35] - \( 35 \equiv 0 \pmod{7} \) 2. [3] - \( 3 \equiv 3 \pmod{7} \) 3. [-7] - \(- 7 \equiv 0 \pmod{7} \) 4. [12] - \( 12 \equiv 5 \pmod{7} \) 5. [0] - \( 0 \equiv 0 \pmod{7} \) 6. [-2] - \( -2 \equiv 5 \pmod{7} \) 7. [17] - \( 17 \equiv 3 \pmod{7} \)
04

Part b: Group equivalence classes with the same remainders modulo 7.

Now that we have the remainders under modulo 7, we can see which equivalence classes are equal: 1. [35], [-7], and [0] are equal because they all have a remainder of 0 modulo 7. 2. [3] and [17] are equal because they both have a remainder of 3 modulo 7. 3. [12] and [-2] are equal because they both have a remainder of 5 modulo 7.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Congruence Modulo
When we talk about congruence modulo, we're exploring how two numbers can relate to each other when divided by a third number, known as the modulus. This relationship is expressed using the notation "\( a \equiv b \pmod{m} \)," which means that \( a \) and \( b \) give the same remainder when divided by \( m \). For instance, \( 7 \equiv 1 \pmod{3} \) tells us that both 7 and 1 have a remainder of 1 when divided by 3.
In essence, congruence modulo helps categorize numbers into groups, known as equivalence classes, based on their remainders. Each of these classes represents numbers that share a common remainder when divided by the modulus, ultimately simplifying many calculations in number theory and algebra.
To apply this in practical terms, if you're asked which numbers are congruent modulo a particular number, you just need to perform simple divisions and check which numbers share the same remainder.
Remainders
The concept of remainders is crucial in understanding modular arithmetic. When you divide any integer \( a \) by another positive integer \( m \), you get a quotient and a remainder. Mathematically, this can be expressed as \( a = qm + r \), where \( q \) is the quotient and \( r \) is the remainder.
The remainder is the part of \( a \) that is left over after dividing \( a \) into as many full parts of \( m \) as possible. Importantly, in modular arithmetic, we are often only concerned with the remainder. For instance, dividing 10 by 3 gives a remainder of 1, meaning \( 10 \equiv 1 \pmod{3} \).
It鈥檚 useful to remember that the remainder \( r \) can range from 0 up to \( m-1 \). Understanding and identifying remainders help link numbers into equivalence classes, furthering our grasp of their relationships under specific moduli.
Modular Arithmetic
Modular arithmetic is sometimes called "clock arithmetic" because it deals with numbers wrapping around once they reach a certain value, much like hours on a clock restarting after 12. This arithmetic system reduces numbers by a modulus, allowing all integers to be condensed into a finite set of values ranging from 0 to \( m-1 \), where \( m \) is the modulus.
In practical applications, modular arithmetic helps simplify computations and solve equations by focusing only on the remainder. Calculating large powers, encrypting data, and even ensuring numbers like library book codes are correct are some instances where this concept shines.
Consider calculating \( 12 + 9 \pmod{8} \). Normally, the sum is 21, but in modulo 8 arithmetic, it is equivalent to \( 5 \) because \( 21 \equiv 5 \pmod{8} \). This simplification shows how modular arithmetic can streamline complex problems into simpler, more manageable pieces. Understanding its rules and patterns can be a powerful tool in mathematical reasoning and problem-solving.

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Most popular questions from this chapter

\(A\) is the set of all students at your college. a. \(R\) is the relation defined on \(A\) as follows: For all \(x\) and \(y\) in \(A\), \(x R y \Leftrightarrow x\) has the same major (or double major) as \(y\). (Assume "undeclared" is a major.) b. \(S\) is the relation defined on \(A\) as follows: For all \(x, y \in A\), \(x S y \Leftrightarrow x\) is the same age as \(y\).

Let \(A\) be the set of points in the rectangle with \(x\) and \(y\) coordinates between 0 and 1 . That is, $$ A=\\{(x, y) \in \mathbf{R} \times \mathbf{R} \mid 0 \leq x \leq 1 \quad \text { and } \quad 0 \leq y \leq 1\\} $$ Define a relation \(R\) on \(A\) as follows: For all \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) in \(A_{1}\) $$ \begin{aligned} \left(x_{1}, y_{1}\right) R\left(x_{2}, y_{2}\right) \Leftrightarrow & \Leftrightarrow \\ \left(x_{1}, y_{1}\right)=\left(x_{2}, y_{2}\right) ; & \text { or } \\ x_{1}=0 & \text { and } x_{2}=1 \quad \text { and } \quad y_{1}=y_{2} ; \quad \text { or } \\ x_{1}=1 & \text { and } x_{2}=0 \quad \text { and } \quad y_{1}=y_{2} ; \quad \text { or } \\ y_{1}=0 & \text { and } y_{2}=1 \quad \text { and } \quad x_{1}=x_{2} ; \quad \text { or } \\ y_{1}=1 & \text { and } y_{2}=0 \quad \text { and } \quad x_{1}=x_{2} . \end{aligned} $$ In other words, all points along the top edge of the rectangle are related to the points along the bottom edge directly beneath them, and all points directly opposite each other along the left and right edges are related to each other. The points in the interior of the rectangle are not related to anything other than themselves. Then \(R\) is an equivalence relation on \(A\). Imagine gluing together all the points that are in the same equivalence class. Describe the resulting figure.

Define a relation \(R\) on the set \(\mathbf{Z}\) of all integers as follows: For all \(m, n \in \mathbf{Z}\), \(\begin{aligned} m R n & \Leftrightarrow & \text { every prime factor of } m \\\ & \text { is a prime factor of } n . \end{aligned}\) Is \(R\) a partial order relation? Prove or give a counterexample.

Each of the following is a relation on \(\\{0,1,2,3\\}\). Draw directed graphs for each relation, and indicate which relations are antisymmetric. a. \(R_{1}=\\{(0,0),(0,2),(1,0),(1,3),(2,2),(3,0),(3,1)\\}\) b. \(R_{2}=\\{(0,1),(0,2),(1,1),(1,2),(1,3),(2,2),(3,2)\\}\) c. \(R_{3}=\\{(0,0),(0,3),(1,0),(1,3),(2,2),(3,3),(3,2)\\}\) d. \(R_{4}=\\{(0,0),(1,0),(1,2),(1,3),(2,0),(2,1),(3,2)\), \((3,0)\\}\)

Let \(P\) be the set of all points in the Cartesian plane except the origin, \(R\) is the relation defined on \(P\) as follows: For all \(p_{1}\) and \(p_{2}\) in \(P\). \(p_{1} R p_{2} \Leftrightarrow p_{1}\) and \(p_{2}\) lie on the same half-line emanating from the origin.
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