Let D be the relation defined on Z as follows: For every m, nEZ, mDn+3\ (m2 – n2). - (a) Prove that D is an equivalence relation. Proof that D is reflexive. Suppose m is any integer. To show that D is reflexive, we must show that ---Select--- ✓ which is true if, and only if, 3 divides ---Select--- ✓ by definition of ---Select--- |---Select--- by definition of D. Therefore, D is reflexive. Proof that D is symmetric. Suppose m and n are any integers such that m Dn. To show that D is symmetric, we must show that ---Select--- Then 3-Select--- by definition of ---Select--- Thus, ---Select--- by definition of --Select--- Note that n² m² --Select--- by basic algebra. Hence, ---Select--- by --Select--- Now, -k is an integer, and so -Select--- ✓ by definition of ---Select--- Thus, --Select--- Proof that D is transitive. by definition of --Select--- and so D is symmetric. Suppose m, n, and p are any integers such that m Dn and n D -Select--- To show that D is transitive, we must show that ---Select--- Then 3 (m² n²) and ---Select--- ✓ by definition of -Select--- Thus, --Select--- by definition of --Select--- And n² - ---Select--- = 3/ for some integer / by definition of ---Select--- Note that m² - p² = (m² - n²) + Hence, ---Select--- (n² - ---Select--- by ---Select--- Now k + / is an integer, and so ---Select--- Thus, --Select--- by basic algebra. by definition of ---Select--- by definition of --Select--- and so D is transitive. ' Proof that D is an equivalence relation. Because D is reflexive, symmetric, and transitive, it is an equivalence relation. Now m² ---Select--- = ---Select--- and 3 divides ---Select--- So 3 divides --Select--- V and thus, (b) What are the equivalence classes for D? Suppose m and n are any integers such that m D n. Then 3-Select--- by definition of ---Select--- Note that m² n² = (mn) --Select--- ✓ by basic algebra. Thus, 3 (m) -n)--Select--- by ---Select--- It follows that 3 | (mn) or 3 | ---Select--- ✓ by the unique factorization of integers theorem because the only positive integer factors of 3 are 1 and 3. Hence, there is an integer, say r, such that m - n = 3r or ---Select--- ✓ = 3r by definition of ---Select--- Then m n + 3r or m = n + 3r by basic algebra. Conversely, if there is an integer r with m = n + 3r or m = n + 3r, then 3 | (mn) or 3 | (m + n), and so 3 | (mn) ---Select--- ✓ Consequently, m Dn if, and only if, m = n + 3r or m = n + 3r for some integer r. Now the equivalence class of 0, [0] is the set of all --Select--- In other words, [0] is the set of all ---Select--- Im such that m D ✓ of the form 0 + 3r = 0 + 3r for some integer r. Since 0 + 3r = -0 + 3r = 3r, [0] = {m EZ | m = 3r for some integer r} = {... -9, -6, -3, 0, 3, 6, 9, ...). Similarly, the equivalence class of 1 is the set of all ---Select--- Im such that m D In other words, m = + 3r or m = -1 + 3r for some integer r. , which implies that 3 | m² - n², and, hence, that m D n. Now if m = -1 + 3r, then m = 1 + 3 + 3r 3 = 2 + 3(r - 1). This shows that m can be written in the form 1 + 3 (some integer) or in the form 2 + 3 (some integer). Thus, the equivalence class of 1 is the set of all integers that have a remainder of 1 or a remainder of 2 when they are divided by 3. [1] = {m EZ | m = 1 + 3r or m = −1 + 3r} = {... -8, -7, -5, -4, -2, -1, 1, 2, 4, 5, 7, 8, 10, 11, ...}. By the quotient-remainder theorem, every integer has a remainder of 0, 1, or 2 when it is divided by 3. Therefore, every integer is in either [0] or and so they form a partition of Z. Thus, these two classes are the distinct equivalence classes of D.

Let D be the relation defined on Z as follows: For every m, nEZ, mDn+3\ (m2 – n2). - (a) Prove that D is an equivalence relation. Proof that D is reflexive. Suppose m is any integer. To show that D is reflexive, we must show that ---Select--- ✓ which is true if, and only if, 3 divides ---Select--- ✓ by definition of ---Select--- |---Select--- by definition of D. Therefore, D is reflexive. Proof that D is symmetric. Suppose m and n are any integers such that m Dn. To show that D is symmetric, we must show that ---Select--- Then 3-Select--- by definition of ---Select--- Thus, ---Select--- by definition of --Select--- Note that n² m² --Select--- by basic algebra. Hence, ---Select--- by --Select--- Now, -k is an integer, and so -Select--- ✓ by definition of ---Select--- Thus, --Select--- Proof that D is transitive. by definition of --Select--- and so D is symmetric. Suppose m, n, and p are any integers such that m Dn and n D -Select--- To show that D is transitive, we must show that ---Select--- Then 3 (m² n²) and ---Select--- ✓ by definition of -Select--- Thus, --Select--- by definition of --Select--- And n² - ---Select--- = 3/ for some integer / by definition of ---Select--- Note that m² - p² = (m² - n²) + Hence, ---Select--- (n² - ---Select--- by ---Select--- Now k + / is an integer, and so ---Select--- Thus, --Select--- by basic algebra. by definition of ---Select--- by definition of --Select--- and so D is transitive. ' Proof that D is an equivalence relation. Because D is reflexive, symmetric, and transitive, it is an equivalence relation. Now m² ---Select--- = ---Select--- and 3 divides ---Select--- So 3 divides --Select--- V and thus, (b) What are the equivalence classes for D? Suppose m and n are any integers such that m D n. Then 3-Select--- by definition of ---Select--- Note that m² n² = (mn) --Select--- ✓ by basic algebra. Thus, 3 (m) -n)--Select--- by ---Select--- It follows that 3 | (mn) or 3 | ---Select--- ✓ by the unique factorization of integers theorem because the only positive integer factors of 3 are 1 and 3. Hence, there is an integer, say r, such that m - n = 3r or ---Select--- ✓ = 3r by definition of ---Select--- Then m n + 3r or m = n + 3r by basic algebra. Conversely, if there is an integer r with m = n + 3r or m = n + 3r, then 3 | (mn) or 3 | (m + n), and so 3 | (mn) ---Select--- ✓ Consequently, m Dn if, and only if, m = n + 3r or m = n + 3r for some integer r. Now the equivalence class of 0, [0] is the set of all --Select--- In other words, [0] is the set of all ---Select--- Im such that m D ✓ of the form 0 + 3r = 0 + 3r for some integer r. Since 0 + 3r = -0 + 3r = 3r, [0] = {m EZ | m = 3r for some integer r} = {... -9, -6, -3, 0, 3, 6, 9, ...). Similarly, the equivalence class of 1 is the set of all ---Select--- Im such that m D In other words, m = + 3r or m = -1 + 3r for some integer r. , which implies that 3 | m² - n², and, hence, that m D n. Now if m = -1 + 3r, then m = 1 + 3 + 3r 3 = 2 + 3(r - 1). This shows that m can be written in the form 1 + 3 (some integer) or in the form 2 + 3 (some integer). Thus, the equivalence class of 1 is the set of all integers that have a remainder of 1 or a remainder of 2 when they are divided by 3. [1] = {m EZ | m = 1 + 3r or m = −1 + 3r} = {... -8, -7, -5, -4, -2, -1, 1, 2, 4, 5, 7, 8, 10, 11, ...}. By the quotient-remainder theorem, every integer has a remainder of 0, 1, or 2 when it is divided by 3. Therefore, every integer is in either [0] or and so they form a partition of Z. Thus, these two classes are the distinct equivalence classes of D.

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter1: Fundamentals

Section1.7: Relations

Problem 21E: 21. A relation on a nonempty set is called irreflexive if for all. Which of the relations in...

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