[K] = {xez/x = k moon] we for these equivalence classes ve define addition as follows: [a] + [b] = [a+b] Example Z3 Z₂ = { [0], [₁], [²]} 3 [0] = {..., −6,-3, 0, 6, 9,... } [1] = {ace, -5, -2, 1, 4, 7, .... Y [2] = {..., -4₁ -1, 2, 5, 8,...3
[K] = {xez/x = k moon] we for these equivalence classes ve define addition as follows: [a] + [b] = [a+b] Example Z3 Z₂ = { [0], [₁], [²]} 3 [0] = {..., −6,-3, 0, 6, 9,... } [1] = {ace, -5, -2, 1, 4, 7, .... Y [2] = {..., -4₁ -1, 2, 5, 8,...3
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I don't understand why [0] equals that. Same for [1] and [2]
![[K] = {xez/x = k modn]
for these equivalence classes ve
define addition as follows:
[a] + [b] = [a+b]
Example Z3
Z₂ = { [0], [1], [²]}
3
[0] = {..., -6, -3, 0, 6, 9,... }
[1] = {ace, -5, -2, 1, 4, 7, .... Y
[2] = {..., -4₁ -1, 2, 5, 8,...3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F30761ad5-6d22-4ff4-adb6-4f166a7ab52a%2Fff1cd51d-32e7-4425-9555-6c74c85330ba%2F6p26y.jpeg&w=3840&q=75)
Transcribed Image Text:[K] = {xez/x = k modn]
for these equivalence classes ve
define addition as follows:
[a] + [b] = [a+b]
Example Z3
Z₂ = { [0], [1], [²]}
3
[0] = {..., -6, -3, 0, 6, 9,... }
[1] = {ace, -5, -2, 1, 4, 7, .... Y
[2] = {..., -4₁ -1, 2, 5, 8,...3
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