Definition 17.5.2. (Modular Equivalence, second definition) a = b (mod n) iff a-b=k-n, where k is an integer (that is, k € Z). A Exercise 17.5.3. Using Definition 17.5.2, show that equivalence mod n is an equivalence relation. (That is, show that equivalence mod n is (a) reflexive, (b) symmetric, and (c) transitive)
Definition 17.5.2. (Modular Equivalence, second definition) a = b (mod n) iff a-b=k-n, where k is an integer (that is, k € Z). A Exercise 17.5.3. Using Definition 17.5.2, show that equivalence mod n is an equivalence relation. (That is, show that equivalence mod n is (a) reflexive, (b) symmetric, and (c) transitive)
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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Please do Exercise 17.5.3 and please show step by step and explain

Transcribed Image Text:Definition 17.5.2. (Modular Equivalence, second definition)
a = b (mod n) iff a-b=k-n, where k is an integer (that is, k € Z). A
Exercise 17.5.3. Using Definition 17.5.2, show that equivalence mod n
is an equivalence relation. (That is, show that equivalence mod n is (a)
reflexive, (b) symmetric, and (c) transitive)
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