The smallest positive solution of the congruence ax = 0( mod n) is called the additiveorder of a modulo n. Find the additive orders of each of the following elements by solvingthe appropriate congruences:(a) 7 modulo 11.(b) 9 modulo 15.(c) 8 modulo 15.

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Answered: SImalest posiuiy OI the n) is order of…

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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22. Evaluation of proofs See the instructions for Exercise (19) on page 100 from Section 3.1. (a) Proposition. For all integers a and b, if (a + 2b) = 0 (mod 3), then (2а + b) %3D 0 (mod 3). Proof. We assume a, b e Z and (a + 2b) = 0 (mod 3). This means that 3 divides a + 2b and, hence, there exists an integer m such that a + 2b — Зт. Hence, a %3 3m – 2b. For (2a +b) = 0 (mod 3), there exists an integer x such that 2a + b = 3x. Hence, 2(Зт — 2b) + b %3 3x бт — 3Ь — 3х 3(2m – b) = 3x 2m — b 3 х. Since (2m – b) is an integer, this proves that 3 divides (2a + b) and hence, (2a + b) = 0 (mod 3). (b) Proposition. For each integer m, 5 divides (m³ – m). Proof. Let m e Z. We will prove that 5 divides (m³ – m) by proving that (m – m) = 0 (mod 5). We will use cases. For the first case, if m = 0 (mod 5), then m³ = 0 (mod 5) and, hence, (m³ – m) = 0 (mod 5). For the second case, if m = 1 (mod 5), then m³ = 1 (mod 5) and, hence, (m³ – m) = (1 – 1) (mod 5), which means that (m³ – m) 0 (mod 5). For…
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SImalest posiuiy OI the n) is order of a modulo n. Find the additive orders of each of the following elements by solving the appropriate congruences: (a) 7 modulo 11. (b) 9 modulo 15. (c) 8 modulo 15.