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5- If a = b mod (n), then ac = bc mod (n). 6- If a = b mod(n), then ak = bkmod(n) for every positive integer k. ас

5- If a = b mod (n), then ac = bc mod (n). 6- If a = b mod(n), then ak = bkmod(n) for every positive integer k. ас

Advanced Engineering Mathematics
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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1- анатоd (п). 2- If a = b mod (n), then b = a mod(n). 3- If a = b mod(n) and b = c mod(n), then a = cmod(n). 4- Ifa = b mod(n) and e = d mod(n), then a +c = b+d mod(n), ac = db mod(n). 5- If a = b mod (n), then ac = be mod(n). 6- If a = b mod(n), then ak = bkmod(n) for every positive integer k. Proof :- Let a, b, c, d e Z 1- Since a - a = 0 = 0.n - k = 0 - a = a mod(n). 2- Let a = b mod(n) → a-b = k.n,k ez multiply (-1) -b-a (-k).n,-k e Z - b = a mod(n). 3- Let a = b mod(n),b = c mod(n) to proof a = c mod(n). Since a = b mod(n) a -b = k,n ,k, € Z and b = c mod(n) b-c = k,n ,kz €Z, take a - c = a - b +b -c = (a – b) + (b - c) = k,n + k,n = (k, + k2)n = k.n 3 k = k, + k2 € Z. So a = cmod(n). 4- Sincea = b mod (n) - a -b = k,n,k, ez, c = d mod(n) c-d = k,n, k2 €Z, : (a – b) + (c - d) = k,n + kan - (a + c) - (b + d) = (k, + k2)n = kn a k = k, + kz € Z. So a -b = c-d mod(n). 5- H.W. 6- H.W.
Transcribed Image Text:1- анатоd (п). 2- If a = b mod (n), then b = a mod(n). 3- If a = b mod(n) and b = c mod(n), then a = cmod(n). 4- Ifa = b mod(n) and e = d mod(n), then a +c = b+d mod(n), ac = db mod(n). 5- If a = b mod (n), then ac = be mod(n). 6- If a = b mod(n), then ak = bkmod(n) for every positive integer k. Proof :- Let a, b, c, d e Z 1- Since a - a = 0 = 0.n - k = 0 - a = a mod(n). 2- Let a = b mod(n) → a-b = k.n,k ez multiply (-1) -b-a (-k).n,-k e Z - b = a mod(n). 3- Let a = b mod(n),b = c mod(n) to proof a = c mod(n). Since a = b mod(n) a -b = k,n ,k, € Z and b = c mod(n) b-c = k,n ,kz €Z, take a - c = a - b +b -c = (a – b) + (b - c) = k,n + k,n = (k, + k2)n = k.n 3 k = k, + k2 € Z. So a = cmod(n). 4- Sincea = b mod (n) - a -b = k,n,k, ez, c = d mod(n) c-d = k,n, k2 €Z, : (a – b) + (c - d) = k,n + kan - (a + c) - (b + d) = (k, + k2)n = kn a k = k, + kz € Z. So a -b = c-d mod(n). 5- H.W. 6- H.W.
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