Prove that the relation of congruence modulo n is transitive. That is, prove that for all integers a, b, c, and n with n> 1, if a =b (mod n) and bc (mod n), then aAC (mod n). Proof: Suppose a, b, c, andn are any integers with n> 1 and a =b (mod n) and b c(mod n). By definition of congruence modulo n, this means that n| (a - b) and n By definition of divisibility, since n| (a - b), and since n there are integers r and s such that a - b= m and b - sn. Now a -c = (a- b)+ Rewriting the difference on the right-hand side of this equation in terms of n, r, and s and factoring the result completely gives that a-c= definition of divisibility. Thus a = c (mod n) by definition of congruence modulo n. Since the sum of any two integers is an integer, it follows that n

Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter2: The Integers
Section2.5: Congruence Of Integers
Problem 58E: a. Prove that 10n(1)n(mod11) for every positive integer n. b. Prove that a positive integer z is...
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Prove that the relation of congruence modulo n is transitive. That is, prove that for all integers a, b, c, and n with n> 1, if a =b (mod n) and bc (mod n), then aA (mod n).
Proof: Suppose a, b, c, andn are any integers with n> 1 and a =b (mod n) and b c (mod n). By definition of congruence modulo n, this means that n| (a - b) and n
By definition of divisibility, since n| (a - b), and since
there are integers rand s such that a - b= m and b -
sn.
Now a - c = (a-b)+
Rewriting the difference on the right-hand side of this equation in terms of n, r, and s and factoring the result completely gives that
Since the sum of any two integers is an integer, it follows that n
) by definition of divisibility. Thus a = c (mod n) by definition of congruence modulo n.
Transcribed Image Text:******* Prove that the relation of congruence modulo n is transitive. That is, prove that for all integers a, b, c, and n with n> 1, if a =b (mod n) and bc (mod n), then aA (mod n). Proof: Suppose a, b, c, andn are any integers with n> 1 and a =b (mod n) and b c (mod n). By definition of congruence modulo n, this means that n| (a - b) and n By definition of divisibility, since n| (a - b), and since there are integers rand s such that a - b= m and b - sn. Now a - c = (a-b)+ Rewriting the difference on the right-hand side of this equation in terms of n, r, and s and factoring the result completely gives that Since the sum of any two integers is an integer, it follows that n ) by definition of divisibility. Thus a = c (mod n) by definition of congruence modulo n.
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