Proof that if a \equiv b (\bmod n), then a+c \equiv b+c(\bmod n) and ac \equiv bc (\bmod n).

Nannie Mack 2021-02-26 Answered

Proof that if ab(bmodn), then a+cb+c(bmodn) and acbc(bmodn).

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Expert Answer

Willie
Answered 2021-02-28 Author has 95 answers

Step 1
Apply the definition of congruence and divisibility, to express ab(bmodn).
From the definition of congruence, ab(bmodn) means that n(ab).
By the definition of divisibility, n(ab) implies that (ab) is a multiple of n.
That is, ab=nk, where k is an integer.
Step 2
To prove a+cb+c(bmodn), add and subtract c on the left hand side of the equation
ab=nk and use the definition of congruence.
ab=nk
ab+cc=nk
(a+c)(b+c)=nk
Since, (a+c)(b+c) is a multiple of n, implies that n[(a+c)(b+c)].
Hence, by the definition of congruence, a+c=≡b+c(bmodn).
Thus, if ab(bmodn) then a+c=≡b+c(bmodn).
Step 3
To prove acbc(bmodn), use the definition of congruence and the properties of divisibility.
ab(bmodn) implies that n(ab).
By the properties of divisibility, if n divides (ab) then n also divides the multiple of (ab).
That is, n|(ab)c=n|(acbc)
Hence, by the definition of congruence, acbc(bmodn).
Therefore, if ab(bmodn) then acbc(bmodn).

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