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Prove that addition modulo n on Zₙ is a binary operation. That is, prove (amodn)+(bmodn)≡ (a+b)modn.

User Farukest
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Final answer:

To prove that addition modulo n on Zₙ is a binary operation, we need to show that (a mod n) + (b mod n) is congruent to (a + b) mod n. Let's consider two numbers, a and b, in Zₙ. Since a and b are integers, they can be written as a = qn + r and b = pn + s, where q, p, r, and s are integers and 0 ≤ r, s < n.

Step-by-step explanation:

To prove that addition modulo n on Zₙ is a binary operation, we need to show that (a mod n) + (b mod n) is congruent to (a + b) mod n.

Let's consider two numbers, a and b, in Zₙ. Since a and b are integers, they can be written as a = qn + r and b = pn + s, where q, p, r, and s are integers and 0 ≤ r, s < n.

Now, let's calculate (a mod n) + (b mod n):

  1. a mod n = r
  2. b mod n = s
  3. (a mod n) + (b mod n) = r + s

On the other hand, let's calculate (a + b) mod n:

  1. (a + b) mod n = (qn + r + pn + s) mod n
  2. = (qn + pn) mod n + (r + s) mod n
  3. = 0 mod n + (r + s) mod n
  4. = (r + s) mod n

Since (a mod n) + (b mod n) = (r + s) mod n and (a + b) mod n = (r + s) mod n, we can conclude that (a mod n) + (b mod n) is congruent to (a + b) mod n. Therefore, addition modulo n on Zₙ is a binary operation.

User Daniel Korn
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