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or all integers a, b, c, m with m > 0 and c > 0, if a ≡ b (mod m) then ac ≡ bc (mod mc)

Question

or all integers a, b, c, m with m > 0 and c > 0, if a ≡ b (mod m) then ac ≡ bc (mod mc)

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Solution

The statement is asking to prove that for all integers a, b, c, m with m > 0 and c > 0, if a is congruent to b modulo m, then ac is congruent to bc modulo mc.

Here's the step-by-step proof:

  1. Start with the given congruence a ≡ b (mod m). By definition of congruence, this means that m divides (a - b), or in other words, there exists some integer k such that a - b = mk.

  2. Multiply both sides of the equation a - b = mk by c to get ac - bc = mck.

  3. This equation can be rewritten as ac ≡ bc (mod mc), because mc divides (ac - bc).

So, we have shown that if a ≡ b (mod m) then ac ≡ bc (mod mc) for all integers a, b, c, m with m > 0 and c > 0.

This problem has been solved

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