Answer
If n|m, where n and m are integers greater than 1, and if a ≡ b (mod m), where a and b are integers, then a ≡ b (mod n).
Work Step by Step
To prove that if n|m, where n and m are integers greater than 1, and if a ≡ b (mod m), where a and b are integers, then a ≡ b (mod n), we can use the following steps:
Given:
n and m are integers greater than 1
n|m (n divides m)
a ≡ b (mod m), where a and b are integers
Step 1: Since n|m, there exists an integer k such that m = nk.
Step 2: Because a ≡ b (mod m), we know that m|(a-b). This means there exists an integer j such that a - b = mj.
Step 3: Substitute m = nk from step 1 into the equation from step 2:
a - b = (nk)j
Step 4: Rearrange the equation from step 3:
a - b = n(kj)
Step 5: Let c = kj. Since k and j are integers, c is also an integer. We can rewrite the equation from step 4 as:
a - b = nc
Step 6: From the equation in step 5, we can see that n|(a-b), which means a ≡ b (mod n).
Therefore, we have shown that if n|m, where n and m are integers greater than 1, and if a ≡ b (mod m), where a and b are integers, then a ≡ b (mod n).
