Question

Suppose that a_j ≡ b_j (mod m) for j = 1, 2,...,n. Use

mathematical induction to prove that
a) ∑_(j=1)^n a_j ≡ ∑_(j=1)^n b_j (mod m)
b) ∏_(j=1)^n a_j ≡ ∏_(j=1)^n b_j (mod m)

Answer 1

To prove both parts a) and b) using mathematical induction, we need to show that the given congruences hold for n=1 and then assume they hold for n=k and prove that they hold for n=k+1. We can prove both parts by applying the principle of mathematical induction.

Step-by-step explanation:

To prove both parts a) and b) using mathematical induction, we need to show that the given congruences hold for n=1 and then assume they hold for n=k and prove that they hold for n=k+1. Let's start with part a):

Base Case: For n=1, we have a_1 ≡ b_1 (mod m), which is true by the given information.

Inductive Hypothesis: Assume that ∑_(j=1)^k a_j ≡ ∑_(j=1)^k b_j (mod m) for some k.

Inductive Step: We want to prove that ∑_(j=1)^(k+1) a_j ≡ ∑_(j=1)^(k+1) b_j (mod m).

Using the inductive hypothesis, we have ∑_(j=1)^(k+1) a_j = (∑_(j=1)^k a_j) + a_(k+1) ≡ (∑_(j=1)^k b_j) + b_(k+1) = ∑_(j=1)^(k+1) b_j (mod m). Therefore, part a) is proved.

Similarly, we can prove part b) using mathematical induction:

Base Case: For n=1, we have a_1 ≡ b_1 (mod m), which is true by the given information.

Inductive Hypothesis: Assume that ∏_(j=1)^k a_j ≡ ∏_(j=1)^k b_j (mod m) for some k.

Inductive Step: We want to prove that ∏_(j=1)^(k+1) a_j ≡ ∏_(j=1)^(k+1) b_j (mod m).

Using the inductive hypothesis, we have ∏_(j=1)^(k+1) a_j = (∏_(j=1)^k a_j) * a_(k+1) ≡ (∏_(j=1)^k b_j) * b_(k+1) = ∏_(j=1)^(k+1) b_j (mod m). Therefore, part b) is proved.