Question

Use mathematical induction to show that if a, b, and c are the lengths of the sides of a right triangle, where c is the length of the hypotenuse, then a^n + b^n < c^n for all integers n with n ≥ 3.

Answer 1

To prove a^n + b^n < c^n for all integers n with n ≥ 3 in a right triangle, we can use mathematical induction. The base case is established by showing that a^3 + b^3 < c^3. In the inductive step, assuming a^k + b^k < c^k for some k ≥ 3, we prove a^(k+1) + b^(k+1) < c^(k+1). Therefore, by mathematical induction, the inequality holds true.

Step-by-step explanation:

To prove that a^n + b^n < c^n for all integers n with n ≥ 3, we will use mathematical induction.

Step 1: Base case: For n = 3, we have a^3 + b^3 < c^3. This is true because a^3 + b^3 = (a + b)(a^2 - ab + b^2) and c^3 = (a^2 + b^2 + 2ab)(a + b). Since a^2 - ab + b^2 > 0 (by the positivity of squares) and a + b < a^2 + b^2 + 2ab, it follows that (a + b)(a^2 - ab + b^2) < (a^2 + b^2 + 2ab)(a + b), which implies a^3 + b^3 < c^3.

Step 2: Inductive step: Assume a^k + b^k < c^k holds true for some positive integer k ≥ 3. We need to prove a^(k+1) + b^(k+1) < c^(k+1). Using the assumption, we have a^(k+1) + b^(k+1) = (a^k + b^k)(a + b) + ab(a^(k-1) + b^(k-1)) < c^k(a + b) + abc^(k-1) by the inductive hypothesis. Since a + b < c and a^(k-1) + b^(k-1) ≤ c^(k-1), it follows that c^k(a + b) + abc^(k-1) < c^k(c) + abc^(k-1) = c^(k+1)(c) + abc^(k-1) = c^(k+1)(a + b) < c^(k+1)(c), proving a^(k+1) + b^(k+1) < c^(k+1).

Therefore, by mathematical induction, we have shown that a^n + b^n < c^n for all integers n with n ≥ 3 in a right triangle with sides a, b, and c.