Question

(a) Solve the inequality x² > 2x + 1

(b) Use mathematical induction to prove that 2^(s+1) > n² for n ∈ ℤ, n > 3.

Answer 1

Solve the inequality x² > 2x + 1 by rearranging the terms and using the quadratic formula to find intervals that satisfy the inequality. Prove that 2^(s+1) > n² for n > 3 using mathematical induction by establishing a base case for n=4 and showing the statement holds for each n=k+1 given it holds for n=k.

Step-by-step explanation:

To solve the inequality x² > 2x + 1, we first bring all terms to one side to get x² - 2x - 1 > 0. This is a quadratic inequality, and the next step is to find the roots of the corresponding quadratic equation x² - 2x - 1 = 0. The roots of this equation might not be integers, so we can use the quadratic formula to find them. Once the roots are found, we test intervals determined by the roots to find where the inequality is satisfied.

Now, to prove the statement 2^(s+1) > n² for n ∈ ℤ, n > 3 using mathematical induction:

1. Base Case: Verify the statement for the initial value of n, which is 4 in this case.
2. Inductive Step: Assume the statement holds for some n = k, where k is a positive integer greater than 3, and show it holds for n = k+1.
3. Finish the proof by verifying that if the statement holds for n = k, it must also hold for n = k+1 using the inductive hypothesis.