Stitz-Zeager_College_Algebra_e-book

In the scenario of example 1015 the period of the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: t P (k + 1) is true. Since (z )k+1 = (z )k z , we can use the induction hypothesis and 576 Sequences and the Binomial Theorem write (z )k = z k . Hence, (z )k+1 = (z )k z = z k z . We now use the product rule for conjugates4 to write z k z = z k z = z k+1 . This establishes (z )k+1 = z k+1 , so that P (k + 1) is true. Hence, by the Principle of Mathematical Induction, (z )n = z n for all n ≥ 1. 3. The first wrinkle we encounter in this problem is that we are asked to prove this formula for n > 5 instead of n ≥ 1. Since n is a natural number, this means our base step occurs at n = 6. We can still use the PMI in this case, but our conclusion will be that the formula is valid for all n ≥ 6. We let P (n) be the inequality 3n > 100n, and check that P (6) is true. Comparing 36 = 729 and 100(6) = 600, we see 36 > 100(6) as required. Next, we assume that P (k ) is true, that is we assume 3k > 100k . We need to show that P (k + 1) is true, that is, we need to show 3k+1 > 100(k + 1). Since...
View Full Document

Ask a homework question - tutors are online