By Jerry R. Shipman

**REA’s Algebra and Trigonometry challenge Solver **

Each

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**Sample text**

Calculate f13 . The Fibonacci numbers obey a surprising formula—assuming for a moment that we √ know about real numbers such as 5. 29 implies that the strange expression on the right-hand side is always an integer. 29. The kth Fibonacci number is given directly by the formula √ k √ k 1 1+ 5 1− 5 fk = √ − . 2 2 5 √ We need to check two base cases, because the recursion formula for fk involves the two previous sequence members. √ Proof. Let a = 1+2 5 and b = 1−2 5 . We prove P(k) : fk = √15 ak − bk by (strong) induction on k ∈ N.

This statement on the right is called the contrapositive of ♥ ⇒ ♣. Contrapositives can be useful for proofs: sometimes it is easier to prove (not ♣) ⇒ (not ♥) than to prove ♥ ⇒ ♣. 4. 6 says that (assuming m, n ∈ Z) m ≤ n ≤ m implies m = n. We could have proved this proposition by showing its contrapositive: if m = n then m > n or n > m. 8. 3. Namely, the negation of m ≤ n ≤ m is m > n or n > m. 5. Re-prove some of the if–then propositions in Chapters 1 and 2 by proving their contrapositives. 3 Negations Here we discuss how to negate mathematical statements.

One can also write x1 x2 · · · xk for ∏kj=1 x j . n (ii) Assuming ∏nj=1 x j defined, we define ∏n+1 j=1 x j := ∏ j=1 x j · xn+1 . We denote the nonnegative integers by Z≥0 := {m ∈ Z : m ≥ 0} . Factorial. As a third example, we define k! (“k factorial”) for all integers k ≥ 0 by: (i) Define 0! := 1. (ii) Assuming n! defined (where n ∈ Z≥0 ), define (n + 1)! ) · (n + 1). In these examples, a new sequence is being defined step by step: the (n+1)th term can be written down only when you already know the nth term, so it may take substantial calculation to actually write down the 1,000,000th term.