Students learn how to read and write proofs by actually reading and writing them, asserts author Joseph J. Rotman, adding that merely *reading* about mathematics is no substitute for *doing* mathematics. In addition to teaching how to interpret and construct proofs, Professor Rotman's introductory text imparts other valuable mathematical tools and illustrates the intrinsic beauty and interest of mathematics.

*Journey into Mathematics *offers a coherent story, with intriguing historical and etymological asides. The three-part treatment begins with the mechanics of writing proofs, including some very elementary mathematics--induction, binomial coefficients, and polygonal areas--that allow students to focus on the proofs without the distraction of absorbing unfamiliar ideas at the same time. Once they have acquired some geometric experience with the simpler classical notion of limit, they proceed to considerations of the area and circumference of circles. The text concludes with examinations of complex numbers and their application, via De Moivre's theorem, to real numbers.

Pythagorean triple. Hint: Let n be even, say, n = 2k. If n = 4, then (3, 4, 5) works; if n ≥ 6, then n < k2 – 1, and (n, k2 – 1, k2 + 1) is a Pythagorean triple. If n is odd, say, n = 2k + 1, then (n, 2k(k + 1), 2k2 + 2k + 1) is a Pythagorean triple. TRIGONOMETRY Since you are now studying geometry and trigonometry, I will give you a problem. A ship sails the ocean. It left Boston with a cargo of wool. It grosses 200 tons. It is bound for Le Havre. The mainmast is broken, the cabin boy

established any connection between the area of a circle and its circumference. It follows that it is not yet legitimate to use radian measure, and so it is premature to use limits of integration for the new definite integral that mention π. Perhaps we should not feel so superior to the Greeks! Exercises 3.9. A pizzeria charges $2.50 for a 10” pizza and $5.00 for a 15” pizza (a 10” pizza is circular with diameter 10 inches). Should four hungry students order four 10” pizzas or two 15” pizzas?

1]). In particular, given e > 0, there is some integer ℓ > 1/ε. Of course, if n ≥ ℓ, then n ≥ ℓ > 1/ε, so that 1/n < ε for all n ≥ ℓ , by Theorem 1.4(iii). Therefore, ∣1/n − 0∣ = 1/n < ε for all n ≥ ℓ, and so 1/n → 0. Example. Here is the reason one defines convergence using tails instead of “small terms.” Define a sequence {an} by a2n−1 = 5 and a2n = 1/n; thus, the sequence begins On intuitive grounds, the recurrence of 5 every other term should prevent {an} from converging to 0. The sequence

does have small terms, however. We have just seen, for any given e > 0, that there is ℓ with 1/ℓ < ε; hence, a2ℓ < ε, and the sequence has small terms. On the other hand, if an → 0, then every open interval (−ε, ε) would contain a tail. But if e = 1, then 5 does not lie in (−1, 1), and so no tail lies within (“half” the terms, those equal to 5, are missing); hence, an ↛ 0. Exercise 3.33 asks you to prove that this sequence diverges. Just as in the applications of the Getting Close Principle, one

Corollary 3.20 (Small Terms Axiom). Let A be a positive number. For every ε > 0, there is an integer ℓ with . Proof. Since , Theorem 3.19 shows that setting c = A in Theorem 3.18 now gives The definition of convergence says that for every ε > 0, there is some integer ℓ with for all n ≥ ℓ. In particular, . The next result shows that the classical Greek notion of approximation is a special case of convergence. Theorem 3.21. If k1 < k2 < k3 < ⋯ < A approximates A from below, then kn → A.