Posted in Algebra

# New PDF release: Glimpses of Algebra and Geometry (Undergraduate Texts in

By Gabor Toth

ISBN-10: 0387953450

ISBN-13: 9780387953458

Prior version offered 2000 copies in three years; Explores the delicate connections among quantity thought, Classical Geometry and glossy Algebra; Over one hundred eighty illustrations, in addition to textual content and Maple records, can be found through the internet facilitate figuring out: http://mathsgi01.rutgers.edu/cgi-bin/wrap/gtoth/; comprises an insert with 4-color illustrations; contains various examples and worked-out difficulties

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Extra resources for Glimpses of Algebra and Geometry (Undergraduate Texts in Mathematics)

Example text

To prove that π and e are irrational, we follow Hermite’s argument, which dates back to 1873. Consider, for ﬁxed n ∈ N, the function f : R → R deﬁned by f(x) xn (1 − x)n n! 1 n! 2n ck xk . k n n n, n + 1, . . , 2n. In Expanding (1 − x) , we see that ck ∈ Z, k n (−1)k n−k (−1)k nk (by the binomial formula), but we fact, ck will not need this. For 0 < x < 1, we have 1 n! 0. Differentiating, we obtain   0, if m < n or m > 2n (m) f (0) cm m!  , if n ≤ m ≤ 2n. n! 0 < f(x) < and f(0) In any case f (m) (0) ∈ Z (since n!

This indeed shows the three basic types discussed earlier! ♦ Let us try our machinery for the Weierstrass form. Since f(x, y) P(x) − y2 , we have ∂f ∂x P (x) and ∂f ∂y −2y. 3. 8 0 and P (x0 ) 0. If, Let (x0 , y0 ) be a critical point of f . Then y0 2 P(x0 ) 0. Thus, a critical furthermore, (x0 , y0 ) ∈ Cf , then y0 P (r) 0. point of f on Cf is of the form (r, 0), where P(r) Differentiating once more, we see that (r, 0) is a saddle point if P (r) > 0. We now look at these conditions on P from an algebraic point of view.

For this, we estimate k! ∞ j k+1 1 j! k! k! k! + + + ··· (k + 1)! (k + 2)! (k + 3)! 1 1 1 + + + ··· k+1 (k + 1)(k + 2) (k + 1)(k + 2)(k + 3) < 1 1 1 + 2 + 3 + ··· 2 2 2 Clearly, k! k j 0 1 −1 1 − (1/2) 1. 1/j! e. But this implies that e is irrational. ¬ Indeed, assume that e a/b, a, b ∈ N. e. ¬ Corollary. eq is irrational for all 0 q ∈ Q. Proof. eqm . Now choose m to If eq is rational, then so is any power (eq )m be the denominator of q (written as a fraction) to get a contradiction to Theorem 1.