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Additional info for Glimpses of Algebra and Geometry (Undergraduate Texts in Mathematics)

Example text

1 + m2 This shows that (x, y) is a rational point on S1 iff m is rational. (Indeed, m ∈ Q clearly implies x, y ∈ Q . ) Varying m in Q , we obtain 1 all rational points on S (except (0, −1), which corresponds to a vertical line intersection point). Setting m −s/t, t 0, s, t ∈ Z, we recover the solution set above. This “method of rational slopes” works for all quadratic curves described by f(x, y) 0, where f is a quadratic polynomial with rational coefﬁcients in the variables x, y, provided that there is at least one rational point on the curve.

There Are No Irrational Numbers at All”—Kronecker 8. Solve Problem 7 along the lines of Problem 6 by building a 3-dimensional pyramid staircase whose volume represents 12 +22 +· · ·+n2 . Cut the staircase into square pyramids and triangular prisms. 9. Show that 13 + 23 + · · · + n3 n(n + 1) 2 2 in the following geometric way (this was known to the Arabs about 1000 years ago): For k 1, . . , n, consider the square Sk ⊂ R2 with vertices (0, 0), (k(k + 1)/2, 0), (k(k + 1)/2, k(k + 1)/2), and (0, k(k + 1)/2).

Berggren, J. Borwein, and P. Borwein, π: A Source Book, Springer, 1997. 6 7 2. “. . ). The Babylonians obtained the ﬁrst value by stating that the ratio of the perimeter of a regular hexagon to the circumference of the circumscribed circle was “equal” to 57/60 + 36/602 . 1). The area of the octagon, 63 units, was rounded up to 82 , and 28 /34 . the approximate value of π turned out to be 82 /(9/2)2 A well-known recorded approximation of π was made by Archimedes (c. ), who applied Eudoxus’s method of exhaustion, the approximation of the circumference of the unit circle by the perimeters of inscribed and circumscribed regular polygons.