Solutions for the Exercises at the End of the Sections 207 28. By hypothesis ða + bÞ 2 = a 2 + b 2 . But, by the rules of algebra ða + bÞ 2 = a 2 + 2ab + b 2 . Thus, we obtain that for all real numbers b, a 2 + 2ab + b 2 = a 2 + b 2 : This implies that 2ab = 0 for all real numbers b. In particular, this equality is true when b 0. Thus we can divide the equality 2ab = 0 by 2b and obtain a = 0. 29. Let n be a four-digit palindrome number (e.g., 1221, 7447, 8998, 1001, etc.). To prove that it is divisible by 11, we need to show that n = 11t for some positive integer t. Since n is a four-digit palindrome number, we can write n = xyyx with x and y integer numbers between 0 and 9, with x 0. We can now try to separate the digits of n. Thus, n = xyyx = 1000x + 100y + 10y + x = 1001x + 110y = 11ð91x + 10yÞ: Because t = 91x + 10y is a positive integer, we proved that n is divisible by 11. 30. We know that x c. So, the denominator of the fraction is a nonzero number, and the fraction is well defined. We want to prove that the number f ðcÞ - f ðxÞ c - x is nonnegative. (Because of the algebraic properties that determine the sign of a fraction, we need to