Math 310: Problem set 8

Instructions: This problem set is due Thursday, November 9, 2006. Your goal is not only to give correct answers but to communicate your ideas well. Make sure you use good English.
  1. Let R be the ring whose elements are continuous functions f:[0,1] -> R (i.e., real valued functions y=f(x) defined on the interval 0 <= x <= 1. Thus x has to be between 0 and 1 inclusive, and for each such x, the value f(x) of f at x has to be real, and f:[0,1] -> R has to be continuous).
  2. Let b, c and d be positive integers bigger than 1. Suppose we try to define a map h:Z/bZ -> Z/cZ x Z/dZ by specifying that h([x]b) = ([x]c,[x]d).
  3. Let b = 210, c = 14 and d = 15. Let h:Z/bZ -> Z/cZ x Z/dZ be defined as in the previous problem. Find an integer n such that h([n]b) = ([5]c,[12]d). Explain how you obtain your answer.
  4. Answer: From the previous problem or from the Chinese Remainder Theorem stated in class, we know h is bijective. Thus there is a solution. To find it, we must solve
    x =  5 (mod 14)
    x = 12 (mod 15)
    
    simultaneously. But from x = 5 (mod 14) we have x = 5 + 14q for some integer q. From x = 12 (mod 15) we have x = 12 + 15r for some integer r. Thus 5 + 14q = x = 12 + 15r or 7 = 14q - 15r. Solve this using Euclid's algorithm. (Alternatively, note that 7 = (14-15)(-7) so we can take q = -7 and r = 7.) We get x = 5 - 7*14 = -93 (which we check with x = 12 - 7*15 = -93). Thus n = -93 works, as does -93 + m210 for any integer m.