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Assignment 4 MATH 1200 SOLUTION Due Monday October 21 Instead of representing real numbers in decimal notation, for any other natural number there is a corresponding base n representation. For n = 10 this is the usual decimal expansion. So, the statement that every real number x in the interval [0, 1] has a base 10 representation means that there are integers ki such that each ki is between 0 and 9 and x= k3 k2 k1 + 3 + ... + 10 102 10 And we write this as x = 0.k1 k2 k3 .... Instead of base 10, let’s consider base 7. It is also the case that every real number has a base 7 representation. Let’s use the following notation for the base 7 representation of a number between 0 and 1: x2 x3 x1 + 2 + 3 + ... where x1 , x2 , ... ∈ {0, 1, ..., 6} 7 7 7 (1) Find the familiar decimal notation for each of the following numbers given in base 7 form: (a) (0.111.....)7 (0.x1 x2 x3 ....)7 = Solution: (0.111.....)7 = 1 1 1 1 1 1 = = 0.1666666.... + + 3 + ... = · 7 72 7 7 (1 − 17 ) 6 (b) (0.12121212.....)7 Solution: 1 2 1 2 1 2 + + 3 + 4 + 5 + 6 + ... = 7 72 7 7 7 7 7 2 7 2 7 2 9 9 9 = 2 + 2 + 4 + 4 + 6 + 6 + ... = 2 + 4 + 6 + ... = 7 7 7 7 7 7 7 7 7 9 49 9 1 1 9 1 9 = = 2 · (1 + 2 + 4 + ...) = 2 · · = = 0.1875 7 7 7 7 (1 − 712 ) 49 48 48 (0.121212.....)7 = (2) Find the first few digits in the base 7 representation of the following numbers. Try to find the full base 7 representation, i.e., all the digits. (a) 17 SOLUTION: 17 is already represented as a sum of reciprocals of powers of 7, so its base 7 representation is (0.1)7 . (b) 1 2 SOLUTION: We want to write 12 = x71 + x722 + x733 + ... where x1 , x2 , ... ∈ {0, 1, ..., 6} I.e., find x1 , x2 , ..... 1/2 lies between 3/7 and 4/7, so the first digit must be 3. 1/2 lies between 3/7 + 3/49 = 24/49 and 3/7 + 4/49 = 25/49 so the second digit is also 3. One can continue in this way but perhaps one can guess that every digit is 3? Well, 3/7 + 3/72 + 3/73 + ... = 3/7 · ( So all the digits are 3. 1 ) = 3/7 · 7/6 = 3/6 = 1/2 1 − 71 2 (c) 3 5 SOLUTION: Proceed as above. Since 4/7 < 3/5 < 5/7. The first digit is 4. Now find x between 0 and 6 such that 4/7 + x/49 ≤ 3/5 < 4/7 + (x + 1)/49. Well since 29/49 < 3/5 < 30/49 and 4/7 + 1/49 = 29/49 the second digit is 1. So, the base 7 expansion of 3/5 is of the form (0.41???????)7 . Hmmmm wonder what the rest of the digits are.....