hw3-s - ECS 20: Homework 3 Solutions Instructor: Prof.Max...

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ECS 20: Homework 3 Solutions Instructor: Prof.Max TA: Yuxi Hu Page 102-104 2. The cubes that might go into the sum are 1, 8, 27,64, 125, 216, 343, 512, and 279. We must show that no two of these sum to a number on this list. If we try the 45 combinations (1+1,1+8,. ..729+729), we see that none of them works. Having exhausted the possibilities, we conclude that no cube less than 1000 is the sum of two cubes. 8. The only perfect squares that differ by 1 are 0 and 1. Therefore these two consecutive integers cannot both be perfect squares. This is a nonconstructive proof. 18.Given x , let n be the greatest integer less than or equal to x , and let ± = x - n . In the notation to be introduced in Section 2.3, n = b x c . Clearly 0 ± < 1, and ± is unique for this n . Any other choice of n would cause the required ± to be less than 0 or greater than or equal to 1, so n is unique as well. 32. Prove by contradiction. Suppose that 3 2 is the rational number p q , where p and q are positive integers with no common factors. Cubing, we see that 2 =
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hw3-s - ECS 20: Homework 3 Solutions Instructor: Prof.Max...

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