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Unformatted text preview: CS 2800 Fall 2008 Homework #2 Solution Section 1.7 Problem 10 [4pts] Since we have three numbers, at least two of them must be positive or at least two of them must be negative (since we have only two signs available for any number, discounting the possibility of any of them being 0). In either case, we can get a positive number by multiplying together the two positive numbers or the two negative numbers. Thus, we have a non-constructive proof (since we did not explicitly state which two numbers to multiply together to get a positive result) that the product of two of the given numbers must be nonnegative. Problem 20 [4pts] We simply follow the hint. Since every square is non-negative, we have ( x- 1 /x ) 2 0. Multiplying out the LHS, we have x 2 + 1 /x 2- 2 0 i.e. x 2 + 1 /x 2 2. Problem 28 [4pts] If | y | 2, then 2 x 2 + 5 y 2 2 x 2 + 20 20, so the only values of y left to try are 0 and 1. In the former case, we are looking for solutions to the equation 2 x 2 = 14 and in the latter case to 2 x 2 = 9. Clearly there are no integer solutions to these equations, so there are no integer solutions to the original equation. Problem 34 [6pts] Let a be a rational number and b be an irrational number such that b > a . Now consider the number c = ( a + b ) 2 . Clearly, since c is the average of a and b , a < c < b . Moreover, c must be irrational. To see why this is true, assume otherwise i.e. c is rational. Then, since c = ( a + b ) 2 , we have b = 2 c- a . Since c and a are both rational and the set of rational numbers is closed under subtraction, b must be rational, which is a contradiction. Hence c is irrational. Thus we have shown the existence of an irrational number between any two arbitrary rational and irrational number. Some people chose to work with an irrational number of a specific form, such as 2 2 b (where b is rational). This is not general enough, since there are many irrational numbers that cannot be written in that form....
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- Fall '07