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Unformatted text preview: ECS 20: Homework 4 Solutions Instructor: Prof.Max TA: Yuxi Hu Page 146149 4a. The domain is the set of nonnegative integers and the range is the set of digits(0 through 9). 4b. The domain is the set of positive integers, and the range is the set of integers greater than 1. 8. a.1; b.2; c.1; 18. If we can find an inverse, the function is a bijection. Otherwise we must explain why the function is not onetoone or not onto. a. This is a bijection since the inverse function is f 1 ( x ) = 4 x 3 b. This is not onetoone since f (17) = f ( 17), for instance. It is also not onto, since the range is the interval (∞ , 7]. For example, 42548 is not in the range. c. This function is a bijection, but not from R to R . To see that the domain and range are not R , note that x = 2 is not in the domain, and x = 1 is not in the range. On the other hand, f is a bijection from R {2 } to R { 1 } , since its inverse is f 1 ( x ) = 1 2 x x 1 . d. It is clear that this continuous function is increasing throughout its entire domain( R ) and it takes on both arbitrarily large values and arbitrarily small ones. So it is a bijection. Its inverse is clearly f 1 ( x ) = 5 √ x 1 30. To clarify the setting, suppose that f : A → B and g : B → C , so that f ◦ g : A → C . We will prove that if f ◦ g is onetoone, then g is also onetoone, so not only is the answer to the question ”yes”, but part of the hypothesis is not even needed. Suppose that g were not onetoone. By definition this means that there are distinct elements a 1 and a 2 in A such that g ( a 1 ) = g ( a 2 ). Then certainly)....
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 Spring '09
 Khoel
 Halting problem, arbitrarily small ones

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