hw4-s - ECS 20: Homework 4 Solutions Instructor: Prof.Max...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECS 20: Homework 4 Solutions Instructor: Prof.Max TA: Yuxi Hu Page 146-149 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 one-to-one or not onto. a. This is a bijection since the inverse function is f- 1 ( x ) = 4- x 3 b. This is not one-to-one 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 one-to-one, then g is also one-to-one, so not only is the answer to the question yes, but part of the hypothesis is not even needed. Suppose that g were not one-to-one. 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)....
View Full Document

Page1 / 3

hw4-s - ECS 20: Homework 4 Solutions Instructor: Prof.Max...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online