HW4-Solutions

# HW4-Solutions - EECS 203 DISCRETE MATHEMATICS Homework 4...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EECS 203: DISCRETE MATHEMATICS Homework 4 Solutions 1. (8 points) Chapter 2.3, Problem 18. Explain your answer to each part. (1)- 3 x + 4: This is a bijection. It is one-to-one because no two elements in the domain map to the same element in the co-domain: if- 3 x + 4 =- 3 y + 4 then x must equal y . The function is onto because every element of the co-domain is in the function’s range: the element in the domain mapping to z is (- z + 4) / 3. (2)- 3 x 2 + 7: This is not a bijection. It is neither one-to-one nor onto: it is not one-to-one because k and- k in the domain map to the same element in the co-domain. The function is not onto because all real numbers greater than 7 are not contained in the function’s range. (3) x +1 x +2 : This is not a bijection from R to R for two reasons. First, there is no x for which f ( x ) = 1, so f is not onto. Second, f is not even a function from R → R because f (- 2) =- 1 / 0 is not in R . (We can make f a function by excluding- 2. If R \{- 2 } is the set of all real numbers except 2 then f is a function from R \{- 2 } → R .) (4) x 5 + 1: This is a bijection, for the same reasons as (1) above. Here, because the variable x is taken to an odd power, the exponent does not cause two elements of the domain to map to the same co-domain element. It is a bijection because every real number has a unique fifth root....
View Full Document

## This note was uploaded on 12/20/2010 for the course EECS 203 taught by Professor Yaoyunshi during the Fall '07 term at University of Michigan.

### Page1 / 3

HW4-Solutions - EECS 203 DISCRETE MATHEMATICS Homework 4...

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

View Full Document
Ask a homework question - tutors are online