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

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

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