hw4 - EECS 203: Homework 4 Solutions Section 2.3 1. (E)...

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EECS 203: Homework 4 Solutions Section 2.3 1. (E) 6acd a.) The domain is Z+ x Z+ and the range is Z+ c.) The domain is the set of all bit strings. The difference between the number of 1’s and 0’s can be any integer and so the range is Z. d.)The domain is Z+ and the range is Z+. 2. (E) 12bcd b.) f(n) = n 2 + 1 A function is one-to-one if f(x) = f(y) ↔ x = y. Let f(x) = f(y), then x 2 + 1 = y 2 + 1 x 2 = y 2 for all real numbers. However in this case, it does not necessarily mean that x = y. For example, when x = 1 and y = -1, the above equation is satisfied. So, the function is not one-to-one. c.) f(n) = n 3 A function is one-to-one if f(x) = f(y) ↔ x = y. Let f(x) = f(y), then x 3 = y 3 x = y for all real numbers (Since cube root is unique). Hence the function is one-to-one. A function is one-to-one if f(x) = f(y) ↔ x = y. Let f(x) = f(y), then . However in this case, it does not necessarily mean that x = y. For example, when x = 3 and y = 4, the above equation is true. So, the function is not one-to-one. 3. (E) 14bcd b.) f(m, n) = m 2 – n 2 No, the range does not include the value 2. To see this, let us assume there is are m and n such that f(m, n) = 2 and seek a contradiction. Then m 2 − n 2 = (m + n)(m − n) = 2. Note that because m 2 − n 2 is even, m 2 and n 2 must either be both even or both odd. As a result, m and n are either both even or both odd (by exercise 3 in section 1.6). Therefore (m+ n) and (m− n) are both even. That is, there are integers s and t such that (m + n)(m − n) = 2 s 2 t = 4 st . If 4 st = 2, then st = 1/2 . But st is an integer, so this is a contradiction and f is not onto. c.) f(m, n) = m + n + 1 Generally, A function is onto if and only if for every element y in the co-domain, there exists an element x in the domain such that f(x) = y. (In short, Range = Co-domain) Here, Let f(m, n) = y, then y = m + n + 1. The function f is onto because for every integer y, there exist integers m and n such that m+ n = y-1. d.) f(m, n) = |m| - |n| Here, we can achieve negative values when we set m to 0 and positive values when n is 0. Therefore range is equal to Co-domain and the function is onto.
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4. (E) 18bd (Briefly justify your answer for problems 2, 3 and 4) b.) f(x) = -3x 2 + 7 By definition, a function is a bijection if it is both one-to-one and onto
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hw4 - EECS 203: Homework 4 Solutions Section 2.3 1. (E)...

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