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Unformatted text preview: Solutions to Homework Set #7 CSE 260, Spring 2007 Section 2.3 2. Determine whether f is a function from Z to R if (a) f ( n ) = n . This is not a function because the rule is not welldefined. We do not know whether f ( n ) = n or f ( n ) = n . For a function, it cannot be both at the same time. (b) f ( n ) = p ( n 2 + 1). This is a function. For all integers n , p ( n 2 + 1) is a welldefined real number. (c) f ( n ) = 1 n 2 4 . This is not a function with domain Z , since for n = 2 and n = 2 the value of f ( n ) is not defined by the given rule. In other words, f (2) and f ( 2) are not specified since division by 0 makes no sense. 5. See textbook. 15. Determine whether the function f : Z Z Z is onto if (a) f ( m,n ) = m + n . Onto. (b) f ( m,n ) = m 2 + n 2 . Not onto. (c) f ( m,n ) = m . Onto. (d) f ( m,n ) =  n  . Not onto. (e) f ( m,n ) = m n . Onto. 19. Determine whether each of these functions is a bijection from R to R ....
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 Spring '08
 SaktiPramanik

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