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Published byEmil Phillips Modified over 9 years ago
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Functions (Mappings)
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Definitions A function (or mapping) from a set A to a set B is a rule that assigns to each element a of A exactly one element b of B. The set A is called the domain of , and B is called the range of . If assigns b to a, then b is called the image of a under . The subset of B comprising all the images of elements of A is called the image of A under
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Notation A –> B means is a mapping from set A to set B. (a) = b or : a –> b means that function maps element a to element b (i.e. the image of a is b).
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Example 1 Let : R –> R be given by (x) = sin(x). The image of /2 under is 1 The image of R under is [-1,1]. domain of is R range of is R (R) = [-1,1]
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Example 2 : Q –> Z given by : p/q –> p+q (1/2) = 1 + 2 = 3 (2/4) = 2 + 4 = 6 Since 1/2 = 2/4, but (1/2) ≠ (2/4) is not a function!
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To prove a rule is a function Assume x 1 = x 2 Show (x 1 ) = (x 2 ) In this case we say that is well-defined.
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Show is well-defined Let :Z –> Z be given by (n) = n 2 mod 2 Suppose n 1 = n 2. (n 1 )– (n 2 ) = n 1 2 mod 2 –n 2 2 mod 2 = (n 1 – n 2 )(n 1 +n 2 ) mod 2 = 0 since (n 1 – n 2 ) = 0 So (n 1 ) = (n 2 ) Therefore, is well-defined.
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Composition of functions Let : A –> B and : B –> C. The composition is the mapping from A to C defined by (a) = ( (a)). AB C a (a) (a)
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Order matters! When we compose and , we must write Unless A = C, does not make sense. AB C a (a) (a)
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One to one functions A function from a set A is called one- to-one if Note: This is the converse to the well- defined condition.
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Show not one-to-one Show :R –> R given by (x) = x 2 is not one–to–one. (–2)= 4 = (2), but –2 ≠ 2 So is not one-to-one.
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Show one-to-one Show :[0,∞) –> R given by (x) = x 2 is one– to–one. Suppose (x 1 )= (x 2 ). Then 0 = (x 1 )– (x 2 ) = x 1 2 –x 2 2 = (x 1 –x 2 )(x 1 +x 2 ) So either (x 1 –x 2 ) = 0 or (x 1 +x 2 ) = 0 If (x 1 –x 2 ) = 0, then x 1 = x 2 If (x 1 +x 2 ) = 0, then x 1 = x 2 = 0 since the domain is [0,∞) In either case, x 1 = x 2, so is one-to-one.
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Onto functions A function from a set A to a set B is said to be onto B if each element of B is the image of at least one element of A.
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Show not onto Show :[0,∞) –> R given by (x) = x 2 is not onto. Suppose –1 = (x) for some x in [0,∞). Then -1 = x 2 ≥ 0 This counterexample shows is not onto.
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Show onto Show :[0,∞) –> [0,∞) given by (x) = x 2 is onto. Proof: Choose any number b ≥ 0. Let a = √b. Then (a) = (√b) 2 = b. So is onto.
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Properties of functions Given :A–>B, :B–>C, and :C–>D, then 1. ( ) = ( ) . (Associativity) 2.If and are one-to-one, then is one-to-one. 3.If and are onto, then is onto. 4.If is one-to-one and onto, then there is a function -1 from B to A such that -1 (a)=a for all a in A and -1 (b)=b for all b in B.
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Proof of Associativity Choose any a in A. Let b = (a), c = (b), and d = (c). Notice that (a) = c and (b) = d. Then ( ) (a) = (b) = d Also, ( )(a) = (c) = d Since ( ) (a) = ( )(a) for all a in A, ( ) = ( )
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Prove one-to-one Suppose (x 1 ) = (x 2 ) Set y 1 = (x 1 ) and y 2 = (x 2 ). Then (y 1 ) = (y 2 ) Since is one-to-one, y 1 =y 2 But then (x 1 ) = (x 2 ). Since is one-to-one, x 1 = x 2. Therefore, is one-to-one.
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Prove onto Choose any c in C Since is onto, there is a b in B with (b)=c. Since is onto, there is an a in A with (a)=b. Then (a) = (b) = c Therefore, is onto.
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Proof of inverse functions The proof consists of three steps. 1.Construct the inverse function. 2.Show that the inverse is well-defined. 3.Show that the inverse function has the required cancellation properties.
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1. Construction Given :A->B is one-to-one and onto. Choose any b in B. Since is onto, there is an a in A with (a)=b. Set -1 (b) = a.
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2. Well-defined Suppose b 1 = b 2 in B. Let a 1 = -1 (b 1 ) and a 2 = -1 (b 2 ) Then (a 1 ) = b 1 = b 2 = (a 2 ) Since is one-to-one, a 1 = a 2 That is, -1 (b 1 ) = -1 (b 2 ) Therefore, -1 is well-defined.
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3. Cancellation Choose any a in A. Set b = f(a) and note that -1 (b) = a. Then -1 (a) = -1 (b) = a, and -1 (b) = (a) = b. Since -1 is well-defined and the cancellation laws hold, we are done.
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