rec0621_key - COT3100 Summer2001 Recitation on functions...

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COT3100 Summer’2001 Recitation on functions (Solutions) 06/21/2001 1. Given g ={(1, b ), (2, c ), (3, a )}, a function from X ={1, 2, 3} to Y ={ a, b, c, d }, and f = {( a , x ), ( b , x ), ( c , z ), ( d , w )}, a function from Y to Z ={ w , x , y , z } a) write f ο g as a set of ordered pairs and draw the arrow diagram of f ο g ; f ο g = {(1, x ), (2, z ), (3, x )} b) Determine if f ο g is surjective or injective. f ο g is neither surjective nor injective. 2. Let A =R - { - 1}, where R is the set of real numbers, and define the function f : A R by the formula 1 2 ) ( + = x x x f . Prove that f is injective but not surjective. 1) To prove that f ( x ) is injective, it is sufficient to show that if f ( x 1 )= f ( x 2 ), then x 1 = x 2 . So, assume f ( x 1 )= f ( x 2 ). Then we have 2 x 1 /( x 1 +1) = 2 x 2 /( x 2 +1). Using the fact that denominators are not zero ( x 1 , x 2 - 1) we get by algebra : 2 x 1 ( x 2 +1) = 2 x 2 ( x 1 +1), or: 2 x 1 x 2 + 2 x 1 = 2 x 2 x 1 +.2 x 2 x 1 = x 2 2) To prove that f ( x ) is not surjective, it is sufficient to find the value y R , that has no pre-image in A . To find such y let y = 2 x /( x +1) and try to solve for x as a function of y . We have by algebra: y ( x +1)=2 x x (2 - y )= y x = y /(2 - y ). From here we see, that if we take y = 2, then there is no real x such that f ( x ) =2. 3. a) Let f be a function, f : S S , where | S | = n . Prove that f is injective iff it is 1 1 2 3 a y z
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surjective.
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rec0621_key - COT3100 Summer2001 Recitation on functions...

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