key4fa00 - COT3100.01, Fall 2000 S. Lang Solution Key to...

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COT3100.01, Fall 2000 S. Lang Solution Key to Assignment #4 (Question 2 corrected) 10/31/2000 First, we define when two functions are considered equal (or identical). Definition. Two functions f : A B and g : A B are said equal , denoted f = g , if f ( x ) = g ( x ) for every x A , where A is the common domain of the two functions. For example, two functions f ( x ) = ( x + 1) 2 and g ( x ) = x 2 + 2 x + 1, both defined from R to R , where R denotes the set of real numbers, are equal because ( x + 1) 2 = x 2 + 2 x + 1 by algebra laws. As a reminder, when there are two functions f : A B and g : B C , they can be composed to form a function denoted g o f : A C , with g precedes f in the composition notation g o f . However, when f and g are considered as relations (because functions are special cases of relations), the notation for composing would be f o g , which as a relation has the property f o g A × C . The context should make it clear which convention (either g o f or f o g ) is used. 1. (11 pts.) Consider a set A = { a , b , c }, a set B = {1, 2}, and a set C = { u , v , w }. Define a relation R A × B with R = {( a , 1), ( b , 1), ( c , 2)}, a relation S B × C with S = {(1, u ), (1, v ), (2, w )}, and a relation T A × C as T = {( a , v ), ( b , w ), ( c , u )}. Now answer each of the following questions with a short explanation. (a) Is R a function? If so, is it an injection? R is a function (because for each element of A , there is one and only one element of B which is related via the relation R ). R is not an injection because both a and b are related to the same image 1. (b) Is S –1 a function? If so, is it a surjection? Since S –1 = {( u , 1), ( v , 1), ( w , 2)} C × B , the relation S –1 defines a function from C to B . It is a surjection because each element of B is the image of some element of C . (More precisely, element 1 is the image of u (and of v ), and element 2 is the image of w .) (c) Is T a function? Is T –1 a function? Since T –1 = {( v , a ), ( w , b ), ( u , c )}, both of the relations T and T –1 are functions. (In fact, they both are bijections and they are inverse functions of each other.) (d) Determine the composed relation S –1 R –1 , that is, find all the pairs in this relation and express the relation as a subset of C × A . Since S –1 = {( u , 1), ( v , 1), ( w , 2)} and R –1 = {(1, a ), (1, b ), (2, c )}, the composed relation S –1 R –1 = {( u , a ), ( u , b ), ( v , a ), ( v , b ), ( w , c )} C × A . (e)
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key4fa00 - COT3100.01, Fall 2000 S. Lang Solution Key to...

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