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Let S be the set of all strings in a's and b's, and define C: S → S by C(s) = as, for each s ∈ S. (C is called concatenation by a on the left.) (b) Show that C is not onto. Counterexample: The string _____ is in S but is not equal to C(s) for any string s because every string in the range of C starts with _____.

Let S be the set of all strings in a's and b's, and define C: S → S by C(s) = as, for each s ∈ S. (C is called concatenation by a on the left.) (b) Show that C is not onto. Counterexample: The string _____ is in S but is not equal to C(s) for any string s because every string in the range of C starts with _____.

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter5: Rings, Integral Domains, And Fields
Section5.2: Integral Domains And Fields
Problem 7E: [Type here] 7. Let be the set of all ordered pairs of integers and . Equality, addition, and...
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Let S be the set of all strings in a's and b's, and define C: S → S by

C(s) = as, for each s ∈ S.
(C is called concatenation by a on the left.)
(b) Show that C is not onto.
Counterexample: The string _____ is in S but is not equal to C(s) for any string s because every string in the range of C starts with _____.
Let S be the set of all strings in a's and b's, and define C: S → S by C(s) = as, for each s E S. (C is called concatenation by a on the left.) (a Is C one-to-one? To answer this question, suppose s, and s, are strings in S such that C(s,) = C(s,). Use the definition of C to write this equation in terms of a, s,, and s, as follows. ds, = as2 Now strings are finite sequences of characters, and since the strings on both sides of the above equation are equal, for each integer n 2 0, the nth character from the left in the left-hand string equals nth character from the left in s, in the right-hand string. It follows that for each integer n 2 0, the in s2. Hence, s. the nth character from the left equals the nth character from the left S21 and so C is one-to-one. (b) Show that C is not onto. Counterexample: The string is in S but is not equal to C(s) for any string s because every string in the range of C starts with
Transcribed Image Text:Let S be the set of all strings in a's and b's, and define C: S → S by C(s) = as, for each s E S. (C is called concatenation by a on the left.) (a Is C one-to-one? To answer this question, suppose s, and s, are strings in S such that C(s,) = C(s,). Use the definition of C to write this equation in terms of a, s,, and s, as follows. ds, = as2 Now strings are finite sequences of characters, and since the strings on both sides of the above equation are equal, for each integer n 2 0, the nth character from the left in the left-hand string equals nth character from the left in s, in the right-hand string. It follows that for each integer n 2 0, the in s2. Hence, s. the nth character from the left equals the nth character from the left S21 and so C is one-to-one. (b) Show that C is not onto. Counterexample: The string is in S but is not equal to C(s) for any string s because every string in the range of C starts with
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