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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 E S. (C is called concatenation by a on the left.) (a) Prove that C is one-to-one. Proof: 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 S2 as follows: as 1 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 is equal to S1 = S21 the nth character from the left in the right-hand string. It follows that for each integer n > 0, the nth character from the left in s, equals the nth character from the left in and so C is one-to-one. Hence (b) Give a counterexample to 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) Prove that C is one-to-one. Proof: 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 S2 as follows: as 1 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 is equal to S1 = S21 the nth character from the left in the right-hand string. It follows that for each integer n > 0, the nth character from the left in s, equals the nth character from the left in and so C is one-to-one. Hence (b) Give a counterexample to 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

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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 E S. (C is called concatenation by a on the left.) (a) Prove that C is one-to-one. Proof: 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 S2 as follows: as 1 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 is equal to S1 = S21 the nth character from the left in the right-hand string. It follows that for each integer n > 0, the nth character from the left in s, equals the nth character from the left in and so C is one-to-one. Hence (b) Give a counterexample to 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) Prove that C is one-to-one. Proof: 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 S2 as follows: as 1 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 is equal to S1 = S21 the nth character from the left in the right-hand string. It follows that for each integer n > 0, the nth character from the left in s, equals the nth character from the left in and so C is one-to-one. Hence (b) Give a counterexample to 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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