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 _____.
Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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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 _____.
![**Title: Understanding Concatenation and Its Characteristics**
**Introduction to Set S and Function C**
Let S be the set of all strings composed of the letters 'a' and 'b'. We define a function C that maps S to S by:
\[ C(s) = as, \text{ for each } s \in S. \]
This operation is referred to as concatenation of 'a' on the left.
### (a) Is C one-to-one?
To explore whether C is one-to-one, assume \( s_1 \) and \( s_2 \) are strings in S such that \( C(s_1) = C(s_2) \). By applying the definition of C, we can express this equation using \( a \), \( s_1 \), and \( s_2 \) as follows:
\[ as_1 = as_2 \]
Since strings are defined as finite sequences of characters, and since both sides of the equation are equal, we conclude the following:
- For each integer \( n \geq 0 \), the nth character from the left in the left-hand string equals the nth character from the left in the right-hand string.
- Therefore, \( s_1 = s_2 \).
Thus, C is demonstrated to be one-to-one.
### (b) Show that C is not onto
To demonstrate that C is not onto, consider the following counterexample:
The string **empty** is in S, however, it does not have any representation in the form \( C(s) \) for **any** string \( s \), because **every string in the range of C starts** with 'a'.
This section of the text employs logical reasoning to explain the characteristics of the function C, focusing on both its injective (one-to-one) and non-surjective (not onto) nature in the realm of string functions over the alphabet consisting of 'a' and 'b'.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fddb57a32-2247-47a5-9d55-8c5fe5ac9cfb%2Fa52b9cdd-eb9c-4213-90be-d1a5c7b35065%2Frb9eac_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Understanding Concatenation and Its Characteristics**
**Introduction to Set S and Function C**
Let S be the set of all strings composed of the letters 'a' and 'b'. We define a function C that maps S to S by:
\[ C(s) = as, \text{ for each } s \in S. \]
This operation is referred to as concatenation of 'a' on the left.
### (a) Is C one-to-one?
To explore whether C is one-to-one, assume \( s_1 \) and \( s_2 \) are strings in S such that \( C(s_1) = C(s_2) \). By applying the definition of C, we can express this equation using \( a \), \( s_1 \), and \( s_2 \) as follows:
\[ as_1 = as_2 \]
Since strings are defined as finite sequences of characters, and since both sides of the equation are equal, we conclude the following:
- For each integer \( n \geq 0 \), the nth character from the left in the left-hand string equals the nth character from the left in the right-hand string.
- Therefore, \( s_1 = s_2 \).
Thus, C is demonstrated to be one-to-one.
### (b) Show that C is not onto
To demonstrate that C is not onto, consider the following counterexample:
The string **empty** is in S, however, it does not have any representation in the form \( C(s) \) for **any** string \( s \), because **every string in the range of C starts** with 'a'.
This section of the text employs logical reasoning to explain the characteristics of the function C, focusing on both its injective (one-to-one) and non-surjective (not onto) nature in the realm of string functions over the alphabet consisting of 'a' and 'b'.
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