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
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](/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: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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