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 _____.

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Chapter2: Second-order Linear Odes
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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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