Let S be the set of all strings in a's and b's, and define C: S - Sby C(s) = as, for each s ES. (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. as, = Now strings are finite sequences of characters, and since the strings on both sides of the above equation are equal, for each integer n2 0, the nth character from the left in the left-hand string equals the nth character from the left Vy in the right-hand string. the nth character from the left It follows that for each integer n 2 0, the nth character from the left ins, equals S,. Hence, s, = V S, and so C is in 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 v 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 - Sby C(s) = as, for each s ES. (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. as, = Now strings are finite sequences of characters, and since the strings on both sides of the above equation are equal, for each integer n2 0, the nth character from the left in the left-hand string equals the nth character from the left Vy in the right-hand string. the nth character from the left It follows that for each integer n 2 0, the nth character from the left ins, equals S,. Hence, s, = V S, and so C is in 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 v string s because every string in the range of C starts with
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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