Question

how do u do 6?

F-'(C-D)= F-'(C)-F-'(D). 4. (10 points) In following questions a function f is defined on a set of real numbers. Determine whether or not f is one-to-one and justify your answers. (a) f(x) = **!, for all real numbers x #0 (6) f(x) = x, for all real numbers x (c) f(x) = 3x=!, for all real numbers x 70 (d) f(x) = **, for all real numbers x 1 (e) f(x) = for all real numbers x 5. (20 points) Let S be the set of all strings of O's and I's, and define D:S- Z as follows: For all SES, D(s) = the number of I's in s minus the number of O's in s. Is D one-to-one? Prove or give a counterexample.Is D onto? Prove of give a counterexample. 6. (20 points) Let S be the set of all strings in a's and b's, and define C:S S by C(s) = as, for all SES. (C is called concatenation by a on the left.) Is Cone-to-one? Prove or give a counterexample.Is C onto? Prove or give a counterexample. 7. (20 points) Define H: RX RX R as follows: H(x, y) = (x + 1,2 - y) for all (x, y) E RX R.IS Hone-to-one? Prove or give a counterexample. Is H onto? Prove or give a counterexample.

Answer 1

Here, we assume that all the strings are non-commutative in nature. That is ab + ba. Now, C is one-one. Since, C(si) = C(s) asi = as2 s1 = 52 C is one-one. But, C is NOT onto, because the element, say, ba has no preimage in S. Suppose the element ba has a preimage, say, s', then, C(s') = as' and that would be equal to ba. This implies, C(s) = as' = ba = s' must contain one a at the right most end (1$' string position) and one b, at the second right most place (2nd string position) and possibly some more string elements in the left say t (with length atleast 1). Then that would again gives, s' = tba atba = ba at must be an empty word which is a contradiction, (since t is of length atleast 1 and so is a and that at is of length atleast 2.) Hence, C is NOT onto.