Question

For each of the functions below, indicate whether the function is onto, one-to-one, neither or both. If the function is not onto or not oneto-one, give an example showing why. (a) f:{0,1}4→(0,1}3. The output of f is obtained by taking the input string and dropping the first bit. For example f(1011)=011 (b) f: {0,1}3→{0,1)3. The output of f is obtained by taking the input string and replacing the first bit by 1, regardless of whether the first bit is a 0 or 1 . For example, f(001)=101 and f(110)=110. (c) f:{0,1}3→{0,1}3. The output of f is obtained by taking the input string and reversing the bits. For example f(011)=110. (d) f: {0,1}3→(0,1)4. The output of f is obtained by taking the input string and adding an extra copy of the first bit to the end of the string. For example, f(100)=1001. (e) Let A be defined to be the set {1,2,3,4,5,6,7,8}, f.T (A)→(0,1,2,3,4,5,6,7,8). For X⊆A,f(X)=∣X∣. Recall that for a finite set A, P(A) denotes the power set of A which is the set of all subsets of A (f) Let A be defined to be the set {1,2,3,4,5,6,7,8),f:P(A)→P(A). For X⊆A,f(X)=A−X. Recall that for a finite set A, P(A) denotes the power set of A which is the set of all subsets of A. (g) Let A be defined to be the set {1,2,3,4,5,6,7,8) and let B=(1), f: P(A)→P(A). For X⊆A,f(X)=X - B. Recall that for a finite set A. P(A) denotes the power set of A which is the set of all subsets of A. (h) A={a,b,c},h,P(A)→P(A). For X∈A,h(X)=X⊕(a) (i) A={a;b,c},h:P(A)→P(A). For X⊆A,h(X)=X∪(a).

Answer 1

The function dropping the first bit is one-to-one but not onto, the function replacing the first bit by 1 is neither, the function reversing bits is both, and the function appending the first bit at the end is one-to-one but not onto.

Step-by-step explanation:

For each of the functions given, we need to determine whether the function is onto, one-to-one, neither, or both:

One-to-one: A function f is called one-to-one, or injective, if every element of the function's codomain is the image of at most one element of its domain.

Onto: A function f is called onto, or surjective, if for every element in the function's codomain there is a preimage in the domain.

For example, consider the function f:{0,1}4→{0,1}3 where f drops the first bit. This function is one-to-one because each element of the domain maps to a unique element in the codomain. However, it is not onto since not all possible 3-bit strings can be the output (e.g. '111').

Now, look at f:{0,1}3→{0,1}3 where f replaces the first bit by 1. This function is not one-to-one as multiple inputs could yield the same output (f(001) = f(101) = 101). It is also not onto as '000' cannot be an output.

The function f:{0,1}3→{0,1}3, which reverses the string, is both one-to-one and onto, since it creates a unique mirror image for each 3-bit string, and every possible 3-bit string can be obtained.

The function f:{0,1}3→{0,1}4, which appends an extra copy of the first bit to the end, is one-to-one as each input maps to a unique 4-bit string, but it is not onto as not all 4-bit strings can be outputs (e.g., '0000' can't be an output).

Answer 2

The functions described in the problem set show different characteristics of being one-to-one, onto, or neither. Each function's definition determines its injective or surjective nature through outputs corresponding to inputs or covering possible outputs.

Step-by-step explanation:

When assessing whether functions are one-to-one (injective), onto (surjective), both (bijective), or neither, it is essential to understand the definitions of these properties. A function is one-to-one if every element of the range corresponds to exactly one element of the domain. It is onto if every element of the codomain has at least one element of the domain mapping to it. For instance:

1. For the function dropping the first bit, it is neither one-to-one nor onto, as different strings can yield the same output, showing it's not one-to-one, and not all possible strings in {0,1}3 are achievable, showing it's not onto.
2. For the function that replaces the first bit with a 1, it is not one-to-one since multiple inputs like 001 and 101 lead to the same output (101), but it is onto since every string in {0,1}3 with a leading 1 can be an output.
3. The bit-reversing function is both one-to-one and onto, as every bit string maps to a unique reverse bit string and vice versa.
4. Adding an extra copy of the first bit creates a function that is neither one-to-one nor onto since not all elements in {0,1}4 can be achieved, and different inputs can yield the same output.

The remaining functions involving sets and power sets need to consider set operations such as taking the complement, the difference, and the symmetric difference with a single element. Each possesses different characteristics, with some being one-to-one, others onto, and some neither.