Consider the function f : {4, 6, 7, 8, 9} → {1, 2, 4, 5, 6, 9} given by the table below: 4 6 7 8 9 f(x) 1 2 5 96 X a. Is f injective? Why or why not? OYes, because every element in the codomain is the image of at most one element of the domain. No, because at least one element in the codomain is the image of two or more elements of the domain. Yes, because every element in the codomain is the image of at least one element of the domain. O No, because at least one element in the codomain is not the image to an element of the domain. b. Is f surjective? Why or why not? Yes, because every element in the codomain is the image of at most one element of the domain. O No, because at least one element in the codomain is the image of two or more elements of the domain. Yes, because every element in the codomain is the image of at least one element of the domain. O No, because at least one element in the codomain is not the image to an element of the domain.

If B is countable and f : P(A) → B is a one to one (i.e., injective) function, then A must be finite. (here, P(A) denotes the power set of A) true false

Let X-(1, 2, 3, 4, 5, 6). Define the functionf from P(X) to (0, 1] as the power set. What is f((2,5|)? O 020050 O 100100 O 01001 O 010010

[D] Consider the function F: {1,5, 8} × {6, 10} → {0, 1, 2, 3, 4, 5} defined as follows: F: (a, b) → (4a + b)mod 6 (D.1) Fill in the ordered pairs in the diagram below to indicate the elements in the domain of F. Then draw in arrows to complete the diagram so that it correspondes to the function. ( • 0 • 1 • 2 • 3 • 4 • 5 (D.2) What is the range of F? (D.3) What is the co-domain of F? 3/5 (D.4) What is the preimage (inverse image) of 3? (D.5) What is the preimage (inverse image) of 0? (D.6) Is F one-to-one? If not provide a counterexample. (D.7) Is F onto? If not provide a counterexample.