Quiz 2 CS 1050 Fall 2001
1. Let A = {10, 12, 14, 16, 18} B = {10,11,12,13,14,15,16} and C =
{14,15,16,17,18,19}. (10pts.)
Find: (2 pts. each)
(A∩B)∩C = {14,16}
(A∩B)∪C = {10,12,14,15,16, 17, 18, 19}
A⊕B = {11,13, 15, 18}
C – A = {15, 17, 19}
(A-B) ⊕ C = {14,15,16,17,19}
2. Let A and B be sets. Prove that (A-B) ⊆ A ∩ B (20pts.)
Note: make sure that each statement is justified and follows a logical sequence that
constitutes a correct proof. Divide the points up more or less equally for each
statement in the proof.
Proof: We must show that (A-B) ⊆ A ∩ B’ .
Let e ∈ (A-B).
Then e∈ A but e ∉ B.
Since e ∉ B, then e∈ B’.
Since e is an element of both A and B’, then e ∈ A∩B’
3. Suppose that g is a function from A to B and f is a function from B to C. Prove
that if both f and g are one-to-one functions, then f ° g is also one-to-one. (20pts.)
Note: make sure that each statement is justified and follows a logical sequence that
constitutes a correct proof. Divide the points up more or less equally for each
statement in the proof.
Proof: We must show that, ∀ x,y∈A, x≠y → (f•g)(x) ≠ (f•g)(y).
Let x,y be distinct elements of A.
Then, since g is one-to-one, g(x) ≠ g(y).
Now, since g(x) ≠ g(y) and f is one-to-one, then f(g(x)) = (f•g)(x) ≠ f(g(y)) = (f•g)(y).
Therefore x≠y → (f•g)(x) ≠ (f•g)(y), so the composite function is one-to-one.
