1. Find the sets A and B if A − B = {1, 5, 7, 8}, B − A = {2, 10}, and A ∩ B = {3, 6, 9}. 2. If B and C are two sets, prove or disprove the identity, B×C=C×B. 3. Prove that if A and B are sets, then A ∩ (A ∪ B) = A. 4. Suppose that A × B = Ø, where A and B are sets. What can you conclude? 5. Let f (x) = 2x where the domain is the set of real numbers. What is f(R)? 6. Determine whether each of these functions is a bijection from R to R. a) f (x) = 2x + 1 b) f (x) = x 3 7. Let f be the function from R to R defined by y=f(m,n)=2m-n. Write a method in C with appropriate return type and parameter list that could be used to realize the function.
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Here's the Solution to this Question
1.
if A − B = {1, 5, 7, 8}, then A contains 1,5,7,8
if B − A = {2, 10}, then B contains 2,10
if A ∩ B = {3, 6, 9}, then A contains 3,6,9 and B contains 3,6,9
So:
A={1,3,5,6,7,8,9}
B={2,3,6,9,10}
2.
For example:
B={1,1}, C={2,2}
So,
3.
(A ∪ B) has all elements of A and elements of B that are not in A. So, A ∩ (A ∪ B) has only all elements of A, i.e. A ∩ (A ∪ B) = A
4.
The Cartesian product of two sets A and B, denoted by A × B, is defined as the set consisting of all ordered pairs (a, b) for which a ∊ A and b ∊ B.
So, if A × B = Ø , then we can conclude that A = Ø and B = Ø
5.
range f(R):
6.
a bijective function f: X → Y is a one-to-one (injective) and onto (surjective), where one-to-one function is a function f that maps distinct elements to distinct elements, onto function is a function f that maps an element x to every element y.
So:
a)
the function is one-to-one
range:
the function is onto
So, this is bijective function.
b)
the function is one-to-one
range:
the function is onto
So, this is bijective function.
bijective function
7.
int func(int m, int n) {
int y;
y=2*m-n;
return y;
}