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.

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}

$B\times C=\{(1,2),(1,2\}$

$C\times B=\{(2,1),(2,1\})$

So, $B\times C\neq C\times B$

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): $x\isin (-\infin,\infin)$

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)

$f (x) = 2x + 1$

$f(x_1)=f(x_2)\implies x_1=x_2$

the function is one-to-one

range: $f(x)\isin (-\infin,\infin)$

the function is onto

So, this is bijective function.

b)

 $f (x) = x^3$

$f(x_1)=f(x_2)\implies x_1=x_2$

the function is one-to-one

range: $f(x)\isin (-\infin,\infin)$

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; 
}