Question

1 Let U = {1, 2, 3, 4, 5, 6, 7} be the universe. Form the set A as follows: Read off your seven digit student number from left to right. For the first digit ni include the number 1 in A if ni is even otherwise omit 1 from A. Now take the second digit n2 and include the number 2 in A otherwise omit 2 from A. Continue in this way until you have read off all digits from your student number. For instance if your student number were 8206811 then A1,2, 3,4,5). If all digits of your student number are odd, set A- U Let B 1,2,7 and C 3,4,5,7). Write down AU B, A- C, A, AnB 2. Let R C A x A be an equivalence relation. For a є A recall [a]-{b A : (a, b) є R). Fix a,be A ard suppose that [a] n固ㄨ Show that [a]-[b] indicating clearly in your arguinent where you use the fact that R is an equivalence relation 3. Let A, B be subsets of a universe U. Use the rules of set manupulation to simplify into expressions involving A, B,A, B only Let D-0, 1,2,3,4,5,6,7,8,9) and S 1, 2,3,4,5,6,7. Define RCS*Dy R = {(1, n), . . . , (7, nr)} where your student number is nı·.. n7. For instance if your student number were 8208611 then R ((1,8), (2, 2), (3,0), (4,8), (5,6), (6,1), (7, 1)) (i) Explain, with reference to the definition why R a function from S to D (ii) What is the domain and range of R (ii) Explain, with reference to the definition, whether or not R is onto. If it is not onto, give an instance where it fails to be. (iv) Explain, with reference to the definition, whether or not R is one-to-one. If it is not one-to-one, give an instance where it fails to be. v) Explain, with reference to the definition, why R is not an equivalence relation.

Could you please answer the question Q1 to Q3. Write the answer clearly and step by step.

Answer 1

1. The set consists of all the elements which are either in set A or in set B or both.

AUB

Hence,

AUB_ {1.2.3.4.5. 6.7

The set will consists of all the elements which are in set A but not in set C. Here since
C 3. 4. 5. 7}

Hence since these elements are in set A but not in set C.

A-C=11.2.6

The set $\bar{A}$ will consists of all the elements which are in universe U but not in set A. Hence $\bar{A} = U - A = \{1,2,3,4,5,6,7\}-\{1,2,3,4,5,6,7\} = \phi$ i.e. an empty set

The set $A \cap \bar{B}$ will contain all the elements which are in set A and in set $\bar{B}$.

Since  

{ 1. 2. 3. 4. 5. 6. 7}-{ 1.2. 7} = {3. 4. 5. 6} B = Ụ-B

Hence  $A \cap \bar{B}$ will be since these are the only elements present in both set A and set B.

3. Here first note that $A \cup \phi = A$ because set $\phi$ is null set and hence all the elements in set A will only be present in set $A \cup \phi$.

Hence, $\overline{A \cup \phi } = \bar{A}$

Then $\overline{A \cap B } = \bar{A} \cup \bar{B}$ by applying De-Morgan's law.

Hence, $\overline{(A \cup \phi)} \cap \overline{(A \cap B) } = \bar{A} \cap (\bar{A} \cup \bar{B})$

Here, $A \cup U = U$ because universal set U contains all the element

Hence $\overline{A \cup U} = \bar{U} = \phi$

Hence, $\overline{(A \cup U)} \cap \bar{A}= \phi \cap {\bar{A}} = \phi$    because any set having intersection with null set will give null set.

Please comment for any clarification.