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Requirement
1. Let A = {1, 2}. Write down each of the following sets:
a) P(A) and |P(A)|.
b) P(P(A)) and |P(P(A))|.
(10 marks)
2. Let A= {1, 2, 3}, B = {2, 3, 4}, and R be a relation from A to B.
a) List the elements of A x B and the elements of R = {(x, y) | x < y}. Write down the domain and range of R.
b) Sketch the graph of R in Z x Z. Would R -1 also be a relation from A to B? Justify your answer. (10 marks)
3. Let X = { 1,2,3,4,5,6,7} and R = {(x,y) / x – y is divisible by 3}.
a) Show that R is an equivalence relation.
b) Draw the digraph of R. (10 marks)
4. a) Let g be the function from the set {a, b, c} to itself such that g(a) = b, g(b) = c, and g(c) = a. Let f be the function from the set {a, b, c} to the set {1, 2, 3} such that f (a) = 3, f (b) = 2, and f (c) = 1. What is the composition of f ? g, and what is thecomposition of g ? f ?
b) Let f and g be the functions from the set of integers to the set of integers definedby f (x) = 2x + 3 and g(x) = 3x + 2. What is the composition of f ? g? What is the composition of g ? f ? (10 marks)
5. Let p and q be the proposition variables denoting
p: It is below freezing.
q: It is snowing.
Write the following propositions using variables, p and q, and logical connectives.
a) It is below freezing but not snowing.
b) It is either below freezing or it is snowing, but it is not snowing if it is below freezing.
Solution
Question 1:
Let A = { 1, 2} . Write down each of the following sets:
A: P(A) and |P(A)|.
Answer :
Given that the set A = {1, 2}. Then the power set of A is set of all subsets.
That is P(A) = {ø , {1} , {2} , {1 , 2}.
The condinality of a set is the total number of elements of the set. Hence the cordinality of P(A) is |P(A)| = 4
B: P(P(A)) and |P(P(A))|
Answer :
From past a) we have
P(A) = {ø, {1}, {2}, {1,2}}
Now, we find the power set of the power set, which is the set at all subsets of P(A).
P(P(A)) = {ø ; {ø} ; {{1}}; {{2}}; {{1,2}}; {{ø}, {1}};
{{ø}, {2}}; {{ø}, {1,2}}; {{1}, {2}}; {{1}, {1,2}};
{{2}, {1,2}}; {{ø}, {1}, {2}}; {{ø},{1}, {1,2}} ;
{{ø}, {2}, {1,2}}; {{1}, {2}, {1,2}}; {{ø}, {1}, {2}, {1,2}}}
Hence, the total number of elements in P(P(A)) are 16,
Therefore |P(P(A))| = 16
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Question 2:
Let A= {1, 2, 3}, B = {2, 3, 4}, and R be a relation from A to B.
A: List the elements of A x B and the elements of R = {(x, y) | x < y}. Write down the domain and range of R.
Answer:
Given A= {1,2,3}, B= {2,3,4} and R is a relation from A to B.
Then, the elements of are.
= {(1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3.2), (3.3), (3,4)}
Now, R is relation ‘is less than’ From A to B, so the ordered paires of R = {(x,y)|x If R be a relation from set A to set B, then the set of all the first components of the ordered pairs belonging to R is called the domain of R.
i.e. Domain (R) = {a∈A : (a,b)∈R For some b∈B} and the set of all second components of the ordered pairs belonging to R is called the range of R.
i.e. Range (R) = {b∈B : (a,b)∈R for some a∈A}.
Therefore, domain of R is {1,2,3} and range of R is {2,3,4}.
B: Sketch the graph of R in Z x Z. Would R-1 also be a relation from A to B? Justify your answer.
Answer :
Question 3:
Let X = { 1,2,3,4,5,6,7} and R={(x,y) / x – y is divisible by 3}.
A: Show that R is an equivalence relation.
Answer:
The element of relation R={(x,y) / x – y is divisible by 3} on ={1,2,3,4,5,6,7} are
R={(1,1), (1,4), (1,7), (2,2), (2,5), (3,3), (3,6), (4,1), (4,4), (4,7), (5,2), (5,5), (6,3), (6,6), (7,1), (7,4), (7,7)}
A relation R defined in a set A, is said to be an equivalence relation if and only if
? R is reflexive, that is, for all
? R is symmetric, that is, ⇒ for all
? R is transitive, that is, and ⇒ for all
Let then 1-1=0 which is divisible by 3.
Therefore, 1R1 holds for all elements in X. The ordered pairs (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) and (7,7) in R are all part of relation. Hence R is reflexive.
Now, we have and holds. Then 1-4 is divisible by 3.
The ordered pairs (1,4) and (4,1) are both part of relation R. Thus, ⇒ .
Similarly, we notice that the numbers paired besides themselves are both part of the relation. Therefore, R is symmetric.
Let and both hold. Then, 1-4 and 4-7 and both divisible by 3.
Therefore, 1-7 = (1-4) + (4+7) is divisible by 3
-6 = -3 + (-3)
-6 = -6
Thus, 1R4 and 4R7 ⇒ 1R7
1 is related to 4, 7, and itself; 4 is related to 1,7, and itself; and 7 is related to 1,4, and itself. The ordered pairs (1,1), (1,4), (1,7), (4,1), (4,4), (4,7), (7,1), (7,4), (7,7) are all part of relation R. Therefore, R is transitive.
Since, R is reflexive, symmetric and transitive, the relation R is an equivalence relation.
B: Draw the digraph of R.
Answer :
The diagraph of relation R given in part (a) is a as follows.
Question 4:
A : Let g be the function from the set {a, b, c} to itself such that g(a) = b, g(b) = c, and g(c) = a. Let f be the function from the set {a, b, c} to the set {1, 2, 3} such that f (a) = 3, f (b) = 2, and f (c) = 1. What is the composition of f g, and what is the composition of g f ?
Answer:
If g be a function from the set A to the set B and be a function from the set B to the set C. Then the composition of the function and , denoted for all a∈A by , is defined by =
Given that g be the function from the set to itself such that , , and , and be the function from the set to the set such that , , and . Let us find the composition of and .
Now, let us find the composition of g and f. Here, the range of function f is {1,2,3} which is not a subset of domain of g. Therefore, the composition is not defined.
B: Let f and g be the functions from the set of integers to the set of integers defined by f (x) = 2x + 3 and g(x) = 3x + 2. What is the composition of f g? What is the composition of g f ?
Answer :
Given that f and g are the functions from the set of integers to the set of integers defined by and . Let us find the composition of and . Both the compositions are defined.
Therefore,
And
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Place Your OrderQuestion 5:
Let p and q be the proposition variables denoting
p: It is below freezing.
q: It is snowing.
Write the following propositions using variables, p and q, and logical connectives.
a) It is below freezing but not snowing.
Answer :
Let p and q be the prepositions.
p: It is below freezing
q: It is not snowing.
In the given statement,
p: It is below freezing.
¬q: It is not snowing.
Note that in logic the word “but” some-times used in place of “and” in conjuction. The statement is conjunction of the prepositions p and . It is denoted by
b) It is either below freezing or it is snowing, but it is not snowing if it is below freezing.
Answer :
In the given statement, p and q be the propositions, then
p: It is either below Freezing.
q: It is snowing.
¬q: It is not snowing.
In the first part of the statement “It is either below Freezing or It is snowing” the logic word “or” is indicates disjunction of p and q, denoted by and in the second part “It is not snowing if It is below Freezing” is a conditional statement that is q is true on the condition that p holds.
Hence the proposition .
Now, the logic word “but” in between the first and second part of the statement indicates the conjunction between the propositions and
Therefore, the logical proposition is
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