IGNOU DYNAMIC SOLVER: MCA Ist sem /MCS-013/Solved Assignment/Discrete Mathematics/2016-2017 New

Q.1(a)

A.1.(a)

A proposition is that part of the meaning of a clause or sentence that is constant, despite changes in such things as the voice or illocutionary force of the clause.

A proposition may be related to other units of its kind through interpropositional relations, such as temporal relations and logical relations.

A proposition is a predicate or an expression which has only two values : True or false.

x - y > 5

We need to substitute the values of x and y in the preposition.

If the value of arithmetic expression (x - y) is more than 5 then the preposition is True.

Otherwise, if the value of (x - y) is less than or equal to 5 then the preposition is false.

Hence, x - y > 5 is a predicate or a preposition.

Another way,

Yes, x-y >5 is a proposition

Because x-y > 5

Always true for x={6,7,8,9,10,……………………} or y ={0,1,2,3,4,…………….. }

And

Always false for { x < y , x = y or y+5=x }

Q.1.(b)

A.1.(b)

(i) p→ (~q ⋁ ~ r) ⋀ (~p ⋁ r)

p	q	r	~p or ~q	p -> (~p or ~q)	~p or r	p→ (~q ⋁ ~ r) ⋀ (~p ⋁ r)
FALSE	FALSE	FALSE	TRUE	FALSE	TRUE	FALSE
FALSE	FALSE	TRUE	TRUE	FALSE	TRUE	FALSE
FALSE	TRUE	FALSE	TRUE	FALSE	TRUE	FALSE
FALSE	TRUE	TRUE	TRUE	FALSE	TRUE	FALSE
TRUE	FALSE	FALSE	TRUE	TRUE	FALSE	FALSE
TRUE	FALSE	TRUE	TRUE	TRUE	TRUE	TRUE
TRUE	TRUE	FALSE	FALSE	TRUE	FALSE	FALSE
TRUE	TRUE	TRUE	FALSE	TRUE	TRUE	TRUE

(ii) p→ (r ⋁ q) ⋀ (~p ⋀ ~q)

p	q	r	r or q	p -> (r or q)	(~p and ~q)	p→ (r ⋁ q) ⋀ (~p ⋀ ~q)
FALSE	FALSE	FALSE	FALSE	TRUE	TRUE	TRUE
FALSE	FALSE	TRUE	TRUE	FALSE	TRUE	FALSE
FALSE	TRUE	FALSE	TRUE	FALSE	FALSE	FALSE
FALSE	TRUE	TRUE	TRUE	FALSE	FALSE	FALSE
TRUE	FALSE	FALSE	FALSE	TRUE	FALSE	FALSE
TRUE	FALSE	TRUE	TRUE	TRUE	FALSE	FALSE
TRUE	TRUE	FALSE	TRUE	TRUE	FALSE	FALSE
TRUE	TRUE	TRUE	TRUE	TRUE	FALSE	FALSE

Q.1.(c)

A.1.(c)

(i) (A U B) U ( B ∩ C )

(ii) (A U B) ∩ ( C ~ A )

Q.1.(d)

A.1.(d)

(i) R x { 4}; where R is a natural number

(ii) {2, 2) x ( 2, -4)

{(2,2),(2,2),(2,-4),(2,-4)}

Q.2.(a)

A.2.(a)

(i) (∃x) ( ∀y) ( ∃z) P

(ii) ( ∀x) ( ∃y) (∃ z) P

Q.2.(b)

A.2.(b)

(i)

(ii)

Q.2.(c)
A.2.(c)

Modus Tollens

In propositional logic, modus tollens (or modus tollendo tollens and also denying the consequent) (Latin for "the way that denies by denying") is a valid argument form and a rule of inference. It is an application of the general truth that if a statement is true, then so is its contra-positive.

The first to explicitly describe the argument form modus tollens were the Stoics.

The inference rule modus tollens validates the inference from

P

implies

Q

and the contradictory of

Q

to the contradictory of

P

The modus tollens rule can be stated formally as:

{\frac {P\to Q,\neg Q}{\therefore \neg P}}

where

P\to Q

stands for the statement "P implies Q".

\neg Q

stands for "it is not the case that Q" (or in brief "not Q"). Then, whenever "

P\to Q

" and "

\neg Q

" each appear by themselves as a line of a proof, then "

\neg P

" can validly be placed on a subsequent line. The history of the inference rule modus tollens goes back to antiquity. Modus tollens is closely related to modus ponens. There are two similar, but invalid, forms of argument: affirming the consequent and denying the antecedent. See also contraposition and proof by contrapositive.

Another rule of inference:

[(p ! q) # ¬q] ! ¬p

Example:

p ! r

r ! s

t $ ¬s

¬t $ u

¬u

% ¬p

Modus Ponens

In propositional logic, modus ponens (Latin for "the way that affirms by affirming"; generally abbreviated to MP or modus ponens) or implication elimination is a rule of inference.^[2] It can be summarized as "P implies Q and P is asserted to be true, so therefore Q must be true." The history of modus ponens goes back to antiquity.

Modus ponens is closely related to another valid form of argument, modus tollens. Both have apparently similar but invalid forms such as affirming the consequent, denying the antecedent, and evidence of absence. Constructive dilemma is the disjunctive version of modus ponens. Hypothetical syllogism is closely related to modus ponens and sometimes thought of as "double modus ponens."

The first rule of inference is Modus Ponens, or the Rule of Detachment:

[p # (p ! q)] ! q

E.g. 1) If Allison vacations in Paris, then she will have to win a scholarship.

2) Allison is vacationing in Paris.

3) Therefore Allison won a scholarship

Q.2.(d)

A.2.(d)

Whenever sets are being discussed, the relationship between the elements of the sets is the next thing that comes up. Relations may exist between objects of the same set or between objects of two or more sets.

Definition and Properties

A binary relation R from set x to y (written as

x R y

R (x, y)

) is a subset of the Cartesian product

x \times y

. If the ordered pair of G is reversed, the relation also changes.

Generally an n-ary relation R between sets

A_{1}, \dots, a n d A_{n}

is a subset of the n-ary product

A_{1} \times \dots \times A_{n}

. The minimum cardinality of a relation R is Zero and maximum is

n^{2}

in this case.

A binary relation R on a single set A is a subset of

A \times A

For two distinct sets, A and B, having cardinalities m and n respectively, the maximum cardinality of a relation R from A to B is mn.

Domain and Range

If there are two sets A and B, and relation R have order pair (x, y), then −

The domain of R, Dom(R), is the set ${x | (x, y) \in R f o r s o m e y i n B}$
The range of R, Ran(R), is the set ${y | (x, y) \in R f o r s o m e x i n A}$

Examples

Let,

A = {1, 2, 9}

and

B = {1, 3, 7}

Case 1 − If relation R is 'equal to' then $R = {(1, 1), (3, 3)}$

Dom(R) = ${1, 3}, R a n (R) = {1, 3}$
Case 2 − If relation R is 'less than' then $R = {(1, 3), (1, 7), (2, 3), (2, 7)}$

Dom(R) = ${1, 2}, R a n (R) = {3, 7}$
Case 3 − If relation R is 'greater than' then $R = {(2, 1), (9, 1), (9, 3), (9, 7)}$

Dom(R) = {2,9},Ran(R)={1,3,7}{2,9},Ran(R)={1,3,7}

Definition(equivalence relation): A binary relation R on a set A is an equivalence relation if and only if
(1) R is reflexive
(2) R is symmetric, and
(3) R is transitive.

Example 1: The equality relation (=) on a set of numbers such as {1, 2, 3} is an equivalence relation.
Example 2: The congruent modulo m relation on the set of integers i.e. {<a, b>| a b (mod m)}, where m is a positive integer greater than 1, is an equivalence relation.

Note that the equivalence relation on hours on a clock is the congruent mod 12, and that when m = 2, i.e. the congruent mod 2, all even numbers are equivalent and all odd numbers are equivalent. Thus the set of integers are divided into two subsets: evens and odds.

Example 3: Taking this discrete structures course together this semester is another equivalence relation.

Equivalence relations can also be represented by a digraph since they are a binary relation on a set. For example the digraph of the equivalence relation congruent mod 3 on {0, 1, 2, 3, 4, 5 , 6} is as shown below. It consists of three connected components.

Q.3.(a)

A.3.(a),(b),(c),(d)

Q.4.(a)A.4.(a),(b)

A.4.(c)

Set theory is an important branch of mathematics. It is the study of mathematical logic and its applications. Set theory defines a term "set". A set is defined as the collection of elements, such as numbers or other objects, which are arranged in a group. The set contains the elements of similar type or category. The set with any numbers can be denoted in the symbol braces { }.
For example, the set of numbers may be represented as {2, 3, 4, 5, 6}. We can also write this set as {x : 1 < x < 7}. A totally bounded set is defined as a set which is having a definite or finite size. The bounded set consists of the numbers which are the set of real numbers. A bounded set has both the upper and lower bounds that exists within a particular interval. The bounded numbers in a set are having a definite or fixed size and it always lies between the given intervals. The bounded set contains a bounded sequence form.
Set theory basically deals with set and set operations. Though, it includes a quite vast study of logic and reasoning. In this page, we are going to focus on introduction of set theory, basic formulas used in set theory, some basic properties and introduction to the Venn diagram. So students, go ahead with us and gain knowledge about set theory in this chapter.

The two basic properties to represent a set are explained below using various examples.

1. The change in order of writing the elements does not make any changes in the set.

In other words the order in which the elements of a set are written is not important. Thus, the set {a, b, c} can also be written as {a, c, b} or {b, c, a} or {b, a, c} or {c, a, b} or {c, b, a}.

For Example:

Set A = {4, 6, 7, 8, 9} is same as set A = {8, 4, 9, 7, 6}

i.e., {4, 6, 7, 8, 9} = {8, 4, 9, 7, 6}

Similarly, {w, x, y, z} = {x, z, w, y} = {z, w, x, y} and so on.

2. If one or many elements of a set are repeated, the set remains the same.

In other words the elements of a set should be distinct. So, if any element of a set is repeated number of times in the set, we consider it as a single element. Thus, {1, 1, 2, 2, 3, 3, 4, 4, 4} = {1, 2, 3, 4}

The set of letters in the word ‘GOOGLE’ = {G, O, L, E}

For Example:

The set A = {5, 6, 7, 6, 8, 5, 9} is same as set A= {5, 6, 7, 8, 9}

i.e., {5, 6, 7, 6, 8, 5, 9} = {5, 6, 7, 8, 9}

In general, the elements of a set are not repeated. Thus,

(i) if T is a set of letters of the word ‘moon’: then T = {m, o, n},

There are two o’s in the word ‘moon’ but it is written in the set only once.

(ii) if U = {letters of the word ‘COMMITTEE’}; then U = {C, O, M, T, E}

Solved examples using the properties of sets:

1. Write the set of vowels used in the word ‘UNIVERSITY’.

Solution:

Set V = {U, I, E}

2. For each statement, given below, state whether it is true or false along with the explanations.

(i) {9, 9, 9, 9, 9, ……..} = {9}

(ii) {p, q, r, s, t} = {t, s, r, q, p}

Solution:

(i) {9, 9, 9, 9, 9, ……..} = {9}

True, since repetition of elements does not change the set.

(ii) {p, q, r, s, t} = {t, s, r, q, p}

True, since the change in order of writing the elements does not change the set.

Q.5.(a)

A.5.(a)

Q.5.(b)(c)

A.5.(b)(c)

Q.5.(d)

A.5.(d)

Proof:

STEP 1: For n=1 (1.1) is true, since

1 =1(1 + 1)/2

.

STEP 2: Suppose (1.1) is true for some n = k ≥ 1, that is

1 + 2 + 3 + . . . + k = k(k + 1)/2

.

STEP 3: Prove that (1.1) is true for n = k + 1, that is

1 + 2 + 3 + . . . + k + (k + 1) ?= (k + 1)(k + 2)/2

.

We have

1 + 2 + 3 + . . . + k + (k + 1) ST.2 =

k(k + 1)/2

+ (k + 1) = (k + 1) k/2+ 1= (k + 1)(k + 2)

Q.6.(a)

A.6.(a)(b)(c)(d)

Q.7.(a)

A.7.(a)(b)(c)(d)

Q.8.(a)

A.8.(a)(b)(c)

3 comments:

Unknown2 April 2017 at 05:25
this is huge !!!!!
IGNOU DYNAMIC SOLVER7 April 2017 at 07:25
It 's a assigment and we have try do write answer with accuracy.And In discrete Mathematics u dont answer in numerical way , u have give answer with some theoritical part. To describe the numerical part
Unknown11 April 2017 at 02:19
This comment has been removed by a blog administrator.

Thursday, 16 March 2017

MCA Ist sem /MCS-013/Solved Assignment/Discrete Mathematics/2016-2017 New﻿