MCS-013 Discrete Mathematics solved assingment by ignousolvedassignments.com

IGNOU Solved Assignments By http://ignousolvedassignments.com Course Code : MCS-013 Course Title : Discrete Mathematics

Assignment Number : MCA(1)/013/Assignment/16-17 MCA(1)/013/Assignment/16-17

Last Dates for Submission : 15th October, 2017 (For July 2017 Session) 15th April, 2018 (For January 2018 Session)

This assignment has twenty questions in all and carries 80 marks. The rest of the 20 marks are for viva-voice. Answer all the questions. All questions carry equal marks (i.e. 4 marks each). Please go through the guidelines regarding assignments given in the Programme Guide for the format of presentation.

Disclaimer: This Assignment is prepared by our students. The Institution and publisher are

not

responsible for any omission and errors.

1. (a) What is proposition? Explain whether, x-y >5 is a proposition or not.

A proposition is that part part of the meaning of a clause or sentence sentence that is constant, despite changes in such things as the voice or illocutionary force of the clause. A proposition may be related related to other units of its kind through through interpropositional interpropositional relations, such as temporal relations and logical relations. A proposition is a predicate predicate or an expression expression which has only two values : True or 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 preposition is True. Otherwise, if the value of (x - y) is less than or equal to 5 then the preposition preposition is false. Hence, x - y > 5 is a predicate or a preposition. 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 }

Course Code :

MCS-013

Course Title : Discrete Mathematics

Assignment Number : MCA(1)/013/Assignment/16-17

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IGNOU Solved Assignments By http://ignousolvedassignments.com

(b) Make truth table for followings. ~ r) (~p r)

(i) p→ (~q

q) (~p ~q)

(ii) p→ (r

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)

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(ii) p→ (r

q) (~p ~q)

q

r

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r or q

p -> (r or q)

(~p and ~q)

p→ (r ⋁ q) ⋀ (~p ⋀ ~q)

p





(i) (A U B) U ( B ∩ C )

(ii) (A U B) ∩ ( C ~ A )

(i)

R x { 4}; where R is a natural number

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(ii)

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

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(i) Some birds can not fly B : are birds F : are fly B->F all birds are flying And at least one bird are flying So Neglect of above sentence (~B) - >F

(ii) Nothing is correct A : all things B : correct A->B All things is correct And neglected of sentence (A) - > (~B) all things are not correct So nothing is correct.

In propositional logic, modus tollens (or modus tollendo tollensand 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.

Consider the following argument...

"If you have a current password, then you can log on to the network" "You have a current password"

Therefore:

"You can log on to the network" This has the form:

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IGNOU Solved Assignments By http://ignousolvedassignments.com p→q p q

This form of argument is calls Modus Ponens (latin for "mode that affirms")

Note that an argument can be valid, even if one of the premises is false. For example, the argument above doesn't say whether you do or don't have a current password. Maybe you do, and maybe you don't . But either way, the argument is still valid.

Consider this argument:

You can't log into the network If you have a current password, then you can log into the network

Therefore

You don't have a current passwo rd. Now, in real life, we might say: that's not a valid argument. For example, maybe you can't log into the network because your Ethernet cable is bad, or because the network is down for maintenance, or any one of a zillion other reasons.

But in fact, this is a valid argument in logic. If we accept the two premises, then the conclusion follows. One of the premises is "If you have a current password, you can log into the network". There are no ifs, ands, or buts.

So, this illustrates an important point: when working with logic problems it is important to take the statements literally and at face value. Don't read things into the problems that aren't there.

We sometimes use problems that seemto be about the real world— world—we do this to make the problems more interesting and relevant, and to give us some insight into what the symbols mean. But the problems, ultimately are not in the real world —they are about amathematical model of the real world, where we make a lot of simplifying assumptions. For example, the hard and fast absolute statement: If you have a current password, then you can log into the network

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IGNOU Solved Assignments By http://ignousolvedassignments.com According to this statement, a current password is sufficient for you to be able to log in.

So, if we assume this to be true, as we do in the argument below, and we assume that you can't log into the network, then we can definitely conclude: you don't have a current password. So here it is again:

You can't log into the network If you have a current password, then you can log into the network

Therefore

You don't have a current passwo rd. This is an argument of the form:

¬q p→q ¬p

This form of argument is called modus tollens (the mode that denies).

Both modus ponens and modus tollens have "universal forms":

Universal modus ponens:

x((P(x)→Q(x)) P(a), where a

{domain of the predicate P}

Q(a)

E.g. All fish have scales. This salmon is a fish. Therefore, this salmon has scales.

Universal modus tollens:

x((P(x)→Q(x)) ¬Q(a), where a

{domain of the predicate P}

¬P(a) Course Code :

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E.g. All surfers are hot. Conrad is not hot. Therefore Conrad is not a surfer.

A relation between two sets is a collection of ordered pairs containing one object from each set. If the object x is from the first set and the object y is from the second set, then the objects are said to be related if the ordered pair (x,y) is in the relation. A function is a type of relation. But, a relation is allowed to have the object x in the first set to be related to more than one object in the second set. So a relation may not be represented by a function machine, because, given the object x to the input of the machine, the machine couldn't spit out a unique output object that is paired to x. Equivalence Relation An equivalence relation on a set is a subset of , i.e., a collection of ordered pairs of elements of , satisfying certain properties. Write "" to mean is an element of , and we say " is related to ," then the properties are 1. Reflexive: for all ,

2. Symmetric: implies for alla,b in X

3. Transitive: and imply for all a,b,c in X,

where these three properties are completely independent. Other notations are often used to indicate a relation, e.g., o

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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} Course Code :

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IGNOU Solved Assignments By http://ignousolvedassignments.com 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.

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b

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 Course Code :

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IGNOU Solved Assignments By http://ignousolvedassignments.com . 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)

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IGNOU Solved Assignments By http://ignousolvedassignments.com vv

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

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

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Course Code : MCS-013 Course Title : Discrete Mathematics

Assignment Number : MCA(1)/013/Assignment/16-17 MCA(1)/013/Assignment/16-17

Last Dates for Submission : 15th October, 2017 (For July 2017 Session) 15th April, 2018 (For January 2018 Session)

Course Code :

MCS-013




MCS-013 Discrete Mathematics solved assingment by ignousolvedassignments.com

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