# Sets,Real Number System and Logic.

Sets Real number System and logic

In mathematics, a set is a collection of distinct objects, considered as an object in its own right. For example, the numbers 2, 4, and 6 are distinct objects when considered separately, but when they are considered collectively they form a single set of size three, written {2,4,6}. Sets are one of the most fundamental concepts in mathematics.

A set can be explained in different ways:

Listing method: A = {a, b, c, .z}

Descriptive method: N = {the natural numbers from 1 to 50}

Set builder method: A – B  = A -  (A∩ B )

Venn – diagram

Universal sets

A universal set is the collection of all objects in a particular context or theory. All other sets in that framework constitute subsets of the universal set, which is denoted as letter U. The objects themselves are known as elements or members of U.

Subsets

The set made by elements of the universal sets is called subsets of the universal sets

For example

U = {1, 2, 3, 4, …………..50}

A = {even integers from 1 to 50}

B   = {odd numbers from 1 to 50}

Here, A and B are the subsets of U

Overlapping sets

Two sets are said to be overlapped if they have same element in common.

A∩ B = {6}

A and B are overlapping sets.

Disjoint sets

Two sets are said to be disjoint sets if there is no element in common.

Cardinality of the sets

The number of the elements in the given sets is known as cardinality of sets.

A = {1, 2, 5,}

B = {5, 3, 4}

AUB = {1, 2, 3, 4, 5}

n(A) = 3

n(B) = 3

n(AUB) = 5

Cardinality of the three sets

Let  A and B and C  represent three sets as a shown in the figure s

n(AUBUC) =  n(A) + n(B) + n(C) - n(A∩ B) -- n(B∩ C) - n(C∩ A) +n(A∩ B∩ C)

Operation of sets

Union of sets

The set which includes elements of A and B is called union of the sets.

A = {1, 2, 5,}

B = {5, 3, 4}

AUB = {1, 2, 3, 4, 5}

Or, AUB = {x: xϵ A or xϵ B}

n(A) = 3

n(B) = 3

n(AUB) = n(A) + n(B) - n(A∩ B)

= 3 +3 -1 = 5

Intersections of sets

If the elements of set belongs to both Sets A and B, it is called intersection of A and B.

A = {1, 2, 5}

B = {5, 3, 4}

A∩ B = {5}

n(A∩ B) = n(A) – n0 (A) = 3 -2 = 1

Or, A∩ B = {x: xϵ A and x ϵ B}

Complement of sets

The set that contains all the elements of universal sets except the given set A is called complement of the set A. It is denoted by  A̅

Difference of sets

If A and B are the two sets , the difference of the dsets is the elements of the set thst includes only in one set .

A – B  = A -  (A∩ B )

B- A    = B  -  (A∩ B )

Example

Worked out examples

Let U = { a, b, c, d, e,f,g,h}  A =  {a,b,c,d } B  = {c,d,e,f }  and C = {d,e,f,g,h }.verify the following relations

a. AUB (B ∩ C) = (AUB)∩(A∩ C)b. A∩ (BUC) = (A∩B) U (A∩ C)

c.${\rm{\: }}\overline {{\rm{AUB}}}$ = A̅ ∩ B̅

d. A – (B U C)= (A – B) ∩ (A – C).

Soln:  Here,

A U B = {a, b, c, d} U {c, d, e, f} = {a, b, c, d, e, f}

A U C = {a, b, c, d} U {d, e, f, g, h} = {a, b, c, d, e, f, g, h}

B ∩ C = {c, d, e, f} U {d, e, f, g, h} = {d, e, f}

a.

L.H.S. = A U (B ∩ C)

= {a, b, c, d, e, f} U {d, e, f} = {a, b, c, d, e, f}

R.H.S. = (A U B) ∩ (A U C)

= {a, b, c, d, e, f} ∩ {a, b, c, d, e, f, g, h} = {a, b, c, d, e, f}

Hence, A U (B ∩ C) = (A U B) ∩ (A U C)

b.

A ∩ B = {a, b, c, d} ∩ {c, d, e, f} = {c, d}

A ∩ C = {a, b, c, d} ∩ {d, e, f, g, h} = {d}

B U C = {c, d, e, f} U {d, e, f, g, h} = {c, d, e, f, g, h}

Now, LHS = A ∩ (B U C) = {a, b, c, d} ∩ {c, d, e, f, g, h} = {c, d}

RHS = (A∩B) U (A ∩ C) = {c, d} U {d} = {c, d}

Hence. A ∩ (B U C) = (A∩B) U (A ∩ C)

c.

soln:

A¯ = U – A = {a, b, c, d, e, f, g, h} – {a, b, c, d} = {e, f, g, h}

B¯ = U – B = {a, b, c, d, e, f, g, h} – {c, d, e, f} = {a, b, g, h}

Now, LHS = AUB¯¯¯¯¯¯¯¯¯¯¯AUB¯

= U – (AUB)

= {a, b, c, d, e, f, g, h} – {a, b, c, d, e, f} = {g, h}

R.H.S. = A¯∩B¯A¯∩B¯ = {e, f, g, h} ∩ {a, b, g, h} = {g, h}

Hence, AUB¯¯¯¯¯¯¯¯¯¯¯=A¯∩B¯AUB¯=A¯∩B¯.

d.

Here,

A – B = {a, b, c, d} – {c, d, e, f} = {a, b}

A – C = {a, b, c, d} – {d, e, f, g, h} = {a, b, c}

Now, LHS = A – (B U C) = {a, b, c, d} – {c, d, e, f, g, h} = {a, b}

RHS = (A – B) ∩ (A – C) = {a, b} ∩ {a, b, c} = {a, b}

Hence, A – (B U C)= (A – B) ∩ (A – C).

In a survey it was found that 8.%  people like oranges ,85% like mangoes and 75% like both But 45 people like none of them  Drawing Venn –diagram , find the number of the people who  were in the survey.

Soln:

Let O and M be the number of people who like oranges and mangoes respectively.

n (U) =  n(OUM) + $\overline {{\rm{n}}\left( {{\rm{OUM}}} \right)}$

100%   =  n (O) + n (M) - n(O∩M) + $\overline {{\rm{n}}\left( {{\rm{OUM}}} \right)}$

100% =  80% + 85% - 75% + $\overline {{\rm{n}}\left( {{\rm{OUM}}} \right)}$

$\overline {{\rm{n}}\left( {{\rm{OUM}}} \right)} {\rm{\: }}$  = 10%

According to the question,

10% of total number (say x)  = 45

Or,  x=  $\frac{{4500}}{{10}}$ = 450

Real number System

The real numbers include all the rational numbers, such as the integer −5 and the fraction 4/3, and all the irrational numbers, such as  √ 2  (1.41421356…, the square root of 2 an irrational algebraic number.

Natural numbers

A natural number is a number that occurs commonly and obviously in nature. As such, it is a whole, non-negative number. The set of natural numbers, denoted N, can be defined in either of two ways: N = {0, 1, 2, 3,}

Integers

An integer is a whole number that can be positive, negative, or zero. Examples of integers are: -5, 1, 5, 8, 97, and 3,043. Examples of numbers that are not integers are: -1.43, 1 3/4, 3.14, .09, and 5,643.1.

Field of axiom

Closure property

If a , b ϵ R and a+b ϵ R

Commutative property

If a, b ϵ R, a + b = b + a

Associative property

a+b  c ϵ R then, a+(b+c)  = (a+b)+ c

a ϵ R  ,there is real number  0  such that a+ 0 = 0 + a =a

a  ϵ R  -a ϵ R a+ (-a) = (-a)  +a  = 0 –a is the additive inverse of a

Multiplicative Axioms

Closure property

If a , b ϵ R and a.b ϵ R

Commutative property

If a , b ϵ R and a.b  = b.a ϵ R

Associative property

If a , b ϵ R  a(bc) =  (ab)c

Multiplicative Inverse

a ϵ R  (a ≠ 0 ) a-1  ϵ R  aa-1  = a-1  a

Distributive property

If a, b, c ϵ R then, a (b+c) = ab +bc

Order axioms

Real number satisfies the order axiom.

If a > 0, b > 0 then ab> 0

Trichotomy property

a, b,  ϵ R a< b   a = b  a > b

Transitivity property

a, b, c ϵ R a >b ,  b > c ⇒   a > c

If a < b a+ c < b+ c

Multiplicative property

If a > b   $\frac{{\rm{a}}}{{\rm{c}}}{\rm{\: }} > \frac{{\rm{b}}}{{\rm{c}}}{\rm{\: \: }}$when c > 0

If a > b   $\frac{{\rm{a}}}{{\rm{c}}}{\rm{\: }} > \frac{{\rm{b}}}{{\rm{c}}}{\rm{\: \: }}$when c < 0

The axiom of the density

If a and b are the two real numbers such that a< b  , then there exists areal number c such that a <  c < b

Example

Solve the following inequalities

(a)

|x + 2| < 4.

=  – 4 < x + 2 < 4

=  – 4  – 2 < x + 2 – 2 < 4 – 2

=  – 6 < x < 2.

So, the solution is {x: –6 < x < 2} and the graph is:

(b)

|x – 1| ≤ 2.

=  – 2 ≤ x – 1 ≤ 2

=  – 2 + 1 ≤ x – 1 + 1≤2 + 1

=  – 1 ≤ x ≤ 3.

So, the solution is {x: –1 ≤ x ≤ 3} and the graph is:

(c)

|2x + 3| ≤ 1.

=  – 1 ≤ 2x + 3 ≤ 1

=  – 1 – 3 ≤ 2x +3 –3 ≤ 1 –3

=  – 4 ≤ 2x ≤  –2.

=  – 2 ≤ x ≤  – 1

So, the solution is {x: –2 ≤ x ≤ –1} and the graph is:

(d)

| x-1|>1

Here, two cases arises,

Case I:

When (x – 1) > 0 then.

= x – 1 + 1 > 1 + 1

So, x > 2, i.e. x ԑ (2, ∞)

Case II:

When (x – 1) < 0 then,

= x – 1 < – 1

= x – 1 + 1 < – 1 + 1

So, x < 0 i.e. x ԑ (–∞, 0)

Hence, the required solution of |x – 1| > 1 is {x:x< 0 or x > 2} i.e. x ԑ (–∞, 0) U (2, ∞)

The graph is:

(e)

|2x + 1| ≥ 3

Here, two cases arises

Case I:

When (2x + 1) > 0 then.

(2x + 1) ≥ 3

= 2x + 1 – 1 ≥ 3 – 1

= 2x ≥ 2

So, x ≥ 1 i.e. x ԑ [1, ∞)

Case II:

When (2x + 1) < 0 then,

– (2x + 1) > 3

= 2x + 1 ≤  – 3

= 2x + 1 – 1 ≤  – 3 – 1

= 2x ≤  – 4

So, x ≤  – 2.i.e. x ԑ (–∞,  – 2]

Thus, the required solution is {x:x ≤  – 2 or x ≥ 1}

I.e. x ԑ(–∞. – 2] U [1, ∞) and the graph is.

Logic

Logic is the process by which we arrive at the conclusion from the given statement with the valid reason.

Logic expressions are written in symbolic form.

Statement

An assertion which is expressed in words, symbols, which is either true or false but not both at same time, is known as a statement.

There are two types of statement:

Simple statement: A statement that declares only one thing

Compound statement: A statement that explains two or more than two things

Logical connectives

Compound statements are made up of simple statement by words or phrase like and or If ….then and   if and only if’’ .They are known as logical connectives or simply connectives.

Conjunctions: If p and q are two statements are joined by the conjunction to make compound statement. The symbol is denoted by   Λ  . I.e. p Λ q

Disjunction: Two simple statements combined by the word or to form a compound statement is known as disconjunction .It is denoted by V.

Negation: The statement which denies the given statement is called negation of the given statement. The symbol for the negation   is ∼. Negation of p is denoted    ∼p.

Conditional (Implication): Two simple statement combined by the word  If ……..then ‘’ to form a compound statement is known as the conditional of the given statement. The symbol used for implication is ⇒  .

Biconditional (Equivalence) :Two simple  statement combined with word   if and only if ‘ to form a compound statement , is known as biconditional .The symbol used for biconditional  is ⇔ .

Tautology: A compound statement which is always true, whatever be the truth value of the components ,is known as tautology .

Laws of logic

Let p, q and r be any three statements

a. Law of excluded middle

Only one statement is true p or ∼ p is true.

b. law of tautology

The disjunction of a statement and its negation is a tautology.

i.e  p V ∼ p

The conjunction of a statement and its negation is a contradiction

i.e  p   Λ ∼ p

d. law of involution

The negation of the negation of the statement is the given statement.

e. Law of Syllogism

If  p ⇒  q and q⇒  r then p⇒  r

( p ⇒  q)  Λ (q⇒  r) ⇒  (P ⇒  r)

f. law of contra positive

( p ⇒  q)   ≡   (∼ q ) ⇒  (∼ p )

g. Inverse Law

(∼ P)  ⇒ (∼ q)  ≡ q ⇒p

The inverse and converse of a conditional are logically equivalent .

Example

If p and q are any two statements, prove that

a. p Λ ~p ≡ c.

b. p v q ≡ p v p.

c. ~[pv(~q)] ≡~p ^ q.

d.. ~p[(~p ^ q)] ≡ p v (~q).

Solution:

(a)

We prove the given relations using truth table.

 p ~p p ^ ~p c = contradiction T T F F F F T T F F F F F F F F

So, from the above truth table, p ^ ~p ≡≡ c.

(b)

We know the given relations using truth table,

 p q pvq qvp T T F F T T T F T T T F T T T F

From the above truth table, p v q ≡ p v p.

(c)

We prove the given relations using truth table.

 p q ~p ~q pv(~q) ~[pv(~q)] ~p^q T T F F T F T F F F T T F T F T T T F T F F T F F F T F

From the above truth table, ~[pv(~q)]≡≡~p ^ q.

(d)

We prove the given relations using truth table.

 p q ~p ~q ~p ^ q ~[(~p)^q] pv(~q) T T F F T F T F F F T T F T F T F F T F T T F T T T F T

From the above truth table, ~p[(~p ^ q)] ≡≡ p v (~q).

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