Grade 11 Mathematics Note
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) – n_{0} (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, nonnegative 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
Addition 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
Additive identity
a ϵ R ,there is real number 0 such that a+ 0 = 0 + a =a
0 is the additive identity.
Additive inverse
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
Addition property
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)
 x1>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
c. Law of contradiction
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).