Mathematics Chapter 1 Sets

Let P(n): 1 + 2 + 3 + ......... + n =
I. For n = 1,
    P(1) : 1 = is true.
II.  Suppose the statement is true for n = m,
      i.e. P(m):           ....(i)
III.    For n = m + 1,
        P(m + 1): 1 + 2 + 3 + ........ + (m + 1) =
or  [1 + 2 + 3 + ...... + m] + (m + 1) =
                                       
                                         [From (i), 1 + 2 + 3 + ...... + m = ]
∴        P (m + 1):

  

    which is true

∴    P(m + 1) is true

∴    P(m) is true P(m + 1) is true
Hence, by mathematical induction
P(n) is true for all

Let
I.      For n = 1,
       is true.
II.    Suppose the statement is true for n = m,
   
          i.e.,     ... (i)
III.     For n = m + 1,
       
or     
     From (i),

∴  



       which is true

∴     P(m + 1) is true

∴     P(m) is true P(m + 1) is true.
Hence, by mathematical induction, P(n) is true for all

 Let P(n) :
I.    For n = 1,
      P(1) : 1

∴     P(1) is true.
II.   Let the statement be true for n = m, 

∴     P(m) :      ... (i)

III.  For   n = m + 1,                      
      P(m + 1) :
or   
  From (i),

∴   

which is true.

∴   P(m + 1) is true.

∴   P(m) is true P (m + 1) is true
Hence, by principle of mathematical induction, P(n) is true for all

Let P(n): 
I.    For n = 1,
     P(1) : 
∴    P(1) is true.
II.   Let the statement be true
       for n = m,

∴          .... (i)

III.   For n = m + 1,
       P(m + 1):
or    
        From (i),
      

∴  
 
      which is true.
∴    P(m + 1) is true
∴    P(m) is true P(m + 1) is true
Hence, by the principal of mathematical induction, P(n) is true for all .

Let P(n) :
I.   For n = 1,
    

∴     P(1) is true
II.  Let the statement be true for n = m, 

∴    P(m):                           ...(i)
III.  For n = m + 1,
      
or 
or       ... (ii)

∴   


        which is true

∴      P (m + 1) is true.

∴      P(m) is true P(m + 1) is true.
Hence, by the principal of mathematical induction, P(n) is true for all

Let
I.     For n = 1,
     

∴    P(1) is true
II.    Suppose the statement is true for n = m,

∴    
III.  For n = m +1,
     
or
  
                                                                                             

∴  



which is true

∴  P(m + 1) is true

∴  P(m) is true P (m + 1) is true
Hence, by the principle of mathematical induction, P(n) is true for all

Let P(n):
I.     For n =1,
     
∴     P(1) is true.
II.   Suppose the statement is true for n = m,

∴   
III.   For n = m + 1,
       
                                                                        
or     1.2.3 + 2.3.4 + 3.4.5 + ........... + m (m + 1) (m + 2) + (m + 1) (m + 2) (m + 3)
                                                                          
        From (i), 1.2.3 + 2.3.4 + 3.4.5 + ......... + m (m + 1) (m + 2)
                                                                       

∴   
                                                                          

       
             which is true.

∴        P(m + 1) is true.

∴        P(m) is true P(m + 1) is true.
Hence, by the principle of mathematical induction, P(n) is true for all

Let P(n):
I.         For n = 1,
         P(1) :

∴       P(1) is true.
II.      Suppose the statement is true for n = m,
         P(m):      ...(i)
III.     For n = (m + 1),
         P(m + 1) :
or       
      From (i),
        

∴  




         Which is true

∴        P(m + 1) is true

∴        P(m) is true P(m + 1) is true
Hence, by the principle of mathematical induction, statement P(n) is true for all

Let
I.      For n = 1,
      

∴      P(1) is true
II.      Let the statement be true for n = m,  
       P(m) :      ...(i)
III.   For n = m + 1,
          
 
or 
    From (i),
       
      Also,  1, 2 , 3,..........., m+1 are in A.P.
 
                                                                          
                                                                              [By using ]

∴    


         which is true.
∴      P(m + 1) is true.

∴      P(m) is true P(m + 1) is true.
Hence, by the principle of mathematical induction, the statement P(n) is true for all

Let P(n):
I.         For n = 1,
           P(1) : 1 =
∴         P(1) is true
II.        Let the statement be true for n = m,
     P(m) : 1 + 3 + 5 + .................... + (2m - 1) = m²                    ...(i)
III.      For n = m + 1,
         P(m + 1) : 1 + 3 + 5 + .......... + [2 (m+1) - 1] = (m + 1)²
or      1 + 3 + 5 + ........... + (2m - 1) + (2m + 1) = (m + 1)²
        From (i),
        

∴       
   

∴       P (m + 1) is true.

∴       P(m) is true. is true.
         Hence, by the principal of mathematical induction, P(n) is true for all

Let P(n) : a + (a + d) + (a + 2d) + .............+ [a + (n - 1)d] =
I.     For n = 1,
     
  

∴      P(1) is true
II.       Suppose the statement is true for n=m,

∴       .... (i)
III.   For n = m + 1,
      
or      
       From (i),
       
∴     




           which is true
∴          P (m + 1) is true

∴          P (m) is true P(m + 1) is true
Hence by the principle of mathematical induction, P(n) is true for all

Let P(n):

I.        For  n = 1,
        P(1) :

∴       P(1) is true
II.      Suppose that the statement P (n) is true for n = m, 

∴      P(m) :                  ...(i)
III.     For n = m + 1,
       
or     
       From (i),

∴  

              which is true
∴             P (m + 1) is true

∴             P(m) is trueP (m + 1) is true
Hence, by the principle of mathematical induction,  P(n) is true for all

Let P(n):  1 + 2 + 3 + .......... + 

I.      For n = 1,
       

∴      P (1) is true
II.    Suppose the statement is true for n = m, 
                            ... (i)

III.    For n = m + 1,
         ... (ii)
or      
          From (i),
         
Adding (m + 1) on both sides, we get
  
1 + 2 + 3 +........+ m + (m + 1)
1 + 2 + 3 +.........+ (m + 1) <
1 + 2 + 3 + ......... + (m + 1)

∴   P (m + 1) is true.

∴   P(m) is true P (m + 1) is true.
Hence, P(n) is true for all

Let
I.       For n = 2(note this step, n>1)
       

        which is true

∴       P(n) is true for n = 2
II.      Suppose the statement is true for n = m,
                         .... (i)
III.      For n = m + 1,
         
or     
or      
         Adding on both sides of (i), we get
 
        
    
     
     
         But,

∴     

∴         P (m + 1) is true
∴         P(m) is true P(m + 1) is true
Hence, P(n) is true for all

Let P(n): n (n + 1) (n + 5) is a multiple of 3.
I.         For n = 1,
         is a multiple of 3
      1 (2) (6) is a multiple of 3 12 is a multiple of 3.
           which is true.

∴          P(n) is true for n = 1.
II.       Suppose P (n) is true for n = m
     P(m) : m(m + 1) (m + 5) is a multiple of 3 m (m + 1) (m + 5) = 3k
    
                                               ...(i)
III.      For  n = m + 1,
           P(m + 1): (m + 1) (m + 1 + 1) (m + 1 + 5) is a multiple of 3.
           (m + 1) (m + 2) (m + 6) is a multiple of 3.
           Now, (m + 1) (m + 2) (m + 6) = (m + 1)() =
                                                                   
           =                  [BY (i)]
           =
               where
        (m + 1) (m + 2) (m + 6) is a multiple of 3.

∴           P(m + 1) is true.

∴           P(m) is true P(m + 1) is true.
            Hence, P(n) is true for all

Let P(n): is a multiple of 27.
I.       For n = 1,
        P(1): is a multiple of 27
   41 - 14 is a multiple of 27 27 is a multiple of 27
         which is true.

∴       P(n) is true for n = 1
II.    Suppose P(n) is true for n = m,
   P(m) : is a multiple of 27   ...(i)
III.   For n = m + 1,
        P (m + 1) : is a multiple of 27
        But,
           
           =        [By (i)]
             where
is a multiple of 27

∴           P(m + 1) is true.

∴           P(m) is true P(m + 1) is true.
Hence, by induction, P(n) is true for all

Let P(n) : is divisible by 11
I.             For n = 1,
            P(1) : is divisible by 11
       10¹ + 1 is divisible by 11  11 is divisible by 11

∴           P(1) is true
II.          Suppose the statement is true for n = m,

∴         P(m) : is divisible by 11.
     
                                                ...(i)
III.         For n = m + 1,
            is divisible by 11.       ...(ii)
            Now,        [By (i)]
            =
                                                                          where k' = 100k -

∴           is divisible by 11

           P (m + 1) is true.

∴              P(m) is true P (m + 1) is true
Hence, by induction, P(n) is true for all

Let is divisible by 8.
I.      For n = 1,
      P(1) : is divisible by 8
  is divisible by 8 81 - 17 is divisible by 8 64 is divisible by 8
      which is true
∴     P(n) is true for n = 1
II.    Let the statement be true for n = m,

∴     is divisible by 8
     ...(i)
III.     For n = m + 1,
        is divisible by 8
Now,   
=                                      [By (i)]
= 72k + 72m + 81 - 8m - 17 = 72k + 64m + 64 = 8(9k + 8m + 8),
= 8k'  where k' = 9k + 8m +

∴     is divisible by 8.
  P(m + 1) is true.

∴   P(m) is true P (m + 1) is true.
Hence, by the principle of mathematical induction, P(n) is true for all

Let n, n+1, n+2 be three consecutive natural numbers.
Let P(n): is divisible by 9.
I.       For n = 1,
       is divisible by 9
  1 + 8 + 27 is divisible by 9 36 is divisible by 9
      which is true

∴     the statement is true for n = 1.
II.     Suppose the statement is true for n = m, 
   P(m) : is divisible by 9.
                              ...(i)
III.   For n = m + 1,                           
       is divisible by 9.
       Now, from (i),
     
             
    
        
         where
     is divisible by 9
       P (m + 1) is true

∴         P (m) is true P (m + 1) is true.
           Hence, by the principal of mathematical induction, P (n) is true for all

Let is divisible by a + b
I.     For n = 1,
      is divisible by a + b           is divisible by a + b
   (a - b) (a + b) is divisible by a + b
       which is true.

∴      P(1) is true.
II.    Suppose the statement is true for n = m,
is divisible by a + b
                          ...(i)
III.   For n = m + 1
      P (m + 1): is divisible by a + b.
     Now,  
   
     From (i),

      where k' =
  is divisible by a + b

∴    P(m + 1) is true

∴    P(m) is true P ( m + 1) is true
Hence, by the principle of mathematical induction, P(n) is true for all

Let P(n) :
I.   For n = 1,
   

∴   P(1) is true
II.   Suppose the statement is true for n = m,
III.  P(m) :           ... (i)
       For n = m + 1,
      
      Now,
           
             =                                                   [By (i)]
             =

∴      P(m + 1) is true.
       P(m) is true P(m + 1) is true
Hence, by the principle of mathematical induction, P(n) is true for all

Let P(n):
I.     For n = 1,

∴    P(1) is true
II.    Suppose the statement P(n) is true for n = m,

∴     P(m) :                                                  ...(i)
III.   For n = m + 1,
                                                    ...(ii)
       From (i),

∴     P (m+1) is true

∴     P(m) is true P (m + 1) is true
Hence, P(n) is true for all

Let P (n):
I.    For n = 1, 

∴    P(1) is true
II.  Suppose the statement is true for n = m,
                                                             ...(i)
III.   For n = m + 1,
      
       From (i),
  
        Also,  

∴      P (m + 1) is true.

∴      P(m) is true P (m + 1) is true.
Hence, statement is true for all

 (i) ‘Yesterday was Sunday’.

This sentence is correct (true) on Monday but not on other day.

Hence, it is not a statement.

(ii) ‘She is a doctor’.

This sentence is not a statement because who ‘she’ is.

(iii) ‘sin θ is always greater than 0’.

It is not a statement because one cannot say anything about the value of sin θ without knowing the value of θ .

(i) Statement.

Reason : This sentence is false because 8 is greater than 6.

(ii) Statement.

Reason : This sentence is false because there are sets which are not finite e.g., N.

(iii) Statement.

Reason : It is a scientifically established fact that sun is a star, so, the sentence is always true.

(iv) Not a statement.

Reason : This sentence is subjective in the sense that for those who like Mathematics, it may be fun but for others it may not be. So, this sentence is not always true.

(v) Statement.

Reason : It is a scientifically established natural phenomenon that cloud is formed before it rains, so, the sentence is always true.

(vi) Not a statement.

Reason : It is a question which also contains the word ‘Here’.

Statements. (i), (ii), (iii), (vii), (viii), (x).

Non-statements, (iv), (v), (vi), (ix).

Statements, (iii), (iv), (v).

Non-Statements, (i), (ii), (vi).

The negation of this statement is:
It is false that Chennai is the capital of Tamil Nadu. Or
Chennai is not the capital of Tamil Nadu.

The negation of this statement is:
It is false that  is not a complex number.   Or
 is a complex number.

The negation of this statement is:
It is false that all triangles are not equilateral triangles.  Or
All triangles are equilateral triangles.

The negation of this statement is:
It is false that the number 2 is greater than 7.  Or
The number 2 is not greater than 7.

The negation of this statement is:
It is false that every natural number is an integer. Or
Every natural number is not an integer.

The statement means that in every rectangle, both the diagonals have the same length.
The negation of this statement is ‘It is false that both the diagonals in a rectangle have the same length’. It means that there is at least one rectangle in which both the diagonals do not have the same length.

(i) The negation of the statement is:
     It is not the case that there does not exist a quadrilateral which has all its sides equal.
                                      Or
     There exists a quadrilateral which has all its sides equal.
     This statement is true because a square is a quadrilateral having all sides equal.
(ii)  The negation of the statement is:
       It is false that the sum of 3 and 4 is 9
                                  Or
        The sum of 3 and 4 is not equal to 9.
        This statement is true because sum of 3 and 4 is 7 which is not equal to 9.
(iii)    The negation of the statement is:
         It is false that Australia is a continent.  Or
         Australia is not a continent.
         This statement is false because Australia is a continent.
(iv)     The negation of the statement is:
          It is false that every natural number is greater than 0. Or
          There exists a natural number which is not greater than 0.
         This statement is false because every natural number is greater than 0.
       

Let  p: The number x is not a rational number.
q: The number x is not an irrational number.
~p: The number x is a rational number.
It means that x is not an irrational number because a number is either rational or irrational both not both.
∴     The statement q is the negation of the statement p.
Hence, both the statements p and q are negations of each other.

(i) Here, p: For every positive real number x, the number x - 1 is also positive.
     ~p: It is false that for every positive real number x, the number x - 1 is also positive.
                                            Or
      ~p: There exists a positive real number x such that x - 1 is not positive.
(ii), Here, q: All cats scratch.
         ~q:  It is false that all cats scratch.
                                             Or
          ~q: There exists a cat which does not scratch.
(iii) Here, r: For every real number x, either x>1 or x<1.
            ~r: It is false that for every real number x, either x>1 or x<1.
                                              Or           
            ~r: There exists a real number x such that neither x>1 nor x<1.
(iv) Here, s: There exists a number x such that 0<x<1.
          ~s: It is false that there exists a number x such that 0<x<1.
                                               Or
           ~s: There does not exist a number x such that 0<x<1.
             
       

The component statements are:
p: Number 3 is a prime.
q: Number 3 is odd.
Here, the connecting words is 'or'.
Also, both the statements are true.

The component statements are:
p: All integers are positive.
q: All integers are negative.
Here, the connecting word is 'or'.
Also, both the statements are false.

The component statements are:
p: 100 is divisible by 3.
q: 100 is divisible by 11.
r: 100 is divisible by 5.
Here, the connecting word is 'and'
Also, statements p, q are false, but statement r is true.

(i) New Delhi is not a city.
(ii) The sky is not blue.
(iii) The sum of the angles of a triangle is not equal to two right angles.
(iv) Kavita is not a hard working girl.
(v) Two plus two is not equal to 6.

Detailed solution not provided.
Ans. (i) ~p: There exists a real number x, such that 
(ii) ~q: For all real numbers x, 
(iii) ~r: There exists a bird which has no wings.
(iv) ~s: There exists a student who does not study Mathematics at the elementary level.

Detailed solution not provided.
Ans. (i)  is a rational number. (False)              (ii)  is an rational number (True)
(iii) 0 is not  a positive number. (True)

Solution not provided.
Ans. Yes.

Solution not provided.
Ans. Yes.

        (i) p: It is raining.
            q: It is cold ; connecting word: 'and'.
       (ii) p: All the rational numbers are real.
            q: All the real numbers are complex connecting word: 'and'.
       (iii) p: 0 is a positive number.
             q: 0 is  a negative number.
            Connecting word : 'or'.

(i) p: A square is a quadrilateral (True)
     q: A square has all its sides equal (True)
     Connecting word : "and"
(ii) p: All prime numbers are odd numbers. (False)
     q: All prime numbers are even numbers. (False)
     Connecting word: "or".
(iii) p: A person who has taken Mathematics can go for MCA (True)
      q: A person who has taken Computer Science can go for MCA. (True)
      Connecting word : " or"
(iv) p: Chandigarh is the capital of Haryana. (True)
      q: Chandigarh is the capital of U.P. (False)
      Connecting word : "and".
(v)   is a rational number. (False)
       is an irrational number. (True)
       Connecting word: "or".
(vi)  p : 24 is a multiple of 2. (True)
       q: 24  is a multiple of 4. (True)
       r: 24 is a multiple of 8. (True)
       Connecting word "and"
      

     

In this statement, the connecting word is 'And'.
Its component statements are:
p: All rational numbers are real.
q: All real numbers are not complex.

In this statement, the connecting word is 'Or'.
Its component statements are:
p: Square of an integer is positive.
q: Square of an integer is negative.

In this statement, the connecting word is 'And'.
Its component statements are:
p: The sand heats up quickly in the Sun.
q: The sand does not cool down fast at night.

In this statement, the connecting word is 'And'.
Its component statements are:
p : x = 2 is the root of equation 3x² - x - 10 = 0
q : x = 3 is the root of equation 3x² - x - 10 = 0

The component statements are:
p : 50 is a multiple of 2
q : 50 is a multiple of 5
Here, both the component statements p and q are true.
Also, the connective is 'And'.
Hence, the compound statement is true.

The component statement are:
 p :  is a rational number.
q :  is an irrational number.
Here, p is false, but q is true.
Also, the connective is 'Or'.
Hence, the compound statement is true.

The component statement are:
p : Mumbai is the capital of Gujarat.
q : Mumbai is the capital of Maharashtra.
Here, p is false and q is true.
Also, the connective is 'Or'.
Hence, the compound statement is true.

The component statement are:
p : A rectangle is a quadrilateral.
q : A rectangle is a 5-sided polygon.
Here, p is true and q is false.
Also, the connective is 'Or'
Hence, the compound statement is true.

The connective ‘or’ used in this statement is inclusive.

You can apply for a driving licence even if you have a ration card as well as passport.

Solution not provided.

Here, the 'Or' is exclusive.
The component statement q is true and r is false. Hence, the statements p is true.

Given statement is
p : For every real number x, x is less than x + 1
Here, the quantifier is 'For every'.
This statement means that if R denotes the set of all real numbers, then for all the members x of the set R, x is less than x + 1.
The negation of given statement is
~p : It is not the case that for every real number x, x is less than x + 1 or
~p : There exists a real number x, such that x is not less than x + 1.

Given statement is
p : There exists a capital for every state in India.
Here, the quantifier is 'There exist'.
The negation of given statement is
~p : It is not the case that there exists a capital for every state in India. Or
~p : There exists a state in india which does not have a captial.

(i)    p : A line is straight.
       q : A line extends indefinitely in both directions; TRUE.
(ii)   p : 0 is less than every positive integer.
       q : 0 is less than every negative integer; FALSE.
(iii)   p : All living things have two legs.
        q : All living things have two eyes; FALSE.
(iv)   p : A point occupies a position.
        q : Its location can be determined; TRUE.
(v)    p : 42 is divisible by 5.
        q : 42 is divisible by 6.
         r : 42 is divisible by 7; FALSE.

Detailed solution not provided.
Ans. Exclusive.

Detailed solution not provided.
Ans. Inclusive.

Detailed solution not provided.
Ans. Inclusive.

Detailed solution not provided.
Ans. Inclusive.

Detailed solution not provided.
Ans. Exclusive.

Detailed solution not provided.
p : Two lines intersect at a point
q : Two lines are parallel; TRUE.

Detailed solution not provided.
p : 125 is a multiple of 7.
q : 125 is a multiple of 8; FALSE.

Detailed solution not provided.
p : School is closed if there is a holiday.
q : School is closed if there is a Sunday; TRUE.

Detailed solution not provided.
p : Mumbai is the capital of Kolkata.
q : Mumbai is the capital of Karnatka; FALSE.

Detailed solution not provided.
(i) Inclusive " Or" ; TRUE.      (ii) Inclusive "Or", TRUE.
(iii) Exclusive "Or", TRUE.

Detailed solution not provided.
Ans. (i) There exists
       (ii) For every.

Detailed solution not provided.
Ans. (i) There exists
       (ii) For every.

Detailed solution not provided.
Ans. (i) Exclusive 'Or' 
       (ii) Exclusive 'Or' .

The above statements in “if-then” form are :

p : If a number is multiple of 9, then it is a multiple of 3.

q : If a number is multiple of 9, then it is a multiple of 3.

The given statement can be written in the following five different ways conveying the same meaning:

(i) A natural number is odd, implies that its square is also odd.

(ii) Knowing that a natural number is odd is sufficient to conclude that its square is also odd.

(iii) A natural number is odd only if its square is also odd.

(iv) When a natural number is odd, it is necessary that its square is odd.

(v) If the square of a natural number is not odd, then the natural number is not odd.

The given statement is:

‘If x is a prime number, then it is odd’. ...(I)

The contrapositive statement of the given statement (I) is ‘If x is not an odd number, then x is not prime’.

The converse of the given statement (I) is:

‘if x is an odd number, then x is prime.’

The given statement is:

If the two lines are parallel, then they do not intersect in the same plane. ...(I)

The contrapositive statement of the given statement (I) is:

If the two lines intersect in the same plane, then they are not parallel.

The converse of the given statement (I) is:

If the two lines do not intersect in the same plane, then they are parallel.

The given statement is:

‘Something is cold implies that it has low temperature’. ...(I)

The contrapositive statement of the given statement (I) is:

if something has not low temperature, then it is not cold’.

The converse of the given statement (I) is:

if something has low temperature, then it is cold’.

The given statement is:
‘x is an even number implies that x is divisible by 4’. ...(I)
The contrapositive statement of the given statement (I) is:
‘If x is not divisible by 4, then x is not an even number.’
The converse of the given statement (I) is:
‘If x is divisible by 4, then x is an even number’.

The above pair of statements using “if and only if” is:
A quadrilateral is a parallelogram if and only if its diagonals bisect each other.

The given statement is:
‘If you live in Delhi, then you have winter clothes.’
The statement (i) is a contrapositive statement of the above statement.
The statement (ii) is converse of the above statement.

(b) The given statement is:
If a quadrilateral is a parallelogram, then its diagonals bisect each other.
The statement (i) is a contrapositive statement of the above statement.
The statement (ii) is converse of the above statement.

(i)  If there is log on to the server, then you have a password.
(ii) If it rains, then there is traffic jam.
(iii) If you can access the website, then you pay a subscription fee.

 (i) A number is a multiple of 9 implies that it is a multiple of 3.
(ii) Knowing that a number is multiple of 9 is sufficient condition to conclude that it is a multiple of 3.
(iii) A number is multiple of 9 only if it is a multiple of 3.
(iv) When a number is multiple of 9, it is necessarily a multiple of 3.
(v) If a number is not a multiple of 3, then it is not a multiple of 9.

Contrapositive:
(i) If you do not have winter clothes, then you do not live in Delhi
(ii) If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram.
Converse:
(i) If you have winter clothes, then you live in Delhi.
(ii) If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

(i) If a number is not divisible by 3, then it is not divisible by 9.
(ii) If you are not a citizen of India, then you were not born in India.
(iii) If a triangle is not isosceles, then it is not equilateral.

(i) If a number n² is even, then n is even.
(ii) If you get an A grade in the class, then you have done all the exercises of the book.
(iii) If two integers a and b are such that a-b is always a positive integer, then a > b.

Detailed solution not provided.
Ans. (i) p : Triangle ABC is equilateral.
           q : Triangle ABC is isosceles; TRUE.
     (ii)  p : a and b are integers.
           q : ab is a rational number; TRUE.

(i) A rectangle is a square if and only if all its four sides are equal.
(ii) A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

If triangle is an obtused angled triangle, then 
where  denotes the obtuse angle of the triangle.
Let 
Since all the angles of the triangle are equal, therefore, each angle = 
∴ Sum of three interior angles of the triangle = 
   which is not possible as sum of interior angles of a triangle is 
∴ Given statement p is not true.

We observe that x = 1 is the root of the equation  which lies between 0 and 2.
∴ Given statement q is not true.

Let a = 1, b = -1
We observe that 
But, 
∴   Given statement is not true.

The component statements of given statement are:
q : x is an integer and x²  is even
r : x is an even integer.
In method of contrapositive,
We assume that ~r is true and we have to prove that ~q is also true.
Now, ~r is true,
Therefore, x is not an even integer or x is an odd integer.
or x = 2n + 1 for some integer n.
or 
or 
or 
or  is an odd integer.
or q is false.
or ~q is true.
Thus, 
Hence, the given statement p is true.

The compound statement with “And” is “25 is a multiple of 5 and 8”.
Since p is true and q is false, therefore, the compound statement “p and q” is not true.
Therefore, the statement “p and q” is not a valid statement.
Now, the compound statement with “or” is
“25 is a multiple of 5 or 8”.
Since p is true and q is false, therefore, the statement “p or q” is true.
Therefore, the statement “p or q” is a valid statement.

Let us assume that p is not true, therefore, sum of an irrational and a rational number is not irrational.
    There exists an irrational number x and a rational number y such that x + y is not irrational.
     x + y is a rational number, say z
     x + y = z
     x = z - y
     x is rational
But, x is irrational.
So, we arrive at a contradiction.
Thus, our supposition is wrong.
Hence, p is true.

Let p be the statement  is irrational.
If possible, let p is not true
          is not irrational.
           is rational
    have no common factors.
      divides 
 11 divides  for some integer c
   
      divides 
  11 divides b. Thus, 11 is a common factor of both a and b which is a contradiction.
Therefore, our supposition is wrong. So,  is irrational.

Detailed solution not provided.
Ans. Statements: (i), (ii)
Non-statements: (iii), (iv), (v), (vi)

(i) True (ii) False.

Converse : If an integer has no divisior other than 1 and itself, then it is prime.
Contrapositive : If an integer has same divisior other than 1 and itself, then it is not prime.

Converse : If I go to a beach, then it is a sunny day.
Contrapositive : If I do not go to beach, then it is not a sunny day.

Converse : If you feel thirsty, then it is hot outside.
Contrapositive : If you do not feel thirsty, then it is not hot outside.

(i) A triangle is equiangular implies that it is an obtuse angled triangle.

(ii) Knowing that a triangle as equiangular is sufficient condition to conclude that it is obtuse angled triangle.

(iii) A triangle is equiangular only if it is obtuse angled triangle.

(iv) When a triangle is equiangular, it is necessarily an obtuse angled triangle.

(v) If triangle is not obtuse angled, then it is not equiangular.

Inclusive;
p : You are wet, when it rains.
q : You are wet, when you are in a river; TRUE.

Detailed solution not provided.
Ans. True.

Solution not provided.

Solution not provided.

The component statements of the given statement are
q : x is a real number such that x³ + 4x = 0
r : x is 0.
To check the validity of statement, we assume that q is true.
Now, we have to show that r is also true.
Since q is true, therefore, x is a real number such that 
i.e, 
i.e, either x = 0 or 
i.e, either x = 0 or 
Now, x is real means 
Therefore, x = 0 i.e, x is 0
Therefore, r is true.
Hence, the given statement is true.

The component statements of the given statement are
q : x is a real number such that x³ + 4x = 0
r : x is 0.
In this method, we assume p is not true.
i.e, If x is a real number such that  then 
i.e., 
i.e, 
i.e,  which is not possible as x is a real number.
Therefore, our supposition is wrong.
Hence, p is true.