TEAM 11  

QUESTIONS-

EXERCISE 2.1

QUESTION 11

A] How many rows are needed for the truth table of the compound statement   (p\/~q) <--> [(~r/\s) --> t], where p,q,r,s,t are primitive statements?

solution]

It requires 32 rows.

Because  if n variables are present  then it requires 2ⁿ rows....

so the given problem requires 2⁵=32 rows.

B]Let p₁,p_2,p_3,.........,p_n denote n primitive statements. Let p be a compound statements that contains atleast one occurence each of p_i, for 1<= i <=n  and p contains no other primitive statement. How many rows are needed to construct the truth table for p?

SOLUTION]

The number of rows required is 2ⁿ rows.

Reason-

as the compound statements contains atleast one occurence of     p_i   where i is 1<= i <=n. So the value of    i   starts from 1 and till n.

it contains no other primitive statements.

So the number of primitive statements is    n.

EXERCISE 2.2

QUESTION 11

Let p,q and r denote primitive statements. Find a form of the contrapositive of  p -> (q -> r) with

a)only one occurence of the connective  -->

b)no occurence of the connective  -->

SOLUTION]

A)   p --> (q --> r)

    =  ~p \/ (q --> r)              [ because p --> q = ~p \/ q]

    contrapositive of the above is

    = (q --> r) \/ ~p            

     solution contains only one  --> connective.

B)    p --> (q --> r)

       = ~p \/ ( q --> r )           [ because  p --> q = ~p \/ q ]

       = ~p \/  ( ~q \/ r )

       Contrapositive of the above is

       =  ( ~q \/ r ) \/ ~p        [ contains no connective -->]

       

EXERCISE 2.4

QUESTION 4

Consider the universe of all polygons with three or four sides and define the following open statements for this universe.

a(x) : all interiors angles of x are equal.

e(x) : x is an equilateral triangle.

h(x) : all sides of  x  are equal.

i(x)  : x is an isoscelus triangle.

p(x) : x has an interior angle that exceeds 180^O.

q(x) : x is a quadrilateral.

r(x)  : x is a rectangle.

s(x) : x is a square.

t(x) : x is a triangle.

Translate each of the following statements into an English sentence  and determine whether the statement is true or false.

SOLUTION]

a) \-/x [q(x)  v_ t(x)]

For all x, either  x is a quadrilateral or x is a triangle  but not both.           TRUE

b) \-/x [i(x) --> e(x)]

For all x, if x is an isocelus triangle then x is an equilateral triangle.         FALSE

c)  -]x [t(x) /\ p(x)]

For some x, x is a triangle and and x has an interior angle that exceeds 180^o.         FALSE

d)  \-/x [ ( a(x) /\  t(x) )  <---> e(x) ]

For all x, all interior angle of x are equal and x is a triangle if and only if x is an equilateral triangle.

FALSE

e)  -]x  [ q(x) /\ ~r(x) ]

For some x, x is a quadrilateral  and x is not a triangle.   TRUE

f)  -]x [ r(x) /\ ~s(x) ]

For some x, x is arectangle and x is not a square.

TRUE 

g)  \-/x [ h(x) --> e(x) ]

For all x, if all sides x are equal then x is an equilateral triangle     TRUE

h)  \-/x [t(x) --> ~p(x)]

For all x, if x is a triangle then x doesnot has an interoir angle that exceeds 180^o   TRUE

i)   \-/x [ s(x) <--> ( a(x) /\ b(x) ) ]

For all x, x is a square if and only if all interiors angles of x are equal and all sides of x are equal.  TRUE

j)  \-/x [ t(x) --> ( a(x) <--> h(x) ) ]

For all x, if x is  a  triangle then all interior angles of x are equal if and only if all sides of x are equal. FALSE

EXERCISE  3.1

QUESTION 11

Let  A = { 1,2,3,4,5,7,8,10,11,14,17,18}

a) How many subsets of A contain six elements?

b) How many six-element subset of A contain four             even integers and two odd integers?

c) How many subsets of A contain only odd integers?

SOLUTION]

a) Total number of subsets of A which contain 6 elements is :: 12C6 = 924

Reason-  total number of elements is 12. and the number of subsets having 6 elements is combination of 6 elements out if 12 elements. Total subset of A is 2¹²=4096. in that 924 contains 6 elements.

2¹²= 12C₀ + 12C₁ + 12C₂ + 12C₃ + 12C₄ + 12C₅ + 12C₆ +  12C₇ + 12C₈ + 12C₉ + 12C₁₀ +12C₁₁ +12C₁₂

b) Subset of A containing 4 even integers and 2 odd integers is

 =  6C₄ * 6C₂

 = 15 * 15

 = 225

Reason - total number of even integers is 6. so 4 elements out of 6 is combination of 6 and 4. similarly out of 6 odd integers 2 should be chosen in 6C2 ways.

c) Subset of A containing only odd integers is 2⁶ -1 = 63

Reason -  total number of odd elements is 6. the power set of 6 elements contain  2⁶ elements. So by removing the null set we get 63 subsets.             

...............END OF ASSIGNMENT.............