Tuesday, November 16, 2010

 TEAM 12

Exercise-2.1: 
12. Determine all the truth value assignments if any for the primitive statement ,q,r,s,t that make each of the following compound statement FALSE.
a)      [(p^q)^r] ( s v t)
b)      [p^(q^r)]   (s ⊻ t)

SOLUTION:
a)    [(p^q)^r] ( s v t)
This statement is false when [(p^q)^r]is TRUE and  ( s v t) is FALSE.
Thus, P = q = r = True
And
Either s =t= False
Or S= True and t = False
Or S= False and t = True.

b)    [p^(q^r)]   (s ⊻ t)
This statement is false when [p^(q^r)] is TRUE and (s ⊻ t) is FALSE.

Thus,
P = q = r = TRUE And s = t = TRUE
Or
P = q = r = TRUE And s = t = FALSE



Exercise-2.2:
12. Show that for primitive statement p,q
p ⊻ q ⇔ [ ( p ^ ~q) v ( ~p ^q ) ⇔ ~(p ↔ q )

SOLUTION:
         Consider   [ ( p ^ q ) v ( ~ p ^ q ) ]
         Now,
   [ ( p ^ q ) v ( ~ p ^ q ) ]
=[ ~(p→q) v( q ^ ~p ) ]       -Commutative law & ~(p→q)(q^~p)
=[~(p→q) v ~(q→p)]        - ~(p→q)(q^~p)
=~[( p→q)^ (q→p)]              -Demorgan’s Law
=~(p ↔ q )                   - ( p→q)^ (q→p)⇔(p ↔ q )
= p ⊻ q                                   - (p ↔ q ) ⇔~(p ⊻ q)



Exercise-2.4:
5. Proffessor Carlson’s class in mechanics is comprised of 29 students of which exactly
1)      Three physics majors are juniors.
2)      Two electrical engineering majors are juniors.
3)      Four mathemaatics maajors are juniors.
4)      Twelve physics majors are seniors.
5)      Four electrical engineering majors are seniors.
6)      Two electrical engineering majors are graduate students ;  and
7)      Two mathematics majors are graduate students.

               Consider the following open statements.
              c(x):  Student x is in the class ( that is , Proffessor Carlson’s mechanics as already described ).
             j(x): Student x is a junior.
             s(x): Student x is a senior.
            g(x): Student x is a graduate student.
            p(x): Student x is a physics major.
            e(x): Student x is an electrical engineering major.
           m(x): Student x is a mathematics major.

Write each of the  following statements in terms of quantifiers and the open statements c(x),j(x),s(x),g(x),p(x),e(x) and m(x) , and determine whether the given statement is ttrue or false. Here the universe comprise of all  12,500 students enrolled at the universitywhere Proffessor Carlson teaches. Furthermore, at this university each studentt haas only one major.


a)      There is a mathemtics major in the class who is a junior.
b)      There is a senior in the class who is not a mathematics major.
c)      Every student in the class is majoring in mathematics.
d)      No graduate student in the class is a physics maajor.
e)      Every senior in the class is majoring in either physics or electrical engineering.
SOLUTION:
a)    ∃x [m(x) ^ c(x) ^j(x) ]   ------------------------TRUE
b)   ∃x[s(x)^c(x)^~m(x)] --------------------------TRUE
c)    ∀x[c(x) →(m(x) p(x))] --------------------------FALSE
d)   ∀x[(g(x) ^c(x)) →~p(x)]--------------------------TRUE
e)    ∀x[(c(x) ^s(x)) →(p(x) e(x))]-----------------TRUE


Exercise-3.1:
12:
Let  S = {1,2,3,…………………………,29,30}.

How many subsets A of S satisfy
a)      |A|=5 ?
b)      |A|=5 and the smallest element in A is 5 ?
c)      |A|=5 and thee smallest element in A is less than 5 ?



SOLUTION:
a)    5 elements can be selected  from 30 elements in 30C5 ways.
Thus, 30C5 subsets of S satisfy the given condition.
b)    S – {1,2,3,4} ={5,6,7………………….,29,30} = B
|B| = 26

Sincce 5 is the smallest number in each case,it is required to select 4 numbers from the remaining 25 numbers.
Thus the 4 numbers  can bbe selected in 25C4 ways.
Thus, 25C4 subsets satisfy the given condition.
c)     Here the smallest element can be 1,2,3 or 4.
S-{5} = B
|B|=29
Now,
Case1:
Let us consider smallest element is 1.
Thus we need 4 more elements for the required subset.
It can be done in 29C4 ways.
Case2:
Let us consider smallest elemnt is 2.
Here 1 has to be omitted from B,  since smallest number in this case is 2..
Thus remaining 4 number can be selected in 28C4 ways.
Case 3:
Let us consider the smallest number is 3.
Thus the remaining 4 number can be selected in 27C4 ways.
Case4:
Let us consider the smallest number is 4.
Thus the remaining 4 numbers can be selected in 26C4ways.

Thus , total number of ways is=
Thus , total number of ways is=
29C4 +28C4 + 27C4 + 26C4ways. 
************************************************************************************************************
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,p2,p3,.........,pn denote n primitive statements. Let p be a compound statements that contains atleast one occurence each of pi, 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     pi   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 180O.
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 180o.         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 180o   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.............

                                                                    TEAM 10


1)Verify that
              [p->q->r)]  ->  [(p->q)->(p->r)]  is a tautology.

Soln:
P     q          r          [p->(q->r)]        [(p->q)->(p->r)]             [p->(q->r)]->[(p->q)->(p->r)]
1      1        1                1                                         1                                                              1
1      1        0                  0                                       0                                                              1
1      0        1                  1                                  1                                                  1
1      0        0                  1                                       1                                                              1
0      1        1                  1                                       1                                                              1
0      1        0                  1                                       1                                                              1
0      0        1                  1                                       1                                                              1
0      0        0                  1                                  1                                                        1



From the above truth table it is verified it is TAUTOLOGY.


1)      Determine whether each of the following is TRUE or FALSE.Here  p , q are arbitrary statements. 
a)      An equivalent way to express the converse of “p is sufficient for q”  is  “p is necessary for                                   q”.
b)      An equivalent way to express the inverse of “p is necessary for q”  is  “~q is sufficient for                                   ~p”.
c)       An equivalent way to express the contrapositive of “p is necessary for q”  is  “~q is necessary for ~p”.
         Soln:    As we know  pàq means that------------
                                p is a sufficient condition for q
                                q is necessary condition for p
                                q  is necessary for p
                                p is sufficient for q
                a)TRUE
                                p is sufficient for q       p->q
                                p is necessary for q      q->p    which is the converse of  p->q.
                b)TRUE
                                p is necessary for q    q->p
                                ~q is sufficient for ~p     ~p-> ~q  which is inverse of p->q
                c)TRUE
                                p is necessary for q     q->p
                                ~q is necessary for ~p    ~q-> ~p  which is contra positive of pàq
2)      Let P(x) be the open statement “ x2 = 2x”,where the universe comprises all integers.Determine whether each of the following statement is TRUE or FALSE.

a)      P(0)
P(x): x2 = 2x
P(0): 0 = 2*0
P(0):0 = 0
 Hence TRUE

b)      P(1)
P(x):x2 = 2x
P(1):1=2
FALSE
c)       P(2)
P(x):x2 = 2x
P(2):4=4
TRUE
d)      P(-2)
P(x):x2 = 2x
P(x):4=-4
FALSE

                e)For some x  p(x)
p(x):x2 =2x
Put x=0
0 = 0 for some P(x) is true.
TRUE


f)       For all  x, p(x)
    Put x=3  belongs to Z
 Hence P(x) becomes
 (3)2= 2*3
9≠6

    FALSE

4)which of the following sets are nonempty?
                a)   {x|x ϵ N,2x+7=3}
                                consider:    2x+7=3;
                                                        2x =3-7
                                                         x=-2
                                since N is a natural number here no value of x satisfy the above condition.
                                Hence EMPTY SET.
                b)   {x ϵ Z | 3x + 5 =9}
                                consider: 3x + 5 =9;
                                                      x = 4/3
since Z is set of integers here no value of x which satisfy the above condition which is integer.Hence EMPTY SET.
c)       {x ϵ Q ,x2 + 4 = 6}
   x2 + 4 =6
   X2=6 – 4
X2 = 2
X =sqrt 2.
Since Q is set of rational nos and sqrt 2 is a irrational no.
Hence EMPTY SET.
d)      {x ϵ R | x2 + 4 = 6 } 
According to previous example
X = sqrt 2.
        Since R is set of real nos and sqrt 2 is a real no.
        Hence NON EMPTY SET.
e)      {x ϵ R | x2 + 3x +3 = 0}
                                  X2 + 3X + 3 = 0
     
     
X = complex roots
Since R is set of real nos.
Hence EMPTY SET.
f)       {x ϵ C | x2 + 3x +3 = 0}
     X2 + 3X + 3 = 0
         
X2 + 3X + 3 = 0
X = complex roots
Since C is set of complex nos.
Hence NON EMPTY SET.