1. Prove that a finite integral domain is a field.
Answer
We know that in the integral domain we have ab=0=>a=0 or b=0. Now it suffices to show that every
non-zero element has a multiplicative inverse. Let D be an integral domain. Now we show
1) There exists 1∈D such that a.1 = a for all a∈D,
2) 0≠a∈D => there exists b∈D such that ab=1.
Let D={x1,x2,…….,xn} and 0≠a∈D. Now x1a, x2a,…,xna are all distinct ( if xia=xja, then (xi-
xj)a=0=>xi-xj=0=>xi=xj(since a≠0)).
Therefore D={x1a,x2a,….,xna}. Since a∈D, a= xka for some 1<=k<=n.
Again since D is commutative, we have xka=a=axk.
We show xk is the identity element. For this, let y∈D then y=xia for some i.
Now consider u.xk=(xia)xk=xi(axk)=xia=y.
Thus yxk=y for all y∈D, therefore xk is the identity element.
For xk∈D={x1a,x2a,…,xna}=>xk=xja for some 1<=j<=n.
Therefore xj is the multiplicative inverse of a. hence D is a field.
Prove that “If S is any subset of a vector space V, then L(S) is a subspace of V”.
Answer
Notation used:
A= alpha
B= Beta
Let V,W∈ L(S) and A,B∈F.
Since V,W∈ L(S), we have that
V=A1V1+…+AnVn and W=B1W1+…..+BmWm for some Vi∈S,Ai∈F for 1<=i<=n and
Wj∈S, Bj∈F for 1<=j<=m.
Now
AV+BW=A(A1V1+….+AnVn)+B(B1W1+…..+BmWm)=AA1V1+AA2V2+…+AANVN+BB1W1+BB
2W2+…+BBMWM , is a linear combination of elements from S.
Hence AV+BV∈L(S). this shows that L(S) is a subspace of V.

2. Show that [P ^ (P v Q)] v ~P is a Tautology.
Answer
P Q PvQ P^(PvQ) ~P [P^(PvQ)]v~P
0 0 0 0 1 1
0 1 1 0 1 1
1 0 1 1 0 1
1 1 1 1 0 1
Hence from above we can conclude that [P ^ (P v Q)] v ~P is always TRUE . Hence it is TAUTOLOGY.
Convert the following statements into symbolic language
(i) All beautiful birds are ornately colored.
(ii) Birds that do not live on honey are dull in color.
Answer
Let
B(x): “x is beautiful”
D(x): x is a bird
O(x): x is ornately colored
H(x): x lives on honey
1) Symbol for “x is beautiful bird” is D(x) ^ B(x).
The statement “All beautiful birds are ornately colored” , is equivalent to
(x)((D(x)^B(x))O(x)) in symbolic way.
2) Given statement is “Birds that do not live on honey are dull in color”
~H(x) , denotes “ x do not have on honey”
Therefore D(x)^~H(x) denotes that “x is a bird and it does not live on honey”.
Also ~O(x) denotes “x is dull in color”.
Therefore the given statement can be written in symbols as
(x)(D(x)^~H(x)~O(x))

3. If A = {2, 3, 4}, B = {4, 5, 6} and C = {6, 7}, then find the following
(i) (A – B) X (B – C)
(ii) (A X B) – (B X C)
Answer
1) (A-B) = {2,3,4}-{4,5,6}
= {2, 3}
(B-C) = {4, 5, 6}-{6, 7}
= {4, 5}
(A – B) X (B – C) = {2,3} X {4,5}
= {(2,4),(2,5),(3,4),(3,5)}
2) A X B = {(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,4),(4,5),(4,6)}
B X C = {(4,6),(4,7),(5,6),(5,7),(6,6),(6,7)}
(A X B) – (B X C) = {(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,4),(4,5)}

4. Find the domain and range of the function x = 1/ 2-cos3x
Answer
Notation used
Theta = @
Domain:
Since cos@ lies between -1 and +1.
i.e -1<=cos@<=1, we have
-1 <= cos3x <= 1, for @= 3x
2-(-1)>=2-cos3x>= 2-1
 3>= 2-cos3x >= 1
1<= 2-cos3x <= 3 for all real x
Hence f(x) is defined for all x ∈ R
Hence domain of f = R
Range:
Let f(x) = y 1/ 2=cos3x
=y2-cos3x = 1/y
cos3x = 2- 1/y
But -1<= cos3x <= 1
Therefore -1 <= 2 – 1/y <= 1
-1-2 <= - 1/y <= 1-2
-3 <= - 1/y <= -1
3 >= 1/y >= 1
1/3 <= y <= 1
y ∈ [1/3 , 1]
So range of f = [1/3 , 1]