Can someone help me solve the problems?
I really dont have any idea of it, the question is in the following link
http://kentechnology.no-ip.org/maths.doc
thanks you
Can someone help me solve the problems?
I really dont have any idea of it, the question is in the following link
http://kentechnology.no-ip.org/maths.doc
thanks you
.......................question (4).........................................
we are given that:
1) f g, and
2) D(g) D(f) and asked to prove :
.................f=g.............................. .......
Since f g it suffices to prove:
g f,or ;
if (x,y)εg then (x,y)εf ..........( if (x,y) belongs to g then (x,y) belongs to f.
So let (x,y)εg........................................... ..................................1
But (x,y)εg =====> xεD(g) & yεR(g). Where D(g) means the domain of g and R(g) the range of g................................................. .................2
And due to (1) and (2) we have :...........xεD(g)................................ .3
But D(g) D(f), and because of (3).....xεD(f).....4
Now since f is a function we have:
for all x, xεD(f) there exists a zεR(f) such that (x,z)εf and because of (4) we have:
..............(x,z)εf............................. ............................................5
But, f g ,and due to (5).........(x,z)eg..............6
And now here is the crucial point about the proof.
Since g is a function:
(x,y)eg from (1) and (x,z)εg from (6) this implies that :
.......................y=z........................ .........................................7
Because no two ordered pairs belonging to g have the same 1st member.
And finally substituting (7) into (5) we have the desired result:
..............................(x,y)εf............. ..............................................8
Thus:.................f=g.....................
.....................QUESTION (2)...........................
Let us 1st show that :
p(A B) = P(A) P(B).
To show that we must show that:
Xε[p(A B)] <====> Χε[ P(A) P(B).].Where "ε" means belongs to.
Thus: Xε[p(A B)] <=====> X A B <======>
.........................(from the definition of power set)............................
<========> ...X A & X B <=======>
<=====>....Xε[P(A) P(B)]. Again from the definition of power set.
Hence ,we have proved Xε[p(A B)] <====> Χε[ P(A) P(B).] ,and :
p(A B) = P(A) P(B).
Note this double implication proof is two proofs at the same time ,one forward and at the same time the converse proof.
If one wishes can separate them by simply following the forward or backward arrows.
ALSO the center of the whole proof is the equivalence :
X A B <======>
<========> ...X A & X B
Now let us show that :
P(AUB) IS not equal in general to P(A)UP(B).
So , let A={1,2} AND B={3},and:
p(A)= P({1,2}) = { {1},{2},{1,2},Φ}, P(B) = P({3})={ {3},Φ },and
P(A)UP(B) = { {1},{2},{3},{1,2},Φ }.............................................1
But AUB = {1,2,3}, AND P(AUB) = {{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3},Φ}......... .................................................. .......................2
And...................(1)=/=(2).
BUT based on the fact that:
X A v X B =====>
X AUB, we can prove that:
P(A)UP(B) P(AUB).A fact that the above counter example confirms as well.