As the title suggests, this section is about orderderd pairs(pairs where ordering is important) and cartisian product of sets.
Cartisian product of two sets A and B is defined as
A X B = {(a,b) | a A and b B}
where (a,b) is the ordered pair. Often a is called the first and b is called the second coordinate of the ordered pair.
One important property we might want to notice is that
A X = X A =
Now, for any statement P(x,y), where x and y are free variables and x ranges over a set A and y ranges over a set B. Then
Truth set(T) of P(x,y) would be defined as {(a,b) | P(a,b)}
clearly if some (a,b) T, then P(a,b)
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Ex-1(a) {(x,y) | x is a parent of y}
Ex-1(b) {(x,y) | someone lives in x and attends y}
Ex-2(a) {(x,y) | x lives in y} for example {(himanshu,bangalore), ...}
Ex-2(b) {(x,y) | population of x is y} for example
{(bangalore, 1 mn), ...}
Ex-3(a) {(x,y) | y = } for example {(0,-2), (1,-2),...}
Ex-3(b) {(x,y) | y < x} for example {(1,0), (2,1), (2,0), ...}
Ex-3(c) {(x,y) | either y = or y = 3x-2 } for example {(0,-2), (1,1), ...}
Ex-3(d) {(x,y) | y < x and either y = or y = 3x-2 }
Ex-4
1.
= {1,2,3} X {4}
= {(1,4), (2,4), (3,4)}
(A X B) (A X C)
= {(1,1), (2,1), (3,1), (1,4), (2,4), (3,4)} {(1,3), (2,3), (3,3), (1,4), (2,4), (3,4)}
= {(1,4), (2,4), (3,4)}
2.
= {1,2,3} X {1,3,4}
= {(1,1), (2,1), (3,1), (1,3), (2,3), (3,3), (1,4), (2,4), (3,4)}
(A X B) (A X C)
= {(1,1), (2,1), (3,1), (1,4), (2,4), (3,4)} {(1,3), (2,3), (3,3), (1,4), (2,4), (3,4)}
= {(1,1), (2,1), (3,1), (1,3), (2,3), (3,3), (1,4), (2,4), (3,4)}
3. (A X B) (C X D)
= {(1,1), (2,1), (3,1), (1,4), (2,4), (3,4)} {(3,5), (4,5)}
=
= {3} X
=
4. (A X B) (C X D)
= {(1,1), (2,1), (3,1), (1,4), (2,4), (3,4)} {(3,5), (4,5)}
= {(1,1), (2,1), (3,1), (1,4), (2,4), (3,4), (3,5), (4,5)}
= {1,2,3,4} X {1,4,5}
= {(1,1), (2,1), (3,1), (4,1), (1,4), (2,4), (3,4), (4,4), (1,5), (2,5), (3,5), (4,5)}
5. A X = X A =
Ex-5
-- Prove = (A X B) (A X C). --
Let p be arbitrary element of . It follows that there exist x and y s.t. p = (x,y) and and . Since , so either or . Let us consider both the cases.
Case#1:
Then . Thus
Case#2:
Then . Thus
Since p is arbitrary, we can conclude that
Now, Let p be arbitrary element of . Then either or . Let us consider both the cases
Case#1:
There exist and s.t. p = (x,y). clearly also, so
Case#2:
By similar argument as above,
Since p is arbitrary, we can conclude that
Thus = (A X B) (A X C)
-- Prove that --
Let p be an arbitrary element of . Then and . So there exists x and y s.t. p = (x,y) and and and and . So and . Hence . Since p is arbitrary, so
Let p be an arbitrary element of . It follows there exist x and y s.t. p = (x,y) and and . So and and and . So and . Thus . Since p is arbitrary, so
Hence
Ex-6 Flaw is that all the cases are not considered. Here is the list of cases that is left out in the proof.
and
and
Ex-7 m * n
Ex-8 It is true. Here is the proof.
Let p be an arbitrary element of (A X B\C). It follows there exist x and y s.t. p = (x,y) and and . So and . Since and , so . Since , so . Hence . Since p is arbitrary, so
Now let p be an arbitrary element of . Then and . It follows that there exist x and y s.t. p = (x,y). Since , so and . Since , so either or .. but we already saw that , so . Since , and . So . Since p is arbitrary, so
Thus
Ex-9 Let p be an arbitrary element of . Then and . Thus there exist x and y s.t. p = (x,y). Since , so and . Since , so either or . Let us consider both the cases.
Case#1:
Since and and , so . Thus
Case#2:
Since and and , so . Thus
Since p is arbitrary,
Again, Let p be an arbitrary element of . Then either or . Let us consider both the cases.
Case#1:
Then there exist x and y s.t. p = (x,y) and and and . Since and , so . Also since , so . Thus
Case#2:
we can use similar argument as above case to prove that
Now since p is arbitrary, so
Thus =
Ex-10 Suppose (A X B) and (C X D) are disjoint. Let (x,y) be an arbitrary ordered pair of (A X B), it follows that . So either or . Since x,y are arbitrary, Thus either A and C are disjoint or B and D are disjoint.
Ex-11
(a) Let (x,y) be an ordered pair in . Then there exists a particular s.t. . It follows that and . Since , so . Also since , so . So . Since x,y is arbitrary, so
(b) Let (x,y) be an arbitrary ordered pair in . Then there exists a particular ordered pair (l,m) in P s.t. . Since , So . Then and . Since , so . Also since , so . Thus . Since (x,y) is arbitrary, so
Again, Let (x,y) be and arbitrary ordered pair in . Then and . Since , so there exists a particular s.t. and similarly there exists a particular s.t. . So . Then . So . Since (x,y) is arbitrary, so .
Thus =
Ex-12 The proof and the theorem, both are incorrect. It starts with assuming elements in A and B. It is nowhere given that A and B, both are non-empty set.
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