Answer to Question #125573 in Discrete Mathematics for kavee

Question

(i) Let A ={a , b , c}, B ={x , y}, and C ={0 , 1}. Find
(a) A×B×C
(b) C ×B×A
(c) C ×A×B
(d) B×B×B

(ii) Let A ={1 , 2 , 3 , 4 , 5}and B ={0 , 3 , 6}. Find
(a) A∪B
(b) A∩B
(c) A−B
(d) B−A

(iii) Let A ={a , b , c , d ,e}and B ={a , b , c , d , e , f , g , h}. Find
(a) A∪B
(b) A∩B
(c) A−B
(d) B−A

Answer 1

(i)

(a) The set A×B×C=$\bigcup_{r\in A, q\in B, s \in C} (r,q,s)$

There are 12 variants:

r=a,q=x,s=0

r=a,q=x,s=1

r=a,q=y,s=0

r=a,q=y,s=1

r=b,q=x,s=0

r=b,q=x,s=1

r=b,q=y,s=0

r=b,q=y,s=1

r=c,q=x,s=0

r=c,q=x,s=1

r=c,q=y,s=0

r=c,q=y,s=1

Then we obtain

A×B×C={(a,x,0),(a,x,1),(a,y,0),(a,y,1),(b,x,0),(b,x,1),(b,y,0),(b,y,1),(c,x,0),(c,x,1),(c,y,0),(c,y,1)}

(b) The set C×B×A=$\bigcup_{r\in C, q\in B, s \in A} (r,q,s)$

There are 12 variants:

r=0,q=x,s=a

r=0,q=x,s=b

r=0,q=x,s=c

r=0,q=y,s=a

r=0,q=y,s=b

r=0,q=y,s=c

r=1,q=x,s=a

r=1,q=x,s=b

r=1,q=x,s=c

r=1,q=y,s=a

r=1,q=y,s=b

r=1,q=y,s=c

Then we obtain

C×B×A={(0,x,a),(0,x,b),(0,x,c),(0,y,a),(0,y,b),(0,y,c),(1,x,a),(1,x,b),(1,x,c),(1,y,a),(1,y,b),(1,y,c)}

(c) The set CxAxB=$\bigcup_{r\in C, q\in A, s \in B} (r,q,s)$

There are 12 variants:

r=0,q=a,s=x

r=0,q=a,s=y

r=0,q=b,s=x

r=0,q=b,s=y

r=0,q=c,s=x

r=0,q=c,s=y

r=1,q=a,s=x

r=1,q=a,s=y

r=1,q=b,s=x

r=1,q=b,s=y

r=1,q=c,s=x

r=1,q=c,s=y

Then we obtain

C×A×B={(0,a,x),(0,a,y),(0,b,x),(0,b,y)(0,c,x),(0,c,y),(1,a,x),(1,a,y),(1,b,x),(1,b,y)(1,c,x),(1,c,y)}

(d) The set BxBxB=$\bigcup_{r\in B, q\in B, s \in B} (r,q,s)$

There are 8 variants:

r=x,q=x,s=x

r=x,q=x,s=y

r=x,q=y,s=x

r=x,q=y,s=y

r=y,q=x,s=x

r=y,q=x,s=y

r=y,q=y,s=x

r=y,q=y,s=y

Then we obtain

B×B×B={(x,x,x),(x,x,y),(x,y,x),(x,y,y),(y,x,x),(y,x,y),(y,y,x),(y,y,y)}

(ii)  Let A ={1 , 2 , 3 , 4 , 5} and B ={0 , 3 , 6}.

(a) The set A$\bigcup$B consists of elements that belong to the set A or to the set B:

0 belongs to B

1 belongs to A

2 belongs to A

3 belongs to A and to B

4 belongs to A

5 belongs to A

6 belongs to B

Then  A∪B={0,1,2,3,4,5,6}

(b) The set A$\bigcap$B consists of elements that belong to the set A and to the set B.

Only element 3 belongs to A and to B.

Then A∩B={3}

(c) The set A-B consists of elements that belong to the set A and not to the set B.

These are the elements 1,2,4 and 5

Then A−B={1,2,4,5}

(d) The set B-A consists of elements that belong to the set B and not to the set A.

These are the elements 0 and 6

Then A−B={0,6}

(iii)

 Let A ={a , b , c , d ,e} and B ={a , b , c , d , e , f , g , h}.

(a) The set AUB consists of elements that belong to the set A or to the set B:

a belongs to A and to B

b belongs to A and to B

c belongs to A and to B

d belongs to A and to B

e belongs to A and to B

f belongs to B

g belongs to B

h belongs to B

Then A∪B={a ,b, c, d, e, f, g, h}

(b) The set A⋂B consists of elements that belong to the set A and to the set B.

Only elements a, b, c, d and e belong to A and to B

Then A∩B={a, b, c, d, e}

(c) The set A-B consists of elements that belong to the set A and not to the set B.

Therу are no such elements.

Then A−B={}. In this case we can write A−B=$\varnothing$ too.

(d) The set B-A consists of elements that belong to the set B and not to the set A.

These are the elements f, g and h.

Then B-A={f,g,h}