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Question: Let \(A = \\{ x \in W|x < 2\\} \), \(B = \\{ x \in N|1 < x \leqslant 4\\} \) and \(C = \\{ 3,5\\} \)...

Let A=xWx<2A = \\{ x \in W|x < 2\\} , B=xN1<x4B = \\{ x \in N|1 < x \leqslant 4\\} and C=3,5C = \\{ 3,5\\}
Verify that A×[BC]=[A×B][A×C]A \times [B \cup C] = [A \times B] \cup [A \times C]

Explanation

Solution

In order to solve such a problem, we must know that what A×BA \times B means and symbolises.
Here this refers to the Cartesian cross product and this includes all the sets of the ordered pairs. This will be clearer by an example. So let us take one example: let us suppose we have the elements in the sets A,BA,B where A=a,b,cA = \\{ a,b,c\\} and B=d,eB = \\{ d,e\\}
So A×B=(a,d),(a,e),(b,d),(b,e),(c,d),(c,e)A \times B = \\{ (a,d),(a,e),(b,d),(b,e),(c,d),(c,e)\\}
And we must also know that the number of elements in the Cartesian product is simply equal to the multiplication of the number of elements in the setAA with the number of elements in the setBB.
n(A×B)=n(A)×n(B)n(A \times B) = n(A) \times n(B)
And we must know the symbol ABA \cup B which includes all the elements of A,BA,B but if the same element is in both the sets then it is taken only once.

Complete step-by-step answer:
Here we are given in the question the three sets which are
A=xWx<2A = \\{ x \in W|x < 2\\}
B=xN1<x4B = \\{ x \in N|1 < x \leqslant 4\\}
C=3,5C = \\{ 3,5\\}
So we can get the elements of these sets as we are given that AA contains the whole number which are less than 22
So
A=0.1A = \\{ 0.1\\}
Set BBcontains the natural number which are greater than one but less than or equal to 44
B=2,3,4B = \\{ 2,3,4\\}
C=3,5C = \\{ 3,5\\}
So by the definition of the Cartesian product we get that
BC=2,3,4,5B \cup C = \\{ 2,3,4,5\\}
A×[BC]A \times [B \cup C] =0,1×2,3,4,5= \\{ 0,1\\} \times \\{ 2,3,4,5\\}
=(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)= \\{ (0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)\\}
So solving for [A×B][A×C][A \times B] \cup [A \times C]
[A×B][A \times B] =0,1×2,3,4= \\{ 0,1\\} \times \\{ 2,3,4\\}
=(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)= \\{ (0,2),(0,3),(0,4),(1,2),(1,3),(1,4)\\}
[A×C][A \times C] =0,1×3,5= \\{ 0,1\\} \times \\{ 3,5\\}
=(0,3),(0,5),(1,3),(1,5)= \\{ (0,3),(0,5),(1,3),(1,5)\\}
[A×B][A×C][A \times B] \cup [A \times C]
=(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)×= \\{ (0,2),(0,3),(0,4),(1,2),(1,3),(1,4)\\} \times (0,3),(0,5),(1,3),(1,5)\\{ (0,3),(0,5),(1,3),(1,5)\\}
The terms which are in both the sets are to be taken only one time according to the definition of union of the two sets.
[A×B][A×C][A \times B] \cup [A \times C] =(0,2),(0,3),(0,4),(1,2),(1,3),(1,4),(0,5),(1,5)= \\{ (0,2),(0,3),(0,4),(1,2),(1,3),(1,4),(0,5),(1,5)\\}
=(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)= \\{ (0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)\\}
=A×[BC]= A \times [B \cup C]

Hence we proved that A×[BC]A \times [B \cup C] =[A×B][A×C] = [A \times B] \cup [A \times C]

Note: Here in these type of questions we must take care that if the two terms are same in both the sets and we take the union of them then we need to take that term only once otherwise the answer may be incorrect and we must know what the following terms signifies in the sets like
AB AB AB BA  A \cup B \\\ A \cap B \\\ A - B \\\ B - A \\\
Because without the knowledge of these terms we will be unable to solve such type of problems in sets.