Kenneth Rosen Edition 6th Exercise 2.1 Question 31 (Page No. 120)

Question

31 Explain why A X B X C and ( A X B ) X C are not same.

Comments

What is A, B and C here?

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These three are sets

Answer 1

Well you can not say $( (a,b),c)$ == $(a,b,c)$, These are not same. I am taking $ * = \times $

$A * B * C$ will generate the set of type $(a,b,c)$ element.

$(A * B)  * C$ will generate the set of type $((a,b),c)$ element.

And the biggest difference is that $A * B * C$ is a triplet cartesion product while $(A * B) * C$ is a binary cartesion product. 

Hence both are not same.

Comments

is there any relation between A X B X C and( A XB )X C?

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No, in the first AxBxC its triplet caetesion operation, while in (AxB)xC) its Binary catesion operation.

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Lets take an eg. A={1,2,3} B={a,b} C={y,z}

Then what will be those two products?

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AxBxC = {(1,a,y),(2,a,y),(3,a,y),.....}

(AxB)xC = { ((1,a),y),((2,a),y), ((3,a),y)... }

Do not think like that.... you have to understand that 

(a,b,c) is not equal to ((a,b),c) because first one is the result of triplet cartesion product, and second one is the result of binary cartesion product.

See I am having international edition of rosen, it has example just above the excercise. You should also check that.. you may see that. 

Answer 2

Let A={1,2} , B=$\phi$, C={3,4}

$(A \times B ) \times C= \phi \times C=\phi$

$A \times (B \times C)= A \times \phi=\phi$

$(A \times B)\times C= A \times (B \times C)$

Now let's take another Example

A={1} , B={2} , C={3}

$(A \times B ) \times C=$ { ((1,2),3) }

$A \times( B \times C)=${ (1,(2,3) ) }

$A \times( B \times C)$ ≠ $(A \times B ) \times C$

Hence we can conclude

The Cartesian product is not associative unless one of the involved sets is empty.

$A \times( B \times C)$ ≠ $(A \times B ) \times C$

"Hence Proved".