defining an operator - 2

• Oct 4th 2009, 09:49 AM
Jones
defining an operator - 2
Hi,

An operator has been defined on $(A \ast B) = (A\setminus{B})\cup(B\setminus{A})$

The part im having difficulties with is the associativity.

$A\ast(B \ast Z) = A\setminus(({B}\setminus{Z})\cup(Z\setminus{B}))\c up (B\setminus{Z})\cup (Z\setminus{B})\setminus{A}$

$(B\ast Z) = (B \setminus{Z})\cup(Z \setminus{B})$

$(A\ast B)\ast Z = (A \setminus{B})\cup(B\setminus{A})\setminus{Z}\cup Z \setminus{((A\setminus{B})\cup(B\setminus{A}))}$

According to my book the operator should be associative, but as far as i see they are not.
• Oct 4th 2009, 10:31 AM
Plato
Quote:

Originally Posted by Jones
An operator has been defined on $(A \ast B) = (A\setminus{B})\cup(B\setminus{A})$
The part im having difficulties with is the associativity.

This is a well known operator. It is usually called the symmetric difference.
And it is the associative property that almost every student hates to prove.
But the operation is indeed associative.

Here is an another and equivalent definition: $A*B = \left( {A \cup B} \right)\backslash \left( {A \cap B} \right)$