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Question

Given a none-empty set X,let:P(X)×P(X)P(X) be defined as A×B=(AB)(BA),A,BP(X). Show that the empty set ϕ is the identity for the operation and all the elements A of P(X) are invertible with A1=A.

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Solution

Given that :P(X)×P(X)P(X) is defined as
AB=(AB)(BA) for A,BP(X)
Let AP(X), Then, we have
Aϕ=(Aϕ)(ϕA)=Aϕ=AϕA=(ϕA)(Aϕ)=ϕA=A
Therefore, Aϕ=A=ϕA for all AP(X)

Thus, ϕ is the identity element for the given operation .
Now, an element AP(X) wil be invertible, if there exists BP(X) such that AB=ϕ=BA. As ϕ is the identity element.
Now, we observed that AA=(AA)(AA)=ϕϕ=ϕAP(X).
Hence, all the elements A of P(X) are invertible with A1=A.


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