document.write( "Question 1163527: An indicator function IA satisfies the condition
\n" ); document.write( "IA=0(X does not belong to A)
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\n" ); document.write( "I（A Union B）=Max(IA,IB)=IA+IB-I（A intersection B） \n" ); document.write( "
Algebra.Com's Answer #787681 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Case I:  I(A U B) = 0\r\n" );
document.write( "If an element is not in the union of two sets, then it is not in either set, hence:\r\n" );
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document.write( "IA = 0, IB = 0, I(A ∩ B) = 0, IA + IB - I(A ∩ B) = 0, and Max(IA,IB) = 0,\r\n" );
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document.write( "therefore I(A U B) = Max(IA,IB) = IA + IB - I(A ∩ B)\r\n" );
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document.write( "Case II: I(A U B) = 1\r\n" );
document.write( "If an element is in the union of two sets, then it is in one or the other or both sets\r\n" );
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document.write( "Case IIa: IA = 1, IB = 0 => I(A ∩ B) = 0, so Max(IA,IB) = IA = 1, IA + IB - I(A ∩ B) = 1 + 0 - 0 = 1\r\n" );
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document.write( "Case IIb: IA = 0, IB = 1 => I(A ∩ B) = 0, so Max(IA,IB) = IB = 1, IA + IB - I(A ∩ B) = 0 + 1 - 0 = 1\r\n" );
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document.write( "Case IIc: IA = 1, IB = 1 => I(A ∩ B) = 1, so Max(IA,IB) = IA or IB = 1, IA + IB - I(A ∩ B) = 1 + 1 - 1 = 1\r\n" );
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document.write( "Q.E.D by exhaustion.\r\n" );
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\n" ); document.write( "John
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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