document.write( "Question 1118504: If 0 < x < 1 , 0 < y < 1 prove that 0 < x + y - xy < 1 \n" ); document.write( "
Algebra.Com's Answer #733881 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "x and y are both positive and less than 1; that means the product xy is less than x and less than y.

\n" ); document.write( "So the expression x + y - xy is always positive; i.e., the inequality

\n" ); document.write( "\"0+%3C+x+%2B+y+-+xy\"

\n" ); document.write( "is always true.

\n" ); document.write( "To prove the other inequality

\n" ); document.write( "\"x+%2B+y+-+xy+%3C+1\"

\n" ); document.write( "rewrite the statement to be proved as

\n" ); document.write( "\"x+%2B+y+-+xy+-+1+%3C+0\"

\n" ); document.write( "Then
\n" ); document.write( "\"x%281-y%29-%281-y%29+%3C+0\"
\n" ); document.write( "\"%28x-1%29%281-y%29+%3C+0\"

\n" ); document.write( "Because x and y are both between 0 and 1, one of those factors is always negative and the other is always positive, so the product is always negative.

\n" ); document.write( "So the second inequality is also always true, making the original compound inequality always true.
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