document.write( "Question 992569: Proof of 2=1
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\n" ); document.write( "a = b
\n" ); document.write( "a2 = ab
\n" ); document.write( "a2 - b2 = ab - b2
\n" ); document.write( "(a-b)(a+b) = b(a-b)
\n" ); document.write( "a+b = b
\n" ); document.write( "b+b = b
\n" ); document.write( "2b = b
\n" ); document.write( "2 = 1
\n" ); document.write( "Can you put these statements in order and give the reasons? \n" ); document.write( "
Algebra.Com's Answer #612151 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "a = b               <--a and b can stand for the same quantity. \r\n" );
document.write( "aČ = ab             <--we can multiply both sides by a.  \r\n" );
document.write( "aČ - bČ = ab - b2   <--we can subtract bČ from both sides.\r\n" );
document.write( "(a-b)(a+b) = b(a-b) <--we factor both sides of the equation.\r\n" );
document.write( "a+b = b             <--we divide both side by (a-b)\r\n" );
document.write( "b+b = b             <--since a = b, we can substitute b for a\r\n" );
document.write( "2b = b              <--combine like terms b+b and get 2b\r\n" );
document.write( "2 = 1               <--we divide both sides by b \r\n" );
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document.write( "The fallacy is in the step colored red. Since a=b, (a-b)=0 and\r\n" );
document.write( "we may never divide by 0, even when it's camouflaged as (a-b).\r\n" );
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document.write( "Edwin
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