document.write( "Question 156458: a@b = a+(b+1)\r
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Algebra.Com's Answer #115275 by Edwin McCravy(20081)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "a@b = a+(b+1) \r\n" );
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document.write( "Since ordinary addition is associative and commutative,\r\n" );
document.write( "we may write \r\n" );
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document.write( "a@b = a+(b+1) = a+b+1\r\n" );
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document.write( "The rule for @ is \"add and then add 1\"\r\n" );
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document.write( "We must prove that  \r\n" );
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document.write( "                 (a @ b) @ c = a @ (b @ c)\r\n" );
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document.write( "We start with the left side and end up with the right side:\r\n" );
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document.write( "                 (a @ b) @ c\r\n" );
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document.write( "Substitute a+b+1 for a @ b\r\n" );
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document.write( "                 (a+b+1) @ c\r\n" );
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document.write( "Add them and then add 1\r\n" );
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document.write( "                 (a+b+1)+c+1\r\n" );
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document.write( "                  a+b+1+c+1 \r\n" );
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document.write( "                  a+b+c+2 \r\n" );
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document.write( "                  a+b+c+1+1\r\n" );
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document.write( "Put in grouping symbols:\r\n" );
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document.write( "                  a + (b+c+1)+1\r\n" );
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document.write( "This is adding a and (b+c+1) and then adding 1, so that is:\r\n" );
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document.write( "                  a @ (b+c+1)\r\n" );
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document.write( "In the parentheses is adding b and c and then adding 1,\r\n" );
document.write( "so that is\r\n" );
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document.write( "                  a @ (b @ c)\r\n" );
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document.write( "So therefore\r\n" );
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document.write( "          (a @ b) @ c = a @ (b @ c)                 \r\n" );
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document.write( "Edwin
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