document.write( "Question 424968: prove that 1+1 is equal to 2 \n" ); document.write( "
Algebra.Com's Answer #296100 by MathLover1(20849)\"\" \"About 
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\n" ); document.write( "A few simple definitions in Peano terms. 1 is the successor of 0, or S0. 2 is the successor of 1, or so on\r
\n" ); document.write( "\n" ); document.write( "Any equation is equal to itself, so we can start with S0 + S0 = S0 + S0 (1 + 1 = 1 + 1)\r
\n" ); document.write( "\n" ); document.write( "Any expression of the form A + SB is equivalent to SA + B, so S0 + S0 is equivalent to SS0 + 0. Thusly we can prove that (\"1+%2B+1+=+2+%2B+0\")\r
\n" ); document.write( "\n" ); document.write( "Since the expressions X + 0 and X are of equal value (by the Peano definition of addition,) SS0 + 0 can be reduced to SS0. Thus, we have proved that 1 + 1 = 2 according to Peano's axioms. (I think... hopefully I haven't skipped over anything or taken an illegal shortcut. It's been a while since I've worked with Peano.)\r
\n" ); document.write( "\n" ); document.write( "Algebraically, we can approach the problem somewhat simpler:\r
\n" ); document.write( "\n" ); document.write( "Take \"1+%2B+1+=+2\"\r
\n" ); document.write( "\n" ); document.write( "subtract \"1\" from both sides, and you see that statement is equivalent to \"1+%2B+1+-+1+=+2+-+1\"\r
\n" ); document.write( "\n" ); document.write( "Simplifying both sides, you can get \"1+=+1\", which is pretty self-evident.\r
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