document.write( "Question 1107353: Explain why 2 + √3 is an irrational number. \n" ); document.write( "
Algebra.Com's Answer #722382 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "Proof By Contradiction:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Let p and q be two integers where q is nonzero. Any rational number is of the form \"p%2Fq\" (ratio of two integers)\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "If \"2%2Bsqrt%283%29\" was rational, then \"2%2Bsqrt%283%29=p%2Fq\" for a certain (p,q) value. \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Multiplying both sides by q leads us to \"q%282%2Bsqrt%283%29%29+=+p\" which becomes \"2q%2Bq%2Asqrt%283%29+=+p\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "The right side of \"2q%2Bq%2Asqrt%283%29+=+p\" is an integer (p is an integer)\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "The left side of \"2q%2Bq%2Asqrt%283%29+=+p\" should be an integer as well; however, it is not an integer. The expression \"2q\" is surely an integer because 2 times any integer is an integer, but the expression \"q%2Asqrt%283%29\" is NOT an integer. It is only an integer if q was of the form \"k%2Asqrt%283%29\", but it is not. Again, q is an integer with no irrational parts. \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "In short, \"2q%2Bq%2Asqrt%283%29\" on the left side is NOT an integer while \"p\" on the right side is an integer.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "So this is where the contradiction lies. This contradiction makes the claim \"2%2Bsqrt%283%29=p%2Fq\" to be false, therefore \"2%2Bsqrt%283%29\" is not rational. The only thing left is that \"2%2Bsqrt%283%29\" must be irrational.
\n" ); document.write( "
\n" ); document.write( "
\n" );