document.write( "Question 708870: prove that root of three is irrational
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Algebra.Com's Answer #436292 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "Assume \r
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\n" ); document.write( "\n" ); document.write( "If is odd, then is odd because the product of any two odd numbers is odd. Similarly, must then also be odd. Consequently, is odd, and since by definition an odd number has no even factor, must be odd.\r
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\n" ); document.write( "\n" ); document.write( "If is even, then , , and must be even. But if and are both even, then is not irreducible. Consequently, and must both be odd.\r
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\n" ); document.write( "\n" ); document.write( "If is the th odd number, . Likewise, can be represented as \r
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\n" ); document.write( "\n" ); document.write( "Go back to and substitute:\r
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\n" ); document.write( "\n" ); document.write( "A little algebra (verification left as an exercise for the student) gives us:\r
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\n" ); document.write( "\n" ); document.write( "Since the LHS is clearly odd and the RHS is clearly even, this equation has no solutions over the integers. Therefore, reductio ad absurdum, the square root of 3 cannot be expressed as a rational number, hence is irrational. \r
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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "Egw to Beta kai to Sigma
\n" ); document.write( "My calculator said it, I believe it, that settles it
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\"The

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