document.write( "Question 91198: Prove that cube root of 7 is an IRRATIONAL number. \n" ); document.write( "
Algebra.Com's Answer #66191 by stanbon(75887)\"\" \"About 
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Prove that cube root of 7 is an IRRATIONAL number ?
\n" ); document.write( "Assume that cube rt 7 is rational.
\n" ); document.write( "Then (7)^(1/3) = a/b where a and b are integers and a/b is reduced to lowest terms.
\n" ); document.write( "Then a=b[7^(1/3)]
\n" ); document.write( "Since a is a multiple of b and a is an integer, b divides a.
\n" ); document.write( "Since b divides a, a = nb and n is an integer.
\n" ); document.write( "Therefore 7^(1/3) = a/b = nb/b, so a/b is not reduced to lowest terms.
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\n" ); document.write( "What led to this contradiction?
\n" ); document.write( "The assumption that 7^(1/3) was rational.
\n" ); document.write( "The assumption must be wrong.
\n" ); document.write( "Therefore 7^(1/3) if irrational.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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