Question 91198
The previous solution is incorrect. Let me correct it:

Assume {{{ root( 3, 7 ) }}} is rational. Then there exists integers {{{c}}} and {{{d}}} such that {{{c/d = root( 3, 7) }}}. Then by Zorn's lemma, there exists a greatest common divisor of {{{c}}} and {{{d}}}, say it is {{{g}}}. Set {{{a}}} such that {{{a*g = c}}} and {{{b}}} such that {{{b*g = d}}}. Then by definition of the rationals, {{{ c/d = (ga)/(gb) = a/b = root(3,7)}}}. But then {{{7}}} divides {{{ a^3 }}}. Then by definition of the rationals, {{{ a/b = root(3,7)}}}.  This is true since if we assume that {{{7}}} divides {{{a}}}; then we can use the unique factorization theorem to conclude that {{{ a = a_1^(m_1)*a_2^(m_2)*...*a_k^(m_k) }}} where all of the {{{a_i}}} are prime and none of them are 7. Thus {{{7}}} divides {{{a^(3) = a_1^(3*m_1)*a_2^(3*m_2)*...*a_k^(3*m_k) }}}  which is a contradiction since it obviously has no factors which are {{{7}}} since {{{7}}} is prime. Thus we have that {{{7}}} divides {{{ a }}}. Since {{{7}}} divides {{{a}}}, there exists {{{k}}} such that {{{ a = 7*k }}}, thus {{{ 7^(2) * k^(3) = b^(3) }}}. Thus by similar logic to the above, we have {{{ 7 ^(2) }}} divides {{{b}}}. Since {{{7^(2)}}} divides {{{b}}} we can use the prime factorization theorem to once again conclude that {{{7}}} divides {{{b}}}. But then {{{a}}} and {{{b}}} must have a common factor. This is a contraction. Thus {{{root(3,7)}}} is irrational.