Question 48331
DEFINITELY DEFINITELY NOT!!!!!!!!


{{{ (sqrt(3))/(sqrt(3) + sqrt(2)) }}} is treated just like {{{ (3)/(3+2) }}} which would be 3/5. You would not say cancel the 3's leaving 1/2.


Treat it as any normal fraction and apply Rules of Fractions and not your own made up rules :-)


Now, with radicals (or surds) ie square roots, we tend to want no square roots on the denominator. Why? because that is the preferred way of writing radical fractions as far as i understand.


So we need to convert the {{{sqrt(3)+sqrt(2)}}} to a rational number. We do this by multiplying by 1 (so the thing we started with is unchanged but we write 1 in this example as {{{ (sqrt(3)-sqrt(2))/(sqrt(3)-sqrt(2)) }}}


So, we have {{{ ((sqrt(3))/(sqrt(3)+sqrt(2)))*((sqrt(3)-sqrt(2))/(sqrt(3)-sqrt(2))) }}}


which is {{{ (sqrt(3)(sqrt(3)-sqrt(2))/((sqrt(3)+sqrt(2))(sqrt(3)-sqrt(2)))) }}}


Now what is the point of doing this? Well multiply out the terms on the denominator and they will simplify to 3-2 which in this case is a lovely 1.


{{{ sqrt(3)(sqrt(3)-sqrt(2))/1 }}}
{{{ sqrt(3)(sqrt(3)-sqrt(2)) }}}
{{{ sqrt(3)sqrt(3)-sqrt(3)sqrt(2) }}}
{{{ 3-sqrt(3)sqrt(2) }}}
{{{ 3-sqrt(6) }}}


As hopefully you can appreciate that this is somewhat simpler than what we started with.


Use your calculator to work out the value of both the original and my answer to show yourself that they are the same value.


Jon.