Question 143431
The process involves multiplying your fraction expression by 1 in a form such that when the denominators are multiplied, the denominator of the result becomes a rational number -- that is to say you have rid yourself of the radical.  The process is also known as "Get that pesky radical OUT of my denominator"


Example 1:


{{{1/sqrt(2)}}}.


The way to make the denominator rational is to multiply it by {{{sqrt(2)}}}, but we aren't allowed to change the value of the fraction.  Fortunately, {{{a*1=a}}} no matter what {{{a}}} is and {{{a/a=1}}} no matter what {{{a}}} is (as long as it isn't zero), so multiplying {{{1/sqrt(2)}}} by {{{sqrt(2)/sqrt(2)}}} (which is just another way to write 1) is allowed.


{{{(1/sqrt(2))(sqrt(2)/sqrt(2))=sqrt(2)/2}}}


Example 2:


{{{1/root(3,2)}}}


This one is a little tricker.  Here we need to multiply the denominator by {{{(root(3,2))^2}}}.  Again, we have to multiply by 1:


{{{(1/root(3,2))(root(3,2)^2/root(3,2)^2)=(root(3,2)^2)/root(3,2)^3=root(3,2)^2/2}}}


Example 3:


{{{(1+sqrt(5))/(2-sqrt(2))}}}


This one is a little trickier still.  We need to take advantage of the 'difference of two squares' factorization, that is: {{{a^2-b^2=(a-b)(a+b)}}}


If we multiply the denominator by what is called its conjugate, {{{2 + sqrt(2)}}} (notice the sign change), then the result will be the difference of the first term squared and the second term squared with no annoying center term containing a radical.  Again, and as always, we have to multiply by 1.


{{{((1+sqrt(5))/(2-sqrt(2)))((2+sqrt(2))/(2+sqrt(2)))=(2+sqrt(2)+2sqrt(5)+sqrt(10))/(4-2)=(2+sqrt(2)+2sqrt(5)+sqrt(10))/2}}}


Ewww! That's a mess, but at least the denominator is rational.  You might want to rephrase the alternate definition I gave earlier for the process to: "Get that pesky radical OUT of my denominator and I don't care how big a mess you make in the numerator"


Write back if you have any questions.