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forced common factor

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Given the fraction $\\dfrac{\\var{questnum}}{\\var{questdenmult}\\sqrt{\\var{questdensurd}}}$ we can rationalise the denominator and rewrite the fraction in the simplified equivalent form 

\n
\n\n\n\n\n\n\n\n\n\n\n
[[0]]   $\\,\\,\\,\\,$[[1]]
[[2]]
\n
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In the above case, to rationalise the denominator we can multiply the top and bottom of the fraction by the surd part. This will rationalise the denominator since $\\sqrt{a}\\times\\sqrt{a}=a$. 

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\dfrac{\\var{questnum}}{\\var{questdenmult}\\sqrt{\\var{questdensurd}}}$$=$$\\dfrac{\\var{questnum}}{\\var{questdenmult}\\sqrt{\\var{questdensurd}}}\\times\\dfrac{\\sqrt{\\var{questdensurd}}}{\\sqrt{\\var{questdensurd}}}$    (multiplying top and bottom by the surd part of the denominator)
$=$$\\dfrac{\\var{questnum}\\sqrt{\\var{questdensurd}}}{\\var{questdenmult}\\times\\var{questdensurd}}$    
$=$$\\dfrac{\\var{questnum}\\sqrt{\\var{questdensurd}}}{\\var{tempden}}$    
$\\,$$=$$\\dfrac{\\var{ansnummult}\\sqrt{\\var{questdensurd}}}{\\var{ansden}}$    (cancelling the common factor of $\\var{cancel}$)
$\\,$$=$$-\\dfrac{\\sqrt{\\var{questdensurd}}}{\\var{ansden}}$    (cancelling the common factor of $\\var{cancel}$)
$\\,$$=$$\\dfrac{\\sqrt{\\var{questdensurd}}}{\\var{ansden}}$    (cancelling the common factor of $\\var{cancel}$)
\n

\n

A nice result of this process is that $\\dfrac{1}{\\sqrt{n}}=\\dfrac{\\sqrt{n}}{n}$ for all $n>0$.

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