Question 293344
<pre>
how would you solve this proof?
{{{ (sqrt(6)-sqrt(2))/4 = sqrt(2-sqrt(3))/2 }}}

This is indeed a NESTED SURD. From the other person's response, let me pick up from this point:
{{{sqrt(8-4*sqrt(3)) = sqrt(6)-sqrt(2) }}} Start with the given equation.

Let's focus on the L.H.S. of the equation.
{{{sqrt(8 - 4sqrt(3))}}}
{{{sqrt(8 - 2*2sqrt(3))}}}
{{{sqrt(8 - 2sqrt(4)sqrt(3))}}} --- Converting 2 to {{{sqrt(4)}}}
{{{sqrt(8 - 2sqrt(4 * 3))}}}
{{{sqrt(8 - 2sqrt(12))}}}
{{{sqrt(6 + 2 - 2sqrt(6 * 2))}}} ---- Changing 8 to 6 + 2, and 12 to 6 * 2
{{{sqrt(6 + 2 - 2sqrt(6)sqrt(2))}}}
{{{sqrt((sqrt(6))^2 + (sqrt(2))^2 - 2sqrt(6)sqrt(2))}}} --- Converting 6 to {{{(sqrt(6))^2}}} and 2 to {{{(sqrt(2))^2}}}
The above is now in the form: {{{(a - b)^2}}}, with a being {{{sqrt(6)}}}, and b being {{{sqrt(2)}}}
So, {{{sqrt((sqrt(6))^2 + (sqrt(2))^2 - 2sqrt(6)sqrt(2))}}} now becomes: {{{sqrt((sqrt(6) - sqrt(2)))^2}}} 
                                              {{{sqrt(6) - sqrt(2)}}} ---- Cancelling square and sqrt
                                              {{{highlight(sqrt(6) - sqrt(2))}}} = {{{highlight(sqrt(6) - sqrt(2))}}}
As seen, L.H.S. = R.H.S.</pre>