Question 1023321
{{{I[1]=sqrt(6)/2}}} is an irrational number,
so {{{a-2b+sqrt(a+b)}}} must be irrational too.
Since {{{a}}} and {{{b}}} are rational,
{{{R=a-2b}}} must be rational.
Since {{{a-2b+sqrt(a+b)}}} is irrational, and
{{{R=a-2b}}} is rational,
{{{I[2]=sqrt(a+b)}}} must be irrational.
What we have in {{{a-2b+sqrt(a+b)=sqrt(6)/2}}} is
{{{R+I[2]=I[1]}}} .
The only way that can be possible is
{{{system(R=0,I[2]=I[1])}}} , which means
{{{system(a-2b=0,sqrt(a+b)=sqrt(6)/2)}}}-->{{{system(a=2b,a+b=6/4)}}}-->{{{system(a=2b,2b+b=3/2)}}}-->
{{{system(a=2b,3b=3/2)}}}-->{{{system(a=2b,b=1/2)}}}-->{{{highlight(system(a=1,b=1/2))}}}