Question 1190710
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Here's the original expression {{{a/(b+sqrt(c)) + d/sqrt(c)}}} in which we are told is rational. 
Any rational number is of the form p/q where p,q are integers and q is nonzero.


Let's rationalize each denominator.
For the first denominator, we'll have to multiply and bottom of the first fraction by {{{b-sqrt(c)}}
{{{a/(b+sqrt(c)) + d/sqrt(c)}}}


{{{(a(b-sqrt(c)))/((b+sqrt(c))(b-sqrt(c))) + d/sqrt(c)}}}


{{{(a(b-sqrt(c)))/(b^2-c) + d/sqrt(c)}}}


Next, we'll rationalize the denominator of the second fraction
{{{(a(b-sqrt(c)))/(b^2-c) + d/sqrt(c)}}}


{{{(a(b-sqrt(c)))/(b^2-c) + (d*sqrt(c))/(sqrt(c)*sqrt(c))}}}


{{{(a(b-sqrt(c)))/(b^2-c) + (d*sqrt(c))/(c)}}}


At this point, both fractions have rational denominators. 
Unfortunately the denominators are different, so we need to get each fraction to the LCD. In this case, the LCD is c(b^2-c)


{{{(a(b-sqrt(c)))/(b^2-c) + (d*sqrt(c))/(c)}}}


{{{(ac(b-sqrt(c)))/(c(b^2-c)) + (d*sqrt(c)*(b^2-c))/(c(b^2-c))}}}


Now we can combine the fractions


{{{(ac(b-sqrt(c)))/(c(b^2-c)) + (d*sqrt(c)*(b^2-c))/(c(b^2-c))}}}


{{{(ac(b-sqrt(c))+d*sqrt(c)*(b^2-c))/(c(b^2-c))}}}


We have one big fraction of the form M/N
{{{M = ac(b-sqrt(c))+d*sqrt(c)*(b^2-c)}}}


{{{N = c(b^2-c)}}}


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Let's focus solely on the numerator for now. I'll come back to the denominator later.


Let's rearrange terms like so
{{{ac(b-sqrt(c))+d*sqrt(c)*(b^2-c)}}}


{{{abc-ac*sqrt(c)+b^2*d*sqrt(c)-cd*sqrt(c)}}}


{{{abc+(-ac*sqrt(c)+b^2*d*sqrt(c)-cd*sqrt(c))}}}


{{{abc+(-ac+b^2*d-cd)*sqrt(c)}}}


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The conclusion of the previous section means this messy fraction 
{{{(ac(b-sqrt(c))+d*sqrt(c)*(b^2-c))/(c(b^2-c))}}}
turns into
{{{(abc+(-ac+b^2*d-cd)*sqrt(c))/(c(b^2-c))}}}


The denominator is an integer (I'm skipping steps a bit but it's probably fairly obvious c(b^2-c) is an integer), so we must have the numerator be an integer as well. 
Otherwise, the entire object isn't a rational number.


To ensure that the numerator {{{abc+(-ac+b^2*d-cd)*sqrt(c)}}} is an integer, we need to have the two pieces {{{abc}}} and {{{(-ac+b^2*d-cd)*sqrt(c)}}} be integers themselves.
Clearly abc is an integer, so there's no need to worry about that too much.


{{{(-ac+b^2*d-cd)*sqrt(c)}}} is only an integer when the coefficient (-ac+b^2*d-cd) is zero.
If it were nonzero, then the sqrt(c) part will mean we have something irrational (since c is a non-perfect square) and thereby breaking the entire object from being rational.


Set that coefficient mentioned equal to zero and rearrange the terms to get what we want like so
{{{-ac+b^2*d-cd = 0}}}


{{{b^2*d = ac+cd}}}


{{{b^2*d = c(a+d)}}}
This concludes the proof.
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