1. ## Surds problem

Hey can you guys help me out with this problem:

It is given that $\displaystyle \frac{a}{b+\sqrt{c}} + \frac{d}{\sqrt{c}}$ is rational, where a,b,c and d are positive integers and c is not a square. Show that as a consequence, $\displaystyle db^2 = c(a+d)$. Use this result to show that $\displaystyle \frac{a}{1+\sqrt{c}} + \frac{d}{\sqrt{c}}$ is not rational.

Thanks!

2. Hello, eskimogenius!

I have a truly "clunky" approach to the first part . . .

$\displaystyle \text{It is given that: }\:\dfrac{a}{b+\sqrt{c}} + \dfrac{d}{\sqrt{c}}\,\text{ is rational,}$
. . $\displaystyle \text{where }a,b,c,d\text{ are positive integers and }c\text{ is not a square.}$

$\displaystyle \text{Show that: }\:db^2 \,=\, c(a+d)$.

We have: .$\displaystyle \dfrac{a}{b +\sqrt{c}} + \dfrac{d}{\sqrt{c}} \;=\;\dfrac{(a+d)\sqrt{c} + db}{b\sqrt{c} + c}$

Long division:
. . $\displaystyle \begin{array}{ccccccc} &&&& \frac{a+d}{b} \\ && ----- & - & ----- \\ b\sqrt{c} + c & \bigg) & (a+d)\sqrt{c} & + & db \\ && (a+d)\sqrt{c} & + & \frac{c(a+d)}{b} \\ && ----- & - & ----- \\ &&&& db - \frac{c(a+d)}{b}\end{array}$

Since the division produces a rational answer, the remainder must be zero.

. . $\displaystyle db - \dfrac{c(a+d)}{b} \;=\;\dfrac{db^2 - c(a+d)}{b} \;=\;0$

Therefore: .$\displaystyle db^2 - c(a+d) \:=\:0$

3. Originally Posted by Soroban
Hello, eskimogenius!

I have a truly "clunky" approach to the first part . . .

We have: .$\displaystyle \dfrac{a}{b +\sqrt{c}} + \dfrac{d}{\sqrt{c}} \;=\;\dfrac{(a+d)\sqrt{c} + db}{b\sqrt{c} + c}$

Long division:
. . $\displaystyle \begin{array}{ccccccc} &&&& \frac{a+d}{b} \\ && ----- & - & ----- \\ b\sqrt{c} + c & \bigg) & (a+d)\sqrt{c} & + & db \\ && (a+d)\sqrt{c} & + & \frac{c(a+d)}{b} \\ && ----- & - & ----- \\ &&&& db - \frac{c(a+d)}{b}\end{array}$

Since the division produces a rational answer, the remainder must be zero.

. . $\displaystyle db - \dfrac{c(a+d)}{b} \;=\;\dfrac{db^2 - c(a+d)}{b} \;=\;0$

Therefore: .$\displaystyle db^2 - c(a+d) \:=\:0$

Wow, truly impressive. I have a question but, how did you know $\displaystyle \dfrac{a+d}{b}$ was the quotient?

Now, there's only the second part of the question

4. Never mind about that second part, my math teacher has solved it. Thank you very much Soroban.

For those who want to know the solution:

The first fraction that was given when compared to the second fraction: it can be seen that b = 1. Now for it to be true, the above proven equation has to hold.

When you sub b=1 into the proven equation above, it can be seen that $\displaystyle c = \dfrac{d}{a+d}$

However, as a, c and d are positive integers, that fraction has to be less than one, and hence this is impossible because c is a positive integer.

Therefore, the second fraction is never rational.

PS. By fraction I mean everything that is need to be proved irrational and everything that is need to be proved irrational.