# Surds problem

• March 6th 2011, 08:12 PM
eskimogenius
Surds problem
Hey can you guys help me out with this problem:

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

Thanks!
• March 7th 2011, 05:18 AM
Soroban
Hello, eskimogenius!

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

Quote:

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

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

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

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

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

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

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

• March 7th 2011, 01:13 PM
eskimogenius
Quote:

Originally Posted by Soroban
Hello, eskimogenius!

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

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

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

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

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

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

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

Now, there's only the second part of the question :D
• March 7th 2011, 11:28 PM
eskimogenius
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 $c = \dfrac{d}{a+d}$

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.