2
$\begingroup$

How does one take radicals out of the denominator of

$$\frac{1}{a + b(\sqrt{2} + \sqrt{3})}$$

assuming $a,b \in \mathbb{Q}$ not both equal to $0$. I know the trick is to multiply this expression by $1$ s.t. the radicals are removed, but I can't think of such $c/d = 1$ to try.

$\endgroup$
6
$\begingroup$

Try

$$ \frac{ a + b \sqrt{2} - b \sqrt{3} } { a + b \sqrt{2} - b \sqrt{3} } \times \frac{ a - b \sqrt{2} + b \sqrt{3} } { a - b \sqrt{2} + b \sqrt{3} } \times \frac{ a - b \sqrt{2} - b \sqrt{3} } { a - b \sqrt{2} - b \sqrt{3} } .$$

The idea is that you have to multiply by all possible conjugates.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

Hint: Multiply numerator and denominator first by $a-b(\sqrt2+\sqrt3)$, to get in the denominator a term of form $c + d\sqrt6$. Now you know the next step.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

This is just like the usual method for rationalizing the denominator, only you need $4$ factors, not two, to get the rational number you are looking for. That is, multiply the top and the bottom by $a+b(\sqrt2-\sqrt3)$, then by $a+b(-\sqrt2+\sqrt3)$ then by $a+b(-\sqrt2-\sqrt3)$.

You can check directly that $$[a+b(\sqrt2+\sqrt3)][a+b(\sqrt2-\sqrt3)][a+b(-\sqrt2+\sqrt3)][a+b(-\sqrt2-\sqrt3)]$$ is a rational number.

Readers who have seen some Galois theory will recognize this as the norm in a Klein-4 extension.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.