# Fractions with roots: voodoo

## Main Question or Discussion Point

Starting with simple fractions, it's known that:

$${{{a \over b}} \over {{c \over d}}} = {{ad} \over {bc}}$$

So when b == d:

$${{{a \over b}} \over {{c \over b}}} = {a \over c}$$

But what if in the case of:

$${{{{1 + \sqrt 2 } \over {\sqrt 2 }}} \over {{{1 - \sqrt 2 } \over {\sqrt 2 }}}}$$

Following the above (for abc), it should result in:

$${{1 + \sqrt 2 } \over {1 - \sqrt 2 }}$$

But if you follow the abcd version:

$${{\left( {1 + \sqrt 2 } \right)\sqrt 2 } \over {\sqrt 2 \left( {1 - \sqrt 2 } \right)}} = {{2 + \sqrt 2 } \over {2 - \sqrt 2 }}$$

...which is not equal to the first expression.

Why?

$${{2 + \sqrt 2 } \over {2 - \sqrt 2 }}$$

...which is not equal to the first expression.

Why?
Are you sure that you did the calculations right, at frist place?????
Check your work, you did not distrubute correctly the $$\sqrt{2}$$

remember $$\sqrt{2}*\sqrt{2}=2 \ \ \ and \ \ \ not \ \ \ \sqrt{2}$$

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Are you sure that you did the calculations right, at frist place?????
Check your work, you did not distrubute correctly the $$\sqrt{2}$$

remember $$\sqrt{2}*\sqrt{2}=2 \ \ \ and \ \ \ not \ \ \ \sqrt{2}$$
Yes.

$$\displaylines{ \sqrt 2 \left( {1 + \sqrt 2 } \right) = \left( {\sqrt 2 \times 1} \right) + \left( {\sqrt 2 \times \sqrt 2 } \right) \cr = \sqrt 2 + 2 \cr}$$

Check the denominator, too.

You got the terms reversed in the demoninator: $${{\left( {1 + \sqrt 2 } \right)\sqrt 2 } \over {\sqrt 2 \left( {1 - \sqrt 2 } \right)}} = {{2 + \sqrt 2 } \over {2 - \sqrt 2 }}$$

Denominator should be $$\sqrt{2}-2$$

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You got the terms reversed in the demoninator: $${{\left( {1 + \sqrt 2 } \right)\sqrt 2 } \over {\sqrt 2 \left( {1 - \sqrt 2 } \right)}} = {{2 + \sqrt 2 } \over {2 - \sqrt 2 }}$$

Denominator should be $$\sqrt{2}-2$$
Ah, that solves it then. Thanks.

As an aside, how can I make this:

$${{dy} \over {dx}} = {{2y - x^2 } \over {y^2 - 2x}}$$

into this:

$${{dy} \over {dx}} = {{x^2 -2y } \over {2x - y^2}}$$

They're meant to be identical, but I can't think how.

Thanks!

Ah, that solves it then. Thanks.

As an aside, how can I make this:

$${{dy} \over {dx}} = {{2y - x^2 } \over {y^2 - 2x}}$$

into this:

$${{dy} \over {dx}} = {{x^2 -2y } \over {2x - y^2}}$$

They're meant to be identical, but I can't think how.

Thanks!
multiply the top and the bottom by -1.

You can do this because in essence it is multiplying the entire expression by 1.

Ah, that solves it then. Thanks.

As an aside, how can I make this:

$${{dy} \over {dx}} = {{2y - x^2 } \over {y^2 - 2x}}$$

into this:

$${{dy} \over {dx}} = {{x^2 -2y } \over {2x - y^2}}$$

They're meant to be identical, but I can't think how.

Thanks!
Are you working on diff. eq. or someone just gave u that expression. Don't tell me that you are!

Are you working on diff. eq. or someone just gave u that expression. Don't tell me that you are!
It's part of a larger exercise on a differential equation. Just a simple example for an implicit function.

It's part of a larger exercise on a differential equation. Just a simple example for an implicit function.
This was someting that i really was affraid i would hear. Damn.!!