What Are Fractional Roots and Why Do They Confuse?

[W3bbo]

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?

 

[sutupidmath]

{{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}

 

[W3bbo]

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{<br /> \sqrt 2 \left( {1 + \sqrt 2 } \right) = \left( {\sqrt 2 \times 1} \right) + \left( {\sqrt 2 \times \sqrt 2 } \right) \cr <br /> = \sqrt 2 + 2 \cr}

 

[Pere Callahan]

Yes.

Check the denominator, too.

 

[robert Ihnot]

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

 

[W3bbo]

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!

 

[Diffy]

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.

 

[sutupidmath]

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!

 

[W3bbo]

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.

 

[sutupidmath]

It's part of a larger exercise on a differential equation. Just a simple example for an implicit function.

This was something that i really was affraid i would hear. Damn.!