Fractions with roots: voodoo

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Main Question or Discussion Point

Starting with simple fractions, it's known that:

[tex]{{{a \over b}} \over {{c \over d}}} = {{ad} \over {bc}}[/tex]

So when b == d:

[tex]{{{a \over b}} \over {{c \over b}}} = {a \over c}[/tex]

But what if in the case of:

[tex]{{{{1 + \sqrt 2 } \over {\sqrt 2 }}} \over {{{1 - \sqrt 2 } \over {\sqrt 2 }}}}[/tex]

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

[tex]{{1 + \sqrt 2 } \over {1 - \sqrt 2 }}[/tex]

But if you follow the abcd version:

[tex]{{\left( {1 + \sqrt 2 } \right)\sqrt 2 } \over {\sqrt 2 \left( {1 - \sqrt 2 } \right)}} = {{2 + \sqrt 2 } \over {2 - \sqrt 2 }}[/tex]

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

Why?
 

Answers and Replies

1,631
4
[tex] {{2 + \sqrt 2 } \over {2 - \sqrt 2 }}[/tex]

...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 [tex]\sqrt{2}[/tex]

remember [tex]\sqrt{2}*\sqrt{2}=2 \ \ \ and \ \ \ not \ \ \ \sqrt{2}[/tex]
 
Last edited:
31
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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 [tex]\sqrt{2}[/tex]

remember [tex]\sqrt{2}*\sqrt{2}=2 \ \ \ and \ \ \ not \ \ \ \sqrt{2}[/tex]
Yes.

[tex]\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}[/tex]
 
1,056
0
You got the terms reversed in the demoninator: [tex]{{\left( {1 + \sqrt 2 } \right)\sqrt 2 } \over {\sqrt 2 \left( {1 - \sqrt 2 } \right)}} = {{2 + \sqrt 2 } \over {2 - \sqrt 2 }}[/tex]

Denominator should be [tex]\sqrt{2}-2[/tex]
 
Last edited:
31
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You got the terms reversed in the demoninator: [tex]{{\left( {1 + \sqrt 2 } \right)\sqrt 2 } \over {\sqrt 2 \left( {1 - \sqrt 2 } \right)}} = {{2 + \sqrt 2 } \over {2 - \sqrt 2 }}[/tex]

Denominator should be [tex]\sqrt{2}-2[/tex]
Ah, that solves it then. Thanks.

As an aside, how can I make this:

[tex]{{dy} \over {dx}} = {{2y - x^2 } \over {y^2 - 2x}}[/tex]

into this:

[tex]{{dy} \over {dx}} = {{x^2 -2y } \over {2x - y^2}}[/tex]

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

Thanks!
 
441
0
Ah, that solves it then. Thanks.

As an aside, how can I make this:

[tex]{{dy} \over {dx}} = {{2y - x^2 } \over {y^2 - 2x}}[/tex]

into this:

[tex]{{dy} \over {dx}} = {{x^2 -2y } \over {2x - y^2}}[/tex]

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.
 
1,631
4
Ah, that solves it then. Thanks.

As an aside, how can I make this:

[tex]{{dy} \over {dx}} = {{2y - x^2 } \over {y^2 - 2x}}[/tex]

into this:

[tex]{{dy} \over {dx}} = {{x^2 -2y } \over {2x - y^2}}[/tex]

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!
 
31
0
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.
 
1,631
4
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.!!
 

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