Exploring the Mysteries of Fractional Roots: A Scientific Investigation

In summary, the conversation discussed the relationship between fractions and the expression {{{a \over b}} \over {{c \over d}}} = {{ad} \over {bc}}. The conversation then explored a specific case where b and d are equal, and determined that the expression {{\left( {1 + \sqrt 2 } \right)\sqrt 2 } \over {\sqrt 2 \left( {1 - \sqrt 2 } \right)}} is not equal to the original expression. The conversation also discussed an exercise involving a differential equation and how to manipulate an expression to make it equivalent to another form.
  • #1
W3bbo
31
0
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?
 
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  • #2
W3bbo said:
[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]
 
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  • #3
sutupidmath said:
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]
 
  • #4
W3bbo said:
Yes.


Check the denominator, too.
 
  • #5
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]
 
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  • #6
robert Ihnot said:
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!
 
  • #7
W3bbo said:
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.
 
  • #8
W3bbo said:
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!
 
  • #9
sutupidmath said:
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.
 
  • #10
W3bbo said:
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.!
 

What are fractions with roots in "voodoo"?

Fractions with roots in "voodoo" refer to mathematical expressions that involve both fractions and square roots, often used in voodoo rituals and practices.

How do you simplify fractions with roots in "voodoo"?

To simplify fractions with roots in "voodoo", you must first simplify the fraction part and then simplify the square root part. This involves finding the greatest common factor and simplifying the square root if possible.

Are there any special rules for adding or subtracting fractions with roots in "voodoo"?

Yes, when adding or subtracting fractions with roots in "voodoo", you must first find a common denominator for both fractions. Then, you can add or subtract the numerators and keep the common denominator.

What about multiplying and dividing fractions with roots in "voodoo"?

Multiplying fractions with roots in "voodoo" is similar to multiplying regular fractions, where you multiply the numerators and denominators separately. When dividing, you must first convert the division sign into a multiplication sign and flip the second fraction. Then, follow the same steps as multiplying.

Can fractions with roots in "voodoo" have negative values?

Yes, fractions with roots in "voodoo" can have negative values. This can occur when the numerator or denominator is negative, or when there is a negative sign in front of the square root. When simplifying, be sure to follow the proper rules for dealing with negative numbers.

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