Continuous fractions for root 2

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Discussion Overview

The discussion revolves around the representation of the square root of 2 as a continued fraction. Participants are exploring the mathematical formulation and transformation of this expression, specifically focusing on the structure and properties of continued fractions.

Discussion Character

  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant asks for guidance on proving that √2 can be expressed as a continued fraction.
  • Another participant proposes a transformation involving the variable x, defined as x = 1 + 1/(2 + 1/(2 + ...)), and questions the derivation of the expression (1/(x-1)) - 1.
  • A correction is made regarding the terminology, noting that the correct term is "continued" fraction rather than "continuous" fraction.
  • Further discussion involves confirming that the transformation leads back to the original variable x, suggesting a standard mathematical technique.

Areas of Agreement / Disagreement

Participants appear to agree on the mathematical transformation involving x, but there is no consensus on the initial proof request or the terminology used.

Contextual Notes

There are assumptions regarding the understanding of continued fractions and the transformations involved, which may not be fully articulated by all participants.

bgwyh_88
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Hi all,

Could anyone guide me on the following prove

√2 = 1+1/(2 + 1/(2+ 1/(2+ 1/(2+···))))
 
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Hi bgwyh_88! :smile:

Let

[tex]x=1+\frac{1}{2+\frac{1}{2+...}}[/tex]

Then what is [itex]\frac{1}{x-1}-1[/itex]?
 
By the way, the term in English is "continued" fraction, not "continuous" fraction.
 
micromass said:
Hi bgwyh_88! :smile:

Let

[tex]x=1+\frac{1}{2+\frac{1}{2+...}}[/tex]

Then what is [itex]\frac{1}{x-1}-1[/itex]?

hey micromass,

You will ultimately get

1+ [itex]\frac{1}{2+\frac{1}{2+...}}[/itex]

Where did you get

[itex]\frac{1}{x-1}-1[/itex] from?
 
bgwyh_88 said:
hey micromass,

You will ultimately get

1+ [itex]\frac{1}{2+\frac{1}{2+...}}[/itex]

Yes, and that is x. So [itex]\frac{1}{x-1}-1=x[/itex]

Where did you get

[itex]\frac{1}{x-1}-1[/itex] from?

You just need to transform x to something that equal x again. It's a standard trick that you had to see once...
 
micromass said:
Yes, and that is x. So [itex]\frac{1}{x-1}-1=x[/itex]



You just need to transform x to something that equal x again. It's a standard trick that you had to see once...

micromass,

Cool. Thanks mate. :approve:
 

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