Continuous fractions for root 2

[bgwyh_88]

Hi all,

Could anyone guide me on the following prove

√2 = 1+1/(2 + 1/(2+ 1/(2+ 1/(2+···))))

 

[micromass]

Hi bgwyh_88!

Let

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

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

 

[HallsofIvy]

By the way, the term in English is "continued" fraction, not "continuous" fraction.

 

[bgwyh_88]

Hi bgwyh_88!

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?

 

[micromass]

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...

 

[bgwyh_88]

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.