1.Let
Then


or
Hence (because p by definition is positive)
Now,

2.Let
Then


or(Neglecting the negative solution)
Now

Could you please explain how to get from

p=(2+(2+(2+(2+...)^{-1})^{-1})^{-1})^{-1}

to

p=(2+p)^{-1}


Thanks -

Since (1) in the above quote can be written as:
where
then since q is clearly the same sequence as p:

1.Let p= (2+(2+(2+(2+----)exp-1)exp-1)exp-1,,,etc ='s 1/(2+p) is what you have to prove! Don't assume it at the start!

I don't understand what you are objecting to!! You are asked to show that the continued fraction is equal to .
Please explain your problem with the above proofs more fully.
Thanks

Vasco you are correct. If p is the inverse infinite sequence
p=(2+(2+(2+(2+....)exp-1)exp-1)exp-1)exp-1 then any place you start your count will give you the same infinite sequence p. Easiest is to start the count again on the second 2 giving p=(2+p)exp-1,, your solution.

The same can be done for 7-sqrt3 = 5+(3+(1+(2+(1+(2+...)exp-1)exp-1 etc
where now we let 2-sqrt3= p=1/(3+q) and q is the repeating series
q=(1+(2+(1+(2+(1+.....)exp-1)exp-1)exp-1)exp-1)exp-1
we can start counting at the second 1+(2+...... to get
q=(1+(2+q)exp-1)exp-1= 1/[1+1/(2+q)] = (2+q)/(3+q)
Multiply out the 3+q
3q+q**2=2+q,,,,,,,or q**2+2q-2=0 then q= -1+sqrt3r, q=-1-sqrt3 now from above
p=1/(3+q), taking q=-1+sqrt3
p=1/(2+sqrt3) multiply top and bottom by 2-sqrt 3
to get
p=2-sqrt3,,,,adding the 5 back to p gives
7-sqrt3=5+(3+(1+(2+(1.........etc

HOW DO I GET A MATH SYMBOL EDITOR! as you can see doing things with my setup quite messy.

There's a built in editor on this forum which is explained in "general->a guide to the forum".
For use on your computer at home you could use the one I do - it's free:
Texnic Center :
http://texniccenter.org/

You can produce pdf files with it and then attach them to your posts on this forum.