I wasn't sure which section this belonged to so I decided to post it here.

Use a method to approximate \sqrt 2 using a sequence of rational numbers. Do the same for \sqrt 3.
I started it like this:

a_1, \ a_2, \ a_3, \ ........... \ a_n \ ...... \ L

\frac{1}{a+a_1}+b, \ \frac{1}{a+a_2}+b \ ............\ \frac{1}{a+a_n}+b \ ..... \ \frac{1}{a+L}+b

where L is the limit.

Both sequences must have the same limit:

\frac{1}{a+L}+b=L

\frac{1+ba+bL}{a+L}=L

1+ba+bL=L^2+aL

L^2+aL-bL-1-ba=0

L^2+L(a-b)-1-ab=0

Since we want L= \sqrt 2:

f(\sqrt 2)=2+ \sqrt {2} (a-b)-1-ab=0

Since we want this to be 0 a=b since the middle term cannot exist.

1-b^2=0

b=1 \ or \ -1

I want the positive \sqrt 2 so b=1.

Therefore the sequence becomes a_{n+1}=\frac{1}{1+a_n}+1. This sequence tends to \sqrt 2.

Using the same method to find one that tends to \sqrt3 gives:

\frac{1}{a+L}+b=L

L^2+L(a-b)-1-ab=0

Once again, the central term cannot be present at the end so a=b:

f(\sqrt3)=3-1-b^2=0

b^2=2

b=\sqrt 2

But b can only be a rational number so it doesn't make sense. I think i'm supposed to use the first answer to find the second one except the two sequences are entirely different!

So am I on the right lines?