There was a correct solution from Andrei Lazanu (School 205, Bucharest). The first part is very clear but I have tried to simplify his solution to the second part for inclusion here. Perhaps someone could improve on this for us. Thank you for your hard work Andrei.

First, I approximated Ö3 using the method given in the problem. I know that Ö3 is between 1 and 2 because 12 < (Ö3)2 < 22 or 1 < 3 < 4.

I know that the approximation of Ö3 correct to five decimal places is:
Ö3 » 1.73205

Now I show each of the approximation steps:

First approximation:


Ö3 » 2

Second approximation:


Ö3 »
3
2
 + 2
2
 =1.75

Third approximation:


Ö3 »
3
1.75
 + 1.75
2
 = 1.732142857

Fourth approximation:


Ö3 »
3
1.732142857
 + 1.732142857
2
 = 1.73205081
So, four approximations are sufficient to approximate Ö3 correct to 5 decimal places.

You could think of the above as


  __
Öa2
 
 »
a2
n
 + n
2
 =m

Where n is the approximation to the root of a 2 (that is "a") and m the next approximation.

The first approximation (n) differs from a by k. I can therefore write n as a + k where k is numerically less than a (k could be negative).

So I have


The next approximation =
a2
a+k
 + a+k
2

But


a2
a+k
 + a+k
2
 = 2a2 + 2ak + k2
2(a+k)

and


2a2 + 2ak + k2
2(a+k)
 = 2a(a+k)+ k2
2(a+k)
 = 2a(a+k)
2(a+k)
 + k2
2(a+k)
 = a + k2
2(a+k)

While a is positive,


k2
2(a+k)
must be positive as k is numerically less than a.

So


a <
a2
a+k
 + a+k
2

But the same equation could be written as:


2a2 + 2ak + k2
2(a+k)
 = a2+ (a+k)2
2(a+k)
 = (a+k)2
2(a+k)
 + a2
2(a+k)
 = a+k
2
 + a2
2(a+k)

The following number is equal to a+k :


a+k
2
 + a2
2(a+k)
 + 2ak+k2
2(a+k)
 = (a+k)2 + a2 + 2ak + k2
2(a+k)
 = a2 + k2 + 2ak + a2 + 2ak + k2
2(a+k)
 = 2(a2 + 2ak + k2)
2(a+k)
 = (a+k)2
(a+k)
 = (a+k)

This means that


a2
a+k
 + a+k
2
 < a+k.

From the two inequalities I obtain that:


a <
a2
a+k
 + a+k
2
 < a+k.

This means that the solution obtained goes closer and closer at each step to the real value of whether k is positive or negative.