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 √3 using the method given in the problem. I know that √3 is between 1 and 2 because 1 ² < (√3) ² < 2 ² or 1 < 3 <4.

I know that the approximation of √3 correct to five decimal places is: .

Now I show each of the approximation steps:

First approximation:

Second approximation:

Third approximation:

Fourth approximation:

So, four approximations are sufficient to approximate √3 correct to 5 decimal places.

You could think of the above as

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

But

and

While a is positive, must be positive as k is numerically less than a.

So

But the same equation could be written as:

The following number is equal to a+k :

This means that

From the two inequalities I obtain that:

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