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

\sqrt{3}

using the method given in the problem. I know that

\sqrt{3}

is between 1 and 2 because

1^{2} < (\sqrt{3})^{2} < 2^{2}

1 < 3 < 4

I know that the approximation of $\sqrt{3}$ correct to five decimal places is:

$\sqrt{3} \approx 1.73205$

Now I show each of the approximation steps:

First approximation:

\sqrt{3} \approx 2

Second approximation:

\sqrt{3} \approx \frac{\frac{3}{2} + 2}{2} = 1.75

Third approximation:

\sqrt{3} \approx \frac{\frac{3}{1.75} + 1.75}{2} = 1.732142857

Fourth approximation:

\sqrt{3} \approx \frac{\frac{3}{1.732142857} + 1.732142857}{2} = 1.73205081

So, four approximations are sufficient to approximate

\sqrt{3}

correct to 5 decimal places.

You could think of the above as

\sqrt{a^{2}} \approx \frac{\frac{a^{2}}{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 = \frac{\frac{a^{2}}{a + k} + a + k}{2}

But

\frac{\frac{a^{2}}{a + k} + a + k}{2} = \frac{2 a^{2} + 2 ak + k^{2}}{2 (a + k)}

and

\frac{2 a^{2} + 2 ak + k^{2}}{2 (a + k)} = \frac{2 a (a + k) + k^{2}}{2 (a + k)} = \frac{2 a (a + k)}{2 (a + k)} + \frac{k^{2}}{2 (a + k)} = a + \frac{k^{2}}{2 (a + k)}

While a is positive,

\frac{k^{2}}{2 (a + k)}

must be positive as k is numerically less than a.

a < \frac{\frac{a^{2}}{a + k} + a + k}{2}

But the same equation could be written as:

\frac{2 a^{2} + 2 ak + k^{2}}{2 (a + k)} = \frac{a^{2} + (a + k)^{2}}{2 (a + k)} = \frac{(a + k)^{2}}{2 (a + k)} + \frac{a^{2}}{2 (a + k)} = \frac{a + k}{2} + \frac{a^{2}}{2 (a + k)}

The following number is equal to a+k :

\frac{a + k}{2} + \frac{a^{2}}{2 (a + k)} + \frac{2 ak + k^{2}}{2 (a + k)} = \frac{(a + k)^{2} + a^{2} + 2 ak + k^{2}}{2 (a + k)} = \frac{a^{2} + k^{2} + 2 ak + a^{2} + 2 ak + k^{2}}{2 (a + k)} = \frac{2 (a^{2} + 2 ak + k^{2})}{2 (a + k)} = \frac{(a + k)^{2}}{(a + k)} = (a + k)

This means that

\frac{\frac{a^{2}}{a + k} + a + k}{2} < a + k .

From the two inequalities I obtain that:

a < \frac{\frac{a^{2}}{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.