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:

${\displaystyle \sqrt{3}\approx 1.73205}$

.

Now I show each of the approximation steps:

First approximation:

${\displaystyle \sqrt{3}\approx 2}$

Second approximation:

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

Third approximation:

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

Fourth approximation:

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

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

You could think of the above as

${\displaystyle \sqrt{a^2}\approx \frac{\frac{a^2}{n}+n}{2}=m}$

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

${\displaystyle \mathrm{The}\mathrm{next}\mathrm{approximation}=\frac{\frac{a^2}{a+k}+a+k}{2}}$

But

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

and

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

While a is positive,

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

must be positive as k is numerically less than a.

So

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

But the same equation could be written as:

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

The following number is equal to a+k :

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

This means that

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

From the two inequalities I obtain that:

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

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