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:

√3 ≈ 1.73205

.

Now I show each of the approximation steps:

First approximation:

√3 ≈ 2

Second approximation:

√3 ≈  3 2 + 2 2 =1.75

Third approximation:

√3 ≈  3 1.75 + 1.75 2 = 1.732142857

Fourth approximation:

√3 ≈  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

√   a2   ≈  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.