### Rationals Between...

What fractions can you find between the square roots of 65 and 67?

### Root to Poly

Find the polynomial p(x) with integer coefficients such that one solution of the equation p(x)=0 is $1+\sqrt 2+\sqrt 3$.

### Consecutive Squares

The squares of any 8 consecutive numbers can be arranged into two sets of four numbers with the same sum. True of false?

# Archimedes and Numerical Roots

##### Stage: 4 Challenge Level:

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 $\sqrt3$ using the method given in the problem. I know that $\sqrt3$ 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: $$\sqrt{3} \approx {1.73205}$$Now I show each of the approximation steps:

First approximation: $$\sqrt{3} \approx {2}$$Second approximation: $$\sqrt{3}\approx {{{3\over{2}} + 2} \over {2}} ={1.75}$$Third approximation: $$\sqrt{3} \approx {{{3\over{1.75}} + 1.75} \over {2}} = {1.732142857}$$ Fourth approximation: $$\sqrt{3} \approx {{{3\over{1.732142857}} + 1.732142857} \over {2}} = {1.73205081}$$ So, four approximations are sufficient to approximate $\sqrt{3}$ correct to 5 decimal places.

You could think of the above as $$\sqrt{a^2}\approx {{{a^2\over{n}} + n} \over {2}} ={m}$$ 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 $$\quad = {{{a^2\over{a+k}} + a+k} \over {2}}$$The next approximation = $${{{a^2\over{a+k}} + a+k} \over {2}}$$But $${{{a^2\over{a+k}} + a+k} \over {2}} = {{2a^2 + 2ak + k^2} \over{2(a+k)}}$$ and $${{2a^2 + 2ak + k^2} \over{2(a+k)}} = {{2a(a+k)+ k^2} \over{2(a+k)}}= {{2a(a+k)} \over{2(a+k)}} + {{k^2} \over{2(a+k)}} = a + {{k^2} \over{2(a+k)}}$$While a is positive, $${{k^2} \over{2(a+k)}}$$must be positive as k is numerically less than a.

So $$a< {{{a^2\over{a+k}} + a+k} \over {2}}$$ But the same equation could be written as: $${{2a^2 + 2ak + k^2} \over{2(a+k)}} = {{a^2+ (a+k)^2} \over{2(a+k)}}= {{(a+k)^2} \over{2(a+k)}} + {{a^2} \over{2(a+k)}} = {{a+k} \over{2}} + {{a^2}\over {2(a+k)}}$$The following number is equal to a+k: $${{a+k} \over{2}} + {{a^2}\over {2(a+k)}} + {{2ak+k^2}\over{2(a+k)}} = {{(a+k)^2 + a^2 + 2ak + k^2}\over{2(a+k)}} = {{a^2 + k^2 + 2ak + a^2 + 2ak + k^2}\over{2(a+k)}} = {{2(a^2 + 2ak + k^2)}\over{2(a+k)}}={{(a+k)^2}\over{(a+k)}} = (a+k)$$ This means that $${{{a^2\over{a+k}} + a+k} \over {2}}< {a+k}.$$From the two inequalities I obtain that: $$a< {{{a^2\over{a+k}} + a+k} \over {2}}< {a+k}.$$ This means that the solution obtained goes closer and closer at each step to the real value of whether k is positive or negative.