# D's Rational Approximations Question from Y!Answers

#### CaptainBlack

##### Well-known member
Some bits are ok but I thought I would include them anyway as it is needed to answer the other parts of the question. I have labelled the parts which I need help with.

(a) Recall why there are no integer solutions $$m, n \in \mathbb{N}$$ to the equation $$m^2 = 2n^2$$.
ANSWER = an irrational number cannot be expressed as a fraction

(b) Show that if $$m, n \in \mathbb{N}$$ are integer solutions to the equation
$$m^2 = 2n^2 + 1$$, (*)
then so are $$M = m^2 + 2n^2$$ and $$N = 2mn$$, namely $$M^2 = 2N^2 + 1$$.
ANSWER = sub them in and then they equal each other

(c)Give a simple solution $$m, n \in \mathbb{N}$$ to equation (*).
ANSWER = When $$m=3$$ and $$n=2$$

(d )Deduce that there are infinitely many pairs of integers $$m, n \in \mathbb{N}$$ satisfying (*).
ANSWER = Help????? A hint is that $$M^2 = 2N^2 + 1$$ is significantly larger than (*)

(e) Let $$m, n \in \mathbb{N}$$ be any pair of integers satisfying equation (*).
Show that if $$p \in \mathbb{N}$$ is a prime number then
$$p | n$$ implies that \p doesn’t divide m.
[Use the fact that if $$p$$ is prime and $$a, b \in \mathbb{N}$$, and $$p | ab$$ then $$p | a$$ or $$p | b$$].
ANSWER = I done this so it’s ok but have included it anyway

( f ) What does it mean for a fraction $$a/b$$ to be in reduced form? ANSWER = done
Explain why if $$m, n \in \mathbb{N}$$ satisfy equation (*), then is $$m/n$$ in reduced form. ANSWER = Help??????

(g) Use the above to generate a fraction which approximates $$\sqrt{2}$$ to 5 decimal places.
ANSWER = Help???????? This is the main bit in which I require help as you have to use the question to generate the approximation.
So mainly g, a bit of f and d is what I need help with thanks!

#### CaptainBlack

##### Well-known member
d) Suppose otherwise, then there are a finite number of solutions to (*). Let $$m,n$$ be the solution with largest $$m$$. Then $$M=m^2+2n^2,\ N=2mn$$ by part (b) is also a solution of (*), but as $$m,n \in \mathbb{N}$$ we have $$M>m$$, a contradiction.

Hence it is not the case that there are a finite number of solutions ..

( e ) $$a/b$$ is in reduced form if $$a$$ and $$b$$ have no common factor greater than $$1$$ (that is are co-prime).

Suppose $$m$$ and $$n$$ have a common factor $$k>1$$, then $$k|m^2$$ and $$k|2n^2$$ in which case $$k$$ does not divide $$2n^2+1$$ a contradiction. That is $$m$$ and $$n$$ have no common factor greater than $$1$$ hence $$m/n$$ is in reduced form.

(g) Consider $$m,n \in \mathbb{N}$$ a solution to (*), then:

$$m^2/n^2 = 2 + 1/n^2$$

So if we have a sequence $$(m_1,n_1), (m_2,n_2) ...$$ of solutions to (*) with $$n_1, n_2, ..$$ increasing then $$(m_k)^2/(n_k)^2$$ tends to $$2$$ as $$k$$ goes to infinity and $$m_k/n_k$$ goes to $$\sqrt{2}$$.

Now we have a recipe for generating such a sequence:

$$m_{k+1}=m_k^2 + 2n_k^2,\ n_{k+1}=2 m_k n_k$$, with $$m_1=3,\ n_1=2$$

So we can now proceed to find an appropriate approximation (you may want to calculate how big $$n_k$$ needs to be to give the required accuracy)

CB

#### chisigma

##### Well-known member
...b) show that if $$m, n \in \mathbb{N}$$ are integer solutions to the equation

$m^{2} = 2\ n^{2} + 1$ (*)

... then so are $$M = m^2 + 2n^2$$ and $$N = 2mn$$, namely...

$$M^2 = 2N^2 + 1$$ (**)...
In my opinion the most insidious task is to demonstrate b). That is my approach: if (*) is true then...

$m^{2}-2\ n^{2} -1 =0$ (1)

... so that is also...

$M^{2}-2\ N^{2} - 1 = m^{4} - 4\ m^{2}\ n^{2} + 4\ n^{4} -1 = (m^{2}-2\ n^{2} -1)\ (m^{2}-2\ n^{2} +1)=0$ (2)

Kind regards

$\chi$ $\sigma$