MHB D's Rational Approximations Question from YAnswers

[CaptainBlack]

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]

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]

...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

$\chi$ $\sigma$