# Golden Ratios (Q Fields)

• Jul 7th 2010, 09:00 AM
Samson
Golden Ratios (Q Fields)
Hello Everybody,

As a continuation to my last thread, there are other examples that I struggled with.

1. There exists a defining equation for the golden ratio: (1+Sqrt(5))/2 , and its norm in Q[Sqrt(5)] can also be found.

Can anyone help me find this equation and the norm? I tried to think of it like normalizing vectors and using conjugates, but I don't think that its getting me anywhere.

2. Assuming a and b are elements of Q[Sqrt(d)],, it can be shown that N(ab)=N(a)N(b) and N(a/b)=N(a)/N(b).

How can this be shown? Should a exist in Q, how can we show that N(a)=(a^2) ?

Any and all help is appreciated!
• Jul 7th 2010, 09:30 AM
undefined
Quote:

Originally Posted by Samson
Hello Everybody,

As a continuation to my last thread, there are other examples that I struggled with.

1. There exists a defining equation for the golden ratio: (1+Sqrt(5))/2 , and its norm in Q[Sqrt(5)] can also be found.

Can anyone help me find this equation and the norm? I tried to think of it like normalizing vectors and using conjugates, but I don't think that its getting me anywhere.

2. Assuming a and b are elements of Q[Sqrt(d)],, it can be shown that N(ab)=N(a)N(b) and N(a/b)=N(a)/N(b).

How can this be shown? Should a exist in Q, how can we show that N(a)=(a^2) ?

Any and all help is appreciated!

(1a) It is well known that the golden ratio is one of the roots of the polynomial $\displaystyle x^2 - x - 1$. It's on Wikipedia and many many websites. You can also do

$\displaystyle \displaystyle \left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)$

(1b) and (2) My Wikipedia-education is a bit slim on norms... It seems though that for $\displaystyle \alpha=a+b\sqrt{D}$, we have norm $\displaystyle N(\alpha)=\alpha\overline{\alpha}$. From this the other properties can probably be shown easily..?
• Jul 7th 2010, 01:24 PM
Samson
Quote:

Originally Posted by undefined
(1a) It is well known that the golden ratio is one of the roots of the polynomial $\displaystyle x^2 - x - 1$. It's on Wikipedia and many many websites. You can also do

$\displaystyle \displaystyle \left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)$

(1b) and (2) My Wikipedia-education is a bit slim on norms... It seems though that for $\displaystyle \alpha=a+b\sqrt{D}$, we have norm $\displaystyle N(\alpha)=\alpha\overline{\alpha}$. From this the other properties can probably be shown easily..?

Thank you undefined, I appreciate all the help! Is anybody else familiar enough with Norms to be able to pick this up at this point? Thank you!
• Jul 10th 2010, 07:17 AM
Samson
Hey all, I just wanted to bump and see if anyone could explain those other properties like undefined had mentioned!

All help is appreciated!
• Jul 10th 2010, 12:51 PM
chiph588@
Quote:

Originally Posted by Samson
Hey all, I just wanted to bump and see if anyone could explain those other properties like undefined had mentioned!

All help is appreciated!

• Jul 13th 2010, 10:16 AM
Samson
Quote:

Originally Posted by chiph588@

Sorry! Does anybody know how we can do number two? I have tried following a few links on google as far as Norms go. Any help is appreciated!
Here is one of the links I've found:

http://www.jstor.org/pss/2324118
• Jul 13th 2010, 10:20 AM
chiph588@
Quote:

Originally Posted by Samson
Sorry! Does anybody know how we can do number two? I have tried following a few links on google as far as Norms go. Any help is appreciated!
Here is one of the links I've found:

Try a direct proof.
• Jul 13th 2010, 10:21 AM
Samson
Quote:

Originally Posted by chiph588@
Try a direct proof.

What do you mean by a "direct proof?"
• Jul 13th 2010, 11:25 AM
undefined
Regarding forum rules: they're here to make everyone's lives easier really; consider post #4, which is the one that went against the rules:

Quote:

Originally Posted by Samson
Hey all, I just wanted to bump and see if anyone could explain those other properties like undefined had mentioned!

All help is appreciated!

This adds nothing to the discussion because it gives us no useful information. All it does is tell us you think your question deserves more attention. From a helper standpoint, a post like this makes us less inclined to help you because then it's like we're working for you rather than with you. (Like we're slaves or something.) In addition it clutters the forum. So it's to everyone's disadvantage.

Contrast this to post #6:

Quote:

Originally Posted by Samson
Sorry! Does anybody know how we can do number two? I have tried following a few links on google as far as Norms go. Any help is appreciated!
Here is one of the links I've found:

Here you give some more information and show effort in moving the discussion forward. This is within the rules. So if you feel the need to "bump" you should consider why you haven't gotten a sufficient response so far, and try to help us help you accordingly (for example by elaborating on any details that might not have been clear, giving us some definitions or references, responding to what we suggested, posting some work and saying where you got stuck, etc.).

Now you haven't responded to my suggestion that $\displaystyle N(\alpha)=\alpha\overline{\alpha}$. I very much suspect it's right, but I was trying to get a definitive source and was piecing together info from here, here, here, here, here, and maybe others (I don't have number theory books easily available at the moment). Presumably there is a definition in your source that would allow you to easily verify that $\displaystyle N(\alpha)=\alpha\overline{\alpha}$.

Now then, $\displaystyle N(\alpha)=\alpha\overline{\alpha} = (a+b\sqrt{D})(a-b\sqrt{D}) = a^2 - b^2D$. So

$\displaystyle N\left(\dfrac{1+\sqrt{5}}{2}\right) = \left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2\cdot5 = -1$

And for your #2 questions, it's just algebra

Let $\displaystyle \alpha = a+b\sqrt{D}, \beta = c+d\sqrt{D}$.

Then $\displaystyle \alpha\beta = (ac + bdD) + (ad + bc)\sqrt{D}$

$\displaystyle N(\alpha\beta) = (ac + bdD)^2 - (ad + bc)^2D$

$\displaystyle = a^2c^2+2abcdD+b^2d^2D^2-D(a^2d^2+2abcd+b^2c^2)$

$\displaystyle = a^2c^2+b^2d^2D^2-a^2d^2D-b^2c^2D$

Compare with

$\displaystyle N(\alpha)N(\beta) = (a^2-b^2D)(c^2-d^2D)$

$\displaystyle = a^2c^2-a^2d^2D-b^2c^2D+b^2d^2D^2$

See that they are the same.

Do similarly for N(a/b)=N(a)/N(b).

Quote:

Originally Posted by Samson
How can this be shown? Should a exist in Q, how can we show that N(a)=(a^2) ?

We have $\displaystyle \alpha = a + 0\sqrt{D}$, so this question is trivial.
• Jul 13th 2010, 11:54 AM
chiph588@
Quote:

Originally Posted by Samson
What do you mean by a "direct proof?"

Direct proof - Wikipedia, the free encyclopedia
• Jul 14th 2010, 06:46 AM
Samson
Thank you both very much! I really appreciate it! :-)