Defining equation for golden ratio; quadratic fields

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September 17th 2008, 11:02 PM
Pn0yS0ld13r

Defining equation for golden ratio; quadratic fields

Definition: If $\alpha$ is an irrational number in $\mathbf{Q}\left(\sqrt{d}\right)$ , then the equation $ax^{2}+bx+c=0$ is called the defining equation for $\alpha$ if $\alpha$ satisfies the equation and a, b, and c are integers, $(a,b,c)=1$ , and $a>0$ .

Find a defining equation for the golden ratio $\dfrac{1+\sqrt{5}}{2}$ .

How do you approach this problem?
September 17th 2008, 11:14 PM
Moo

Hello,

Quote:

Originally Posted by Pn0yS0ld13r

Definition: If $\alpha$ is an irrational number in $\mathbf{Q}\left(\sqrt{d}\right)$ , then the equation $ax^{2}+bx+c=0$ is called the defining equation for $\alpha$ if $\alpha$ satisfies the equation and a, b, and c are integers, $(a,b,c)=1$ , and $a>0$ .

Find a defining equation for the golden ratio $\dfrac{1+\sqrt{5}}{2}$ .

How do you approach this problem?

The solutions to such an equation are :

$x=\frac{-b {\color{red}\pm} \sqrt{b^2-4ac}}{2a}$

Since a and b are integers, the only way to get this $\sqrt{5}$ is $\sqrt{b^2-4ac}$ .
So you can see that $\frac{1{\color{red}-}\sqrt{5}}{2}$ will also be a root. (The reasoning may be similar to the ones involving complex roots).

Develop $\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)$ and identify a,b and c :p
September 19th 2008, 01:19 AM
Pn0yS0ld13r

Thank you Moo.

I can't believe I didn't get this before... (Headbang)

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