Find a Defining Equation for the Golden Ratio: $(1 + \sqrt{5})/2$

Question

Find a defining equation for the golden ratio: $(1 + \sqrt{5})/2$. Furthermore, find its norm in $\mathbb{Q}[\sqrt{5}]$. It seems like it would be simple, but I am still new to quadratic field. Could I get some aid on how to start and solve this? Thanks!

Answer 1

$$x= (1 + √5)/2$$ $$2x= 1 + √5$$ $$2x-1= √5$$ $$4x^2-4x +1= 5$$ $$4x^2-4x -4= 0$$ $$x^2-x -1= 0$$ gives one possibility, though the squaring has introduced another root, namely $x= (1 - √5)/2$. If you multiply these two roots together you will get $-1$, the norm. See https://en.wikipedia.org/wiki/Field_norm for more on finding norms

Answer 2

The golden ratio is one of the solutions to $x^2-x-1=0$. (I presume this is what you mean by "defining equation.")

Its norm would be $\left(\frac{1}{2}\right)^2 - 5 \cdot \left(\frac{1}{2}\right)^2 = -1$.

Answer 3

Suppose you split a length in two. You split it according to the golden ratio if the ratio of the larger length $L$ to the smaller length $l$ is equal to the ratio of the total length $L+l$ to the larger length $L$.

In formulae, if $L=xl$, $$x=\frac{(x+1)l}{lx}\iff x^2=x+1.$$

The norm of an element in $\mathbf Q(\sqrt 5)$ is the product of all the conjugates of this element. If $x=a+b\sqrt 5$ ($a,b\in\mathbf Q$), we have $N(x)=a^2-5b^2$. Thus the norm of the Golden ratio is equal to $-1$.

Answer 4

$\!\!\!\begin{align}{\rm Recall}\ \ w = a + b\sqrt{n}\rm \ \ has\ \ {\bf norm}\ &=\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2\! - n\: b^2\\[4pt] {\rm and,\ furthermore,\ }w\rm \ \ has\ \ {\bf trace}\ &=\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2a\end{align}$

Now apply Vieta $\ (x\!-\!w)(x\!-\!w') = x^2 - (w\!+\!w')x + ww' =\, x^2 - 2a\, x + a^2\!-nb^2$

Answer 5

How about the original construction from the Greeks?

Let $a, b, c$ be the sides of a right triangle with $c$ as hypotenuse and $b:a=2:1$. Then:

$b^2=c^2-a^2=(c+a)(c-a)$ (the difference of squares factorization is easily proved by the geometric methods used by the Greeks).

$c-a=(c+a)-b$ (from $b=2a$, hypothesis)

So

$b^2=(c+a)(c+a-b)$

From the cross product rule for proportions (again known since ancient times):

$(c+a)/b=b/(c+a-b)$

This represents whatcwe now call a "golden division" (a term invented later), and the ratio in the last equatyin is both $(c+a)/b$ and $b/(c-a)$. These formulas are commonly used to geometrically construct the golden ratio, for example in Ptolemy's construction of the regular pentagon/pentagram.