Golden Ratio

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Golden Ratio

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Golden Ratio

The golden ratio, approximately 1.618, is called golden because many geometric figures involving this ratio are often said to possess special beauty. Be that true or not, the ratio has many beautiful and surprising mathematical properties.

1 Basic Description
2 A More Mathematical Explanation
- 2.1 Continued Fraction Representation and Fibonacci Sequences
- 2.2 Proof of the Golden Ratio's Irrationality
3 For More Information
4 Future Directions for this Page

Basic Description

History

The Greeks were aware of the golden ratio, but did not consider it particularly significant with respect to aesthetics. It was not called the "divine" proportion until the 15th century, and was not called "golden" ratio until the 18th century.

Since then, it has been claimed that the golden ratio is the most aesthetically pleasing ratio, and claimed that this ratio has appeared in architecture and art throughout history. Among the most common such claims are that the Parthenon and Leonardo Da Vinci's Mona Lisa's use the golden ratio. Even more esoteric claims propose that the golden ratio can be found in the human facial structure, the behavior of the stock market, and the Great Pyramids.

However, such claims have been criticized in scholarly journals (see references at the end of the page) as wishful thinking or sloppy mathematical analysis. Additionally, there is no solid evidence that supports the claim that the golden rectangle is the most aesthetically pleasing rectangle.

Nonetheless, some artists have used it explicitly, such as Salvador Dalí in his painting The Sacrament of the Last Supper, shown below.

A Geometric Representation

From a geometric sense, the golden ratio can be defined using a line segment divided into two sections, of lengths a and b, respectively. If a and b are appropriately chosen, the ratio of b to a is the same as the ratio of a + b to b.

The value of this ratio, denoted $\varphi$ , turns out not to depend on the particular values of a and b, as long as they satisfy the proportion above.

An Algebraic Representation

We may algebraically solve for the ratio ( $\varphi$ ) by observing that ratio satisfies the following property by definition:

$\frac{b}{a} = \frac{a+b}{b} = \varphi \,$

Our goal is to obtain an expression that includes $\varphi$ and known constants. We want to eliminate a and b from our equations so that we can solve for $\varphi$ .

We can write our ratio as.

$\varphi = \frac{b}{a} \,$

First, let's eliminate the variable a. Because $a = \frac{b}{\varphi}$ , we can substitute $\frac{b}{\varphi}$ for $a$ in the equation $\frac{a+b}{b} = \varphi$ .

This gives us

$\frac{\frac{b}{\varphi}+b}{b} = \varphi \,$

Multiplying the top and bottom of the fraction on the left by $\varphi$ , we get,

$\frac{b+b\varphi}{b\varphi}=\varphi \,.$

Dividing the top and the bottom of the fraction by $b$ gives:

$\frac{1+\varphi}{\varphi}=\varphi \,.$

Finally, multiplying both sides of the equation by $\varphi$ and then moving everything to the left side of the equation gives:

${\varphi}^2-{\varphi}-1=0\,.$

Using the quadratic formula , we are able to find the positive solution, $\textstyle\frac{1 + \sqrt{5} }{2} \approx 1.61803399...$ . This algebra shows that if two positive numbers a and b satisfy the proportion described earlier, then this is the only value the ratio can have.

The quadratic formula does yield a second value from the equation. Since it is negative, we discard it, because the ratio of two positive lengths a and b must also be positive.

Applications

Triangles

The Golden Ratio $\varphi$ is used to construct the golden triangle, an isoceles triangle that has legs of length $\varphi$ and base length of 1. It is above and to the left. Similarly, the golden gnomon has base ${\varphi}$ and legs of length 1. It is shown above and to the right.

These triangles can be used to from a pentagram, which has several golden ratio proportions.

$\frac{\mathrm{red} }{\mathrm{green} } = \frac{\mathrm{green} }{\mathrm{blue} } = \frac{\mathrm{blue} }{\mathrm{magenta} } = \varphi .$

These triangles can be used to form fractals and are one of the only ways to tile a plane using pentagonal symmetry.

A More Mathematical Explanation

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Continued Fraction Representation and Fibonacci Sequences

The golden ratio can also be written as what is called a continued fraction by using recursion.

We have already solved for $\varphi$ using the following equation:

${\varphi}^2-{\varphi}-1=0$ .

We can add one to both sides of the equation to get

${\varphi}^2-{\varphi}=1$ .

Factoring this gives

$\varphi(\varphi-1)=1$ .

Dividing by $\varphi$ gives us

$\varphi -1= \cfrac{1}{\varphi }$ .

Solving for $\varphi$ gives

$\varphi =1+ \cfrac{1}{\varphi }$ .

Now use recursion and substitute in the entire right side of the equation for $\varphi$ in the bottom of the fraction.

$\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{\varphi } }$

Substituting in again,

$\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\varphi}}}$

$\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\cdots}}}$

This last infinite form is a continued fraction

If we evaluate truncations of the continued fraction by evaluating only part of the continued fraction (the finite displays above it), replacing $\varphi$ by 1, we produce the ratios between consecutive terms in the Fibonacci sequence.

$\varphi \approx 1 + \cfrac{1}{1} = 2$

$\varphi \approx 1 + \cfrac{1}{1+\cfrac{1}{1}} = 3/2$

$\varphi \approx 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1} } } = 5/3$

$\varphi \approx 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1+\cfrac{1}{1}}}} = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{2}}} =1 + \cfrac{1}{1 + \cfrac{2}{3}} = 8/5$

Thus we discover that the golden ratio is approximated in the Fibonacci sequence.

$1,1,2,3,5,8,13,21,34,55,89,144...\,$

$1/1$	$=$	$1$
$2/1$	$=$	$2$
$3/2$	$=$	$1.5$
$8/5$	$=$	$1.6$
$13/8$	$=$	$1.625$
$21/13$	$=$	$1.61538462...$
$34/21$	$=$	$1.61904762...$
$55/34$	$=$	$1.61764706...$
$89/55$	$=$	$1.61818182...$

$\varphi = 1.61803399...\,$

As you go farther along in the Fibonacci sequence, the ratio between the consecutive terms approaches the golden ratio.

In fact, we can prove this relationship using mathematical Induction.

Since we have already shown that

$\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\cdots}}}$ ,

we only need to show that each of the terms in the continued fraction is the ratio of Fibonacci numbers as shown above.

First, let $x_1=1$ , $x_2=1+\frac{1}{1}=1+\frac{1}{x_1}$ , $x_3= 1+\frac{1}{1+\frac{1}{1}}=1+\frac{1}{x_2}$ and so on so that $x_n=1+\frac{1}{x_{n-1}}$ .

These are just the same truncated terms as listed above. Let's also denote the terms of the Fibonacci sequence as $f_n=f_{n-1}+f_{n-2}$ where $f_1=1$ , $f_2=1$ , and so $f_3=1+1=2$ , $f_4=1+2=3$ and so on.

We want to show that $x_n=\frac{f_{n+1}}{f_n}$ for all n.

First, we establish our base case. We see that $x_1=1=\frac{1}{1}=\frac{f_2}{f_1}$ , and so the relationship holds for the base case.

Now we assume that $x_k=\frac{f_{k+1}}{f_{k}}$ for some $1 \leq k < n$ (This step is the inductive hypothesis). We will show that this implies that $x_{k+1}=\frac{f_{(k+1)+1}}{f_{k+1}}=\frac{f_{k+2}}{f_{k+1}}$ .

By our definition of $x_n$ , we have

$x_{k+1}=1+\frac{1}{x_k}$ .

By our inductive hypothesis, this is equivalent to

$x_{k+1}=1+\frac{1}{\frac{f_{k+1}}{f_{k}}}$ .

Now we only need to complete some simple algebra to see

$x_{k+1}=1+\frac{f_k}{f_{k+1}}$

$x_{k+1}=\frac{f_{k+1}+f_k}{f_{k+1}}$

Noting the definition of $f_n=f_{n-1}+f_{n-2}$ , we see that we have

$x_{k+1}=\frac{f_{k+2}}{f_{k+1}}$

Since that was what we wanted to show, we see that the terms in our continued fraction are represented by ratios of Fibonacci numbers.

The exact continued fraction is $x_{\infty} = \lim_{n\rightarrow \infty}\frac{f_{n+1}}{f_n} =\varphi$ .

Proof of the Golden Ratio's Irrationality

Remarkably, the Golden Ratio is irrational, despite the fact that we just proved that is approximated by a ratio of Fibonacci numbers.

We will use the method of contradiction to prove that the golden ratio is irrational.

Suppose $\varphi$ is rational. Then it can be written as fraction in lowest terms $\varphi = b/a$ , where a and b are integers.

Our goal is to find a different fraction that is equal to $\varphi$ and is in lower terms. This will be our contradiction that will show that $\varphi$ is irrational.

First note that the definition of $\varphi = \frac{b}{a}=\frac{a+b}{b}$ implies that $b > a$ since clearly $b+a>b$ and the two fractions must be equal.

Now, since we know

$\frac{b}{a}=\frac{a+b}{b}$

we see that $b^2=a(a+b)$ by cross multiplication. Writing this all the way out gives us $b^2=a^2+ab$ .

Rearranging this gives us $b^2-ab=a^2$ , which is the same as $b(b-a)=a^2$ .

Dividing both sides of the equation by $(b-a)$ and $a$ gives us that

$\frac{b}{a}=\frac{a}{b-a}$ .

Since $\varphi=\frac{b}{a}$ , we can see that $\varphi=\frac{a}{b-a}$ .

Since we have assumed that a and b are integers, we know that b-a must also be an integer. Furthermore, since $a<b$ , we know that $\frac{a}{b-a}$ must be in lower terms than $\frac{b}{a}$ .

Since we have found a fraction of integers that is equal to $\varphi$ , but is in lower terms than $\frac{b}{a}$ , we have a contradiction: $\frac{b}{a}$ cannot be a fraction of integers in lowest terms. Therefore $\varphi$ cannot be expressed as a fraction of integers and is irrational.

For More Information

Markowsky. “Misconceptions about the Golden Ratio.” College Mathematics Journal. Vol 23, No 1 (1992). pp 2-19.

http://www.contracosta.edu/math/pentagrm.htm

A citation note: Proof that hides originally appeared in User:AnnaP's discrete math class homework assignment.

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- A animation for the main image that shows the specific line segments being removed and compared to each other.

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Golden Ratio

From Math Images

Contents

Basic Description

History

A Geometric Representation

An Algebraic Representation

Applications

Triangles

A More Mathematical Explanation

Continued Fraction Representation and Fibonacci Sequences

Proof of the Golden Ratio's Irrationality

For More Information

Future Directions for this Page

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