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 ===Vitruvian Man===   ===Vitruvian Man=== 
  :[[Image:Vitruvian man mixed.jpeg]]  +  :[[Image:Vitruvian man mixed.jpeg350px]] 
 Another instance in which the Golden ratio appears is in Leonardo Da Vinci’s drawing of the Vitruvian Man. Da Vinci’s picture of man’s body fits the approximation of the golden ratio very closely. This picture is considered to a depiction of a perfectly proportioned human body. The Golden Ratio in this picture is the distance from the naval to the top of his head, divided by the distance from the soles of the feet to the top of the head.   Another instance in which the Golden ratio appears is in Leonardo Da Vinci’s drawing of the Vitruvian Man. Da Vinci’s picture of man’s body fits the approximation of the golden ratio very closely. This picture is considered to a depiction of a perfectly proportioned human body. The Golden Ratio in this picture is the distance from the naval to the top of his head, divided by the distance from the soles of the feet to the top of the head. 
 ===Music===   ===Music=== 
 +  :[[Image:Chromaticscales.gif350px]] 
 +  The Golden Ratio manifests in the idea that music is proportionally balanced. However, it is an area of debate, because some scholars claim that these appearances are purely coincidental, or a result of number juggling from aficionados. However, many contemporary artists have intentionally used the Golden Ratio in their pieces. Composer, mathematician and teacher Joseph Schillinger believed that music could be based entirely on mathematical formulation. He developed a System of Musical Composition in which successive notes in a melody were followed by Fibonacci intervals when counted in half steps. Half steps are the smallest intervals possible, and it is the closest note that can be played higher or lower. 
 +  For example, if C were the first note in a composition, the following note would be half a step higher, because one is a Fibonacci number. So, the second note would be a D flat. Schillinger would alternate between moving up and down in intervals. Thus, the third note would be two half steps (because two is the next Fibonacci number), down from the D flat, which would be B flat. The next Fibonacci number after two is three, so the next note would be three half steps higher than the last. Three half steps higher than B flat is E flat. Schillinger believed that these notes convey the same sense of harmony as the phyllotactic ratios found in leaves. 
   
 +  ImageDesc===Fibonacci Numbers== 
 +  One of the greatest breakthroughs regarding the Golden Ratio came when its relation to Fibonacci numbers, also known as the Fibonacci sequence, was discovered. Fibonacci numbers can also be found in the arts, and nature. The Fibonacci sequence is the series of numbers, 
 +  0,1,1,2,3,5,8,13,21,34,55,89,144,233… 
 +  The next term in the Fibonacci sequence, starting from the third, is determined by adding the previous two terms together. The Fibonacci sequence is related to the Golden Ratio because as the sequence grows, the ratio of consecutive terms gradually approaches the Golden Ratio. For example here are the ratios of the successive numbers in the Fibonacci sequence: <br/> 
 +  1/1=1.000000 <br/> 
 +  2/1=2.000000 <br/> 
 +  3/2=1.500000 <br/> 
 +  5/3=1.666666 <br/> 
 +  8/5=1.600000 <br/> 
 +  13/8=1.625000 <br/> 
 +  21/13=1.615385 <br/> 
 +  34/21=1.619048 <br/> 
 +  55/34=1.617647 <br/> 
 +  89/55=1.618182 <br/> 
 +  144/89=1.717978 <br/> 
 +  233/144=1.618056 <br/> 
 +  377/233=1.618026 <br/> 
 +  610/377=1.618037 <br/> 
 +  987/610=1.618033 <br/> 
   
 +  ==The Golden Ratio in Nature== 
   
  ==A Geometric Representation==  +  ===Spirals & Phyllotaxis=== 
   +  AuthorName=Joyce Han 
  ===The Golden Ratio in a Line Segment===
 +  
  [[Image:Golden_segment.jpg400px]][[Image:Animation2.gif]]
 +  
   +  
 
 +  
  The golden number can be defined using a line segment divided into two sections of lengths ''a'' and ''b''. If ''a'' and ''b'' are appropriately chosen, the ratio of ''a'' to ''b'' is the same as the ratio of ''a'' + ''b'' to ''a'' and both ratios are equal to <math>\varphi</math>. The line segment above (left) exhibits the golden proportion. The line segments above (right) are also examples of the golden ratio. In each case,
 +  
   +  
  <math>\frac{{\color{Red}\mathrm{red}}+\color{Blue}\mathrm{blue}}{{\color{Blue}\mathrm{blue}} }= \frac{{\color{Blue}\mathrm{blue}} }{{\color{Red}\mathrm{red}} }= \varphi . </math>
 +  
   +  
  <div id="Jump2"></div>
 +  
  ===The Golden Rectangle===
 +  
  A '''golden rectangle''' is any rectangle where the ratio between the sides is equal to phi. When the sides lengths are proportioned in the golden ratio, the rectangle is said to possess the '''golden proportions.''' A golden rectangle has sides of length <math>\varphi \times r</math> and <math>1 \times r</math> where <math>r</math> can be any constant. Remarkably, when a square with side length equal to the shorter side of the rectangle is cut off from one side of the golden rectangle, the remaining rectangle also exhibits the golden proportions. This continuing pattern is visible in the golden rectangle below.
 +  
  :[[Image:Coloredfinalrectangle1.jpg]]
 +  
   +  
   +  
   +  
  ===Triangles===
 +  
  [[Image:1byrrectangle1.jpg500px]][[Image:Pentagon_final.jpg300px]]
 +  
   +  
  The golden number, <math>\varphi</math>, is used to construct the '''golden triangle,''' an isoceles triangle that has legs of length <math>\varphi \times r</math> and base length of <math>1 \times r</math> where <math>r</math> can be any constant. It is above and to the left. Similarly, the '''golden gnomon''' has base <math>\varphi \times r</math> and legs of length <math>1 \times r</math>. It is shown above and to the right. These triangles can be used to form regular pentagons (pictured above) and <balloon title="A pentagram is a five pointed star made with 5 straight strokes">pentagrams.</balloon>
 +  
   +  
  The pentgram below, generated by the golden triangle and the golden gnomon, has many side lengths proportioned in the golden ratio.
 +  
  :[[Image:Star1.jpg]]
 +  
  :::<math>\frac{{\color{SkyBlue}\mathrm{blue}} }{{\color{Red}\mathrm{red}} } = \frac{{\color{Red}\mathrm{red}} }{{\color{Green}\mathrm{green}} } = \frac{{\color{Green}\mathrm{green}} }{{\color{Magenta}\mathrm{pink}} } = \varphi . </math>
 +  
   +  
  These triangles can be used to form [[Field:Fractals fractals]] and are one of the only ways to tile a plane using '''pentagonal symmetry'''. Pentagonal symmetry is best explained through example. Below, we have two fractal examples of pentagonal symmetry. Images that exhibit pentagonal symmetry have five symmetry axes. This means that we can draw five lines from the image's center, and all resulting divisions are identical.
 +  
   +  
  :[[Image:Penta1.jpg400px]]
 +  
  :[[Image:Pent111.jpg400px]]
 +  
  ImageDesc==An Algebraic Derivation of Phi=
 +  
   +  
  {{SwitchPreviewShowMessage=Click to expandhideMessage=Click to hidePreviewText=How can we derive the value of <math>\varphi</math> from its characteristics as a ratio? We may algebraically solve for the ratio (<math>\varphi</math>) by observing that ratio satisfies the following property by definition:
 +  
  :<math>\frac{b}{a} = \frac{a+b}{b} = \varphi</math>FullText=
 +  
  Let <math> r </math> denote the ratio :
 +  
  :<math>r=\frac{a}{b}=\frac{a+b}{a}</math>.
 +  
   +  
  So
 +  
  :<math>r=\frac{a+b}{a}=1+\frac{b}{a}</math> which can be rewritten as
 +  
   +  
  :<math>1+\cfrac{1}{a/b}=1+\frac{1}{r}</math> thus,
 +  
   +  
  :<math>r=1+\frac{1}{r}</math>
 +  
   +  
  Multiplying both sides by <math>r</math>, we get
 +  
   +  
  :<math> {r}^2=r+1</math>
 +  
   +  
  which can be written as:
 +  
  :<math>r^2  r  1 = 0 </math>.
 +  
   +  
  Applying the <balloon title="load:myContent">quadratic formula</balloon>
 +  
  <span id="myContent" style="display:none">
 +  
  An equation, <math>\frac{b \pm \sqrt {b^24ac}}{2a}</math>, which produces the solutions for equations of form <math>ax^2+bx+c=0</math>
 +  
  </span>, we get <math>r = \frac{1 \pm \sqrt{5}} {2}</math>.
 +  
   +  
  The ratio must be positive because we can not have negative line segments or side lengths. Because the ratio has to be a positive value,
 +  
   +  
  :<math>r=\frac{1 + \sqrt{5}}{2} = 1.61803399... =\varphi</math>.
 +  
  NumChars=500}}
 +  
   +  
   +  
   +  
   +  
  ==Continued Fraction Representation and [[Fibonacci sequenceFibonacci Sequences]]==
 +  
  The golden ratio can also be written as what is called a '''continued fraction,'''a fraction of infinite length whose denominator is a quantity plus a fraction, which latter fraction has a similar denominator, and so on. This is done by using <balloon title="Recursion is the method of substituting an equation into itself">recursion</balloon>.
 +  
   +  
  {{SwitchPreviewShowMessage=Click to expandhideMessage=Click to hidePreviewText= FullText=We have already solved for <math>\varphi</math> using the following equation:
 +  
   +  
  <math>{\varphi}^2{\varphi}1=0</math>.
 +  
   +  
  We can add one to both sides of the equation to get
 +  
   +  
  <math>{\varphi}^2{\varphi}=1</math>.
 +  
   +  
  Factoring this gives
 +  
   +  
  <math> \varphi(\varphi1)=1 </math>.
 +  
   +  
  Dividing by <math>\varphi</math> gives us
 +  
   +  
  <math>\varphi 1= \cfrac{1}{\varphi }</math>.
 +  
   +  
  Adding 1 to both sides gives
 +  
   +  
  <math>\varphi =1+ \cfrac{1}{\varphi }</math>.
 +  
   +  
  Substitute in the entire right side of the equation for <math> \varphi </math> in the bottom of the fraction.
 +  
   +  
  <math>\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{\varphi } }</math>
 +  
   +  
  Substituting in again,
 +  
   +  
  <math>\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\varphi}}}</math>
 +  
   +  
   +  
   +  
  <math>\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\cdots}}}</math>
 +  
   +  
  Continuing this substitution forever, the last infinite form is a continued fraction
 +  
   +  
  If we evaluate truncations of the continued fraction by evaluating only part of the continued fraction, replacing <math>\varphi</math> with 1, we produce the ratios between consecutive terms in the [[Fibonacci sequence]].
 +  
   +  
  <math>\varphi \approx 1 + \cfrac{1}{1} = 2</math>
 +  
   +  
  <math>\varphi \approx 1 + \cfrac{1}{1+\cfrac{1}{1}} = 3/2</math>
 +  
   +  
  <math>\varphi \approx 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1} } } = 5/3</math>
 +  
   +  
  <math>\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</math>
 +  
   +  
  Thus we discover that the golden ratio is approximated in the Fibonacci sequence.
 +  
   +  
  <math>1,1,2,3,5,8,13,21,34,55,89,144...\,</math>
 +  
  <table style="position:relative;left:50px">
 +  
  <tr>
 +  
  <td align="right"><math>1/1</math></td>
 +  
  <td align="center"><math>=</math></td>
 +  
  <td align="left"><math>1</math></td>
 +  
  </tr>
 +  
  <tr>
 +  
  <td align="right"><math>2/1</math></td>
 +  
  <td align="center"><math>=</math></td>
 +  
  <td align="left"><math>2</math></td>
 +  
  </tr>
 +  
  <tr>
 +  
  <td align="right"><math>3/2</math></td>
 +  
  <td align="center"><math>=</math></td>
 +  
  <td align="left"><math>1.5</math></td>
 +  
  </tr>
 +  
  <tr>
 +  
  <td align="right"><math>8/5</math></td>
 +  
  <td align="center"><math>=</math></td>
 +  
  <td align="left"><math>1.6</math></td>
 +  
  </tr>
 +  
  <tr>
 +  
  <td align="right"><math>13/8</math></td>
 +  
  <td align="center"><math>=</math></td>
 +  
  <td align="left"><math>1.625</math></td>
 +  
  </tr>
 +  
  <tr>
 +  
  <td align="right"><math>21/13</math></td>
 +  
  <td align="center"><math>=</math></td>
 +  
  <td align="left"><math>1.61538462...</math></td>
 +  
  </tr>
 +  
  <tr>
 +  
  <td align="right"><math>34/21</math></td>
 +  
  <td align="center"><math>=</math></td>
 +  
  <td align="left"><math>1.61904762...</math></td>
 +  
  </tr>
 +  
  <tr>
 +  
  <td align="right"><math>55/34</math></td>
 +  
  <td align="center"><math>=</math></td>
 +  
  <td align="left"><math>1.61764706...</math></td>
 +  
  </tr>
 +  
  <tr>
 +  
  <td align="right"><math>89/55</math></td>
 +  
  <td align="center"><math>=</math></td>
 +  
  <td align="left"><math>1.61818182...</math></td>
 +  
  </tr>
 +  
  </table>
 +  
   +  
   +  
  <!Alan, here's your old stuff just in case I messed up the numbers or something. Maria
 +  
   +  
  <math>2/1 = 2, 3/2 = 1.5, 8/5 = 1.6, 13/8 = 1.625, 21/13 = 1.61538462...,\,</math>
 +  
  <math>34/21 = 1.61904762..., 55/34=1.61764706..., 89/55 = 1.61818182..., ...\,</math>
 +  
  >
 +  
  <math>\varphi = 1.61803399...\,</math>
 +  
   +  
  As you go farther along in the Fibonacci sequence, the ratio between the consecutive terms approaches the golden ratio. Many realworld applications of the golden ratio are related to the Fibonacci sequence. For more realworld applications of the golden ratio [[Fibonacci Numbersclick here!]]
 +  
   +  
  In fact, we can prove that the ratio between terms in the Fibonacci sequence approaches the golden ratio by using mathematical [[Induction]].
 +  
   +  
  {{Switchlink1=Click to show prooflink2=Click to hide proof1= 2=
 +  
   +  
  Since we have already shown that
 +  
   +  
  <math> \varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\cdots}}} </math>,
 +  
   +  
  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
 +  
  :<math> x_1=1</math>,
 +  
  :<math> x_2=1+\frac{1}{1}=1+\frac{1}{x_1} </math>,
 +  
  :<math> x_3= 1+\frac{1}{1+\frac{1}{1}}=1+\frac{1}{x_2} </math> and so on so that
 +  
  :<math> x_n=1+\frac{1}{x_{n1}} </math>.
 +  
   +  
  These are just the same truncated terms as listed above. Let's also denote the terms of the Fibonacci sequence as
 +  
  :<math> s_n=s_{n1}+s_{n2} </math> where <math>s_1=1</math>,<math>s_2=1</math>,<math>s_3=2</math>,<math>s_4=3</math> etc.
 +  
  <br>
 +  
   +  
  We want to show that
 +  
  :<math> x_n=\frac{s_{n+1}}{s_n} </math> for all n.
 +  
   +  
  First, we establish our [[Inductionbase case]]. We see that
 +  
  :<math> x_1=1=\frac{1}{1}=\frac{s_2}{s_1} </math>, and so the relationship holds for the base case.
 +  
   +  
  Now we assume that
 +  
  :<math> x_k=\frac{s_{k+1}}{s_{k}} </math> for some <math> 1 \leq k < n </math> (This step is the [[Inductioninductive hypothesis]]). We will show that this implies that
 +  
  :<math> x_{k+1}=\frac{s_{(k+1)+1}}{s_{k+1}}=\frac{s_{k+2}}{s_{k+1}} </math>.
 +  
   +  
  <br><br>
 +  
   +  
  By our assumptions about ''x<sub>1</sub>'',''x<sub>2</sub>''...''x<sub>n</sub>'', we have
 +  
   +  
  :<math> x_{k+1}=1+\frac{1}{x_k} </math>.
 +  
   +  
  By our inductive hypothesis, this is equivalent to
 +  
   +  
  :<math>x_{k+1}=1+\frac{1}{\frac{s_{k+1}}{s_{k}}}</math>.
 +  
   +  
  Now we only need to complete some simple algebra to see
 +  
   +  
  :<math> x_{k+1}=1+\frac{s_k}{s_{k+1}} </math>
 +  
   +  
  :<math> x_{k+1}=\frac{s_{k+1}+s_k}{s_{k+1}} </math>
 +  
   +  
  Noting the definition of <math>s_n=s_{n1}+s_{n2}</math>, we see that we have
 +  
   +  
  <math> x_{k+1}=\frac{s_{k+2}}{s_{k+1}} </math>
 +  
   +  
  So by the principle of mathematical induction, we have shown that the terms in our continued fraction are represented by ratios of consecutive Fibonacci numbers.
 +  
   +  
  The exact continued fraction is
 +  
  :<math> x_{\infty} = \lim_{n\rightarrow \infty}\frac{s_{n+1}}{s_n} =\varphi </math>.
 +  
   +  
  }}NumChars=75}}
 +  
   +  
   +  
   +  
  ==Proof of the Golden Ratio's Irrationality==
 +  
   +  
  {{SwitchPreviewShowMessage=Click to expandhideMessage=Click to hidePreviewText= FullText=
 +  
  Remarkably, the Golden Ratio is <balloon title="A number is irrational if it cannot be expressed as the fraction between two integers.">irrational</balloon>, despite the fact that we just proved that is approximated by a ratio of Fibonacci numbers.
 +  
  We will use the method of <balloon title="A method of proving a statement true by assuming that it's false and showing this assumption would logically lead to a statement that is already known to be untrue.">contradiction</balloon> to prove that the golden ratio is irrational.
 +  
   +  
  Suppose <math>\varphi </math> is rational. Then it can be written as fraction in lowest terms <math> \varphi = b/a</math>, where ''a'' and ''b'' are integers.
 +  
   +  
  Our goal is to find a different fraction that is equal to <math> \varphi </math> and is in lower terms. This will be our contradiction that will show that <math> \varphi </math> is irrational.
 +  
   +  
  First note that the definition of <math> \varphi = \frac{b}{a}=\frac{a+b}{b} </math> implies that <math> b > a </math> since clearly <math> b+a>b </math> and the two fractions must be equal.
 +  
   +  
  <br>
 +  
   +  
  Now, since we know
 +  
   +  
  :<math> \frac{b}{a}=\frac{a+b}{b} </math>
 +  
   +  
  we see that <math> b^2=a(a+b) </math> by cross multiplication. Foiling this expression gives us <math> b^2=a^2+ab </math>.
 +  
   +  
  Rearranging this gives us <math> b^2ab=a^2 </math>, which is the same as :<math> b(ba)=a^2 </math>.
 +  
   +  
  Dividing both sides of the equation by ''a(ba)'' gives us
 +  
   +  
  :<math> \frac{b}{a}=\frac{a}{ba} </math>.
 +  
   +  
  Since <math> \varphi=\frac{b}{a} </math>, this means <math> \varphi=\frac{a}{ba} </math>.
 +  
   +  
  Since we have assumed that a and b are integers, we know that ba must also be an integer. Furthermore, since <math> a<b </math>, we know that <math> \frac{a}{ba} </math> must be in lower terms than <math> \frac{b}{a} </math>.
 +  
   +  
  Since we have found a fraction of integers that is equal to <math> \varphi </math>, but is in lower terms than <math> \frac{b}{a} </math>, we have a contradiction: <math> \frac{b}{a} </math> cannot be a fraction of integers in lowest terms. Therefore <math> \varphi </math> cannot be expressed as a fraction of integers and is irrational.
 +  
  NumChars=75}}
 +  
   +  
   +  
  =For More Information=
 +  
   +  
  :*Markowsky. “Misconceptions about the Golden Ratio.” College Mathematics Journal. Vol 23, No 1 (1992). pp 219.
 +  
   +  
   +  
  other=Algebra, Geometry
 +  
  AuthorName=azavez1  +  
  SiteName=The Math Forum
 +  
  SiteURL=http://mathforum.org/mathimages/index.php/Image:180pxPentagramphi.svg.png
 +  
 Field=Algebra   Field=Algebra 
 Field2=Geometry   Field2=Geometry 
  References=<references />
 
  ToDo=animation?
 
 
 
  http://www.metaphorical.net/note/on/golden_ratio
 
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  ImageDesc=
 
  ==Fibonacci Numbers==
 
  ==The Golden Ratio in Nature==
 
  ===Spirals & Phyllotaxis===
 
  Field=Algebra
 
  InProgress=No
 
 }}   }} 