An image of a rabbit representing the Chinese Zodiac symbol.

Fibonacci Sequence: Lesson 4

 

The Golden Ratio

You’ve seen how to generate the terms of the Fibonacci sequence:

 

Number

Fibonacci Term

1

1

2

1

3

2

4

3

5

5

6

8

7

13

8

21

9

34

10

55

11

89

12

144

Each term in the Fibonacci sequence is the sum of the previous two terms. In this lesson we will be looking at the ratio of consecutive terms to reveal an interesting property of the Fibonacci sequence.

Let’s add a third column to our table to track this ratio.

 

Number

Term

Ratio

1

1

 

2

1

1

3

2

2

4

3

1.5

5

5

1.6667

6

8

1.6000

7

13

1.6250

8

21

1.6154

9

34

1.6190

10

55

1.6176

11

89

1.6182

12

144

1.6180

 

Do you notice that the numbers in the third column approach a particular decimal value? A graph of the data will show this better.

 

As you can see, the ratios converge on a value close to 1.618. This value represent the irrational number phi (pronounced FIE), whose symbol is ɸ. Phi is also known as the Golden Ratio, and it has its origin in geometry. Let’s take a look.

The values for the 11th and 12th terms are 89 and 144. Suppose there’s a rectangle whose sides have those lengths.

Let’s create an 89 x 89 square from the rectangle, like this.


 

Now let’s look at two ratios, based on the lengths, remembering that the longer side of the rectangle is 144.

Notice that the two ratios are nearly identical. So, Phi is defined as the lengths a and b that result in the Golden Ratio.

The Algebraic Solution to the Golden Ratio

The Golden Ratio ɸ is an irrational number. Although it is often written as a non-terminating decimal, we can also write it as a radical expression. Here’s how. 

From the illustration above we know the following:

{"mathml":"<math style=\"font-family:Arial;font-size:36px;\" xmlns=\"http://www.w3.org/1998/Math/MathML\"><mstyle mathsize=\"36px\"><mfrac><mi mathvariant=\"bold\">a</mi><mi mathvariant=\"bold\">b</mi></mfrac><mo mathvariant=\"bold\">=</mo><mfrac><mrow><mi mathvariant=\"bold\">a</mi><mo mathvariant=\"bold\">+</mo><mi mathvariant=\"bold\">b</mi></mrow><mi mathvariant=\"bold\">a</mi></mfrac></mstyle></math>","truncated":false} 

We can turn this into an algebra equation, as shown below.

 

{"mathml":"<math style=\"font-family:Arial;font-size:36px;\" xmlns=\"http://www.w3.org/1998/Math/MathML\"><mstyle mathsize=\"36px\"><mtable columnspacing=\"0px\" columnalign=\"right center left\"><mtr><mtd><msup><mi mathvariant=\"bold\">a</mi><mn mathvariant=\"bold\">2</mn></msup></mtd><mtd><mo mathvariant=\"bold\">=</mo></mtd><mtd><mi mathvariant=\"bold\">b</mi><mstyle mathvariant=\"bold\"><mo stretchy=\"true\">(</mo><mrow><mi>a</mi><mo>+</mo><mi>b</mi></mrow><mo stretchy=\"true\">)</mo></mstyle></mtd></mtr><mtr><mtd><msup><mi mathvariant=\"bold\">a</mi><mn mathvariant=\"bold\">2</mn></msup></mtd><mtd><mo mathvariant=\"bold\">=</mo></mtd><mtd><mi mathvariant=\"bold\">ab</mi><mo mathvariant=\"bold\">+</mo><msup><mi mathvariant=\"bold\">b</mi><mn mathvariant=\"bold\">2</mn></msup></mtd></mtr><mtr><mtd><msup><mi mathvariant=\"bold\">a</mi><mn mathvariant=\"bold\">2</mn></msup><mo mathvariant=\"bold\">-</mo><mi mathvariant=\"bold\">ab</mi><mo mathvariant=\"bold\">-</mo><msup><mi mathvariant=\"bold\">b</mi><mn mathvariant=\"bold\">2</mn></msup></mtd><mtd><mo mathvariant=\"bold\">=</mo></mtd><mtd><mn mathvariant=\"bold\">0</mn></mtd></mtr></mtable></mstyle></math>","truncated":false}

To simplify solving this equation, suppose we let b = 1. Here’s the solution to the equation:

{"mathml":"<math style=\"font-family:Arial;font-size:36px;\" xmlns=\"http://www.w3.org/1998/Math/MathML\"><mstyle mathsize=\"36px\"><mstyle indentalign=\"left\"><msup><mi mathvariant=\"bold\">a</mi><mn mathvariant=\"bold\">2</mn></msup><mo mathvariant=\"bold\">-</mo><mi mathvariant=\"bold\">a</mi><mo mathvariant=\"bold\">-</mo><mn mathvariant=\"bold\">1</mn><mo mathvariant=\"bold\">=</mo><mn mathvariant=\"bold\">0</mn><mspace linebreak=\"newline\"/><mi mathvariant=\"bold\">Use</mi><mo mathvariant=\"bold\">&#xA0;</mo><mi mathvariant=\"bold\">the</mi><mo mathvariant=\"bold\">&#xA0;</mo><mi mathvariant=\"bold\">quadratic</mi><mo mathvariant=\"bold\">&#xA0;</mo><mi mathvariant=\"bold\">formula</mi><mspace linebreak=\"newline\"/><mi mathvariant=\"bold\">a</mi><mo mathvariant=\"bold\">=</mo><mfrac><mrow><mo mathvariant=\"bold\">-</mo><mstyle mathvariant=\"bold\"><mo stretchy=\"true\">(</mo><mrow><mo>-</mo><mn>1</mn></mrow><mo stretchy=\"true\">)</mo></mstyle><mo mathvariant=\"bold\">&#xB1;</mo><msqrt><msup><mstyle mathvariant=\"bold\"><mo stretchy=\"true\">(</mo><mrow><mo>-</mo><mn>1</mn></mrow><mo stretchy=\"true\">)</mo></mstyle><mn mathvariant=\"bold\">2</mn></msup><mo mathvariant=\"bold\">-</mo><mn mathvariant=\"bold\">4</mn><mstyle mathvariant=\"bold\"><mo stretchy=\"true\">(</mo><mn>1</mn><mo stretchy=\"true\">)</mo></mstyle><mstyle mathvariant=\"bold\"><mo stretchy=\"true\">(</mo><mrow><mo>-</mo><mn>1</mn></mrow><mo stretchy=\"true\">)</mo></mstyle></msqrt></mrow><mn mathvariant=\"bold\">2</mn></mfrac><mspace linebreak=\"newline\"/><mo mathvariant=\"bold\">=</mo><mfrac><mrow><mn mathvariant=\"bold\">1</mn><mo mathvariant=\"bold\">&#xB1;</mo><msqrt><mn mathvariant=\"bold\">5</mn></msqrt></mrow><mn mathvariant=\"bold\">2</mn></mfrac></mstyle></mstyle></math>","truncated":false}

We only use the positive root, which means that:

{"mathml":"<math style=\"font-family:stix;font-size:36px;\" xmlns=\"http://www.w3.org/1998/Math/MathML\"><mstyle mathsize=\"36px\"><mi>&#x3D5;</mi><mo>=</mo><mfrac><mrow><mn>1</mn><mo>+</mo><msqrt><mn>5</mn></msqrt></mrow><mn>2</mn></mfrac></mstyle></math>","truncated":false}

Plug this expression into a calculator or spreadsheet and you’ll see the non-terminating decimal version of ɸ.