Phi

Basic Properties

Phi, denoted \(\varphi\), can be represented as follows: \[\varphi - \frac{1}{\varphi} = 1\] In other words, phi differs from its reciprocal by 1. It happens to be the only number for which this is true, which leads to some interesting behavior. For example, if \(\varphi = 1 + 1/\varphi\), we can substitute this identity wherever it occurs. So \[ 1 + \frac{1}{\varphi} = 1 + \cfrac{1}{1 + \cfrac{1}{\varphi}}\] And you can continue this substitution for as long as you’d like. What this means is that \(\varphi\) can be represented as an infinite continuing fraction: \[1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\dots}}}}}\]

All infinite continuing fractions are irrational, so this implies that \(\varphi\) is irrational. Now, take the original identity from the first paragraph and multiply it by \(\varphi\), yielding \(\varphi^{2}- 1 = \varphi\). We rearrange terms and take the square root of both sides. Then, we once again perform substitutions given what we already know. \[\begin{align}\varphi &= \sqrt{1 + \varphi}\\ &= \sqrt{1 + \sqrt{1 + \varphi}}\\ &= \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{\ldots }}}}\end{align}\]

And we could similarly show that \[ \frac{1}{\varphi} = \sqrt{1 - \sqrt{1 - \sqrt{1 - \sqrt{\ldots}}}} \]

Phi is commonly called the “golden ratio”. Geometrically, it is the ratio obtained between a shorter segment and a longer segment such that the shorter segment is related to the longer segment in the same way that the longer segment is related to the whole. Symbolically: \[\frac{L}{S} = \frac{L+S}{L} = 1 + \frac{S}{L}\]

Displayed below, you will see what is referred to as a “golden rectangle”. Assume the first square has a side of \(1\), and that the golden rectangle is of dimensions \(1 \times 1\cdot\varphi\). Hover your mouse right and left to control the number of iterations. Each additional square drawn will take a side of \(1/\varphi^{n}\), where \(n\) is the iteration. You will see many self-similar golden rectangles appear, and arcs of radius \(1/\varphi^{n}\) will be drawn in each square, generating what is called a logarithmic spiral. Click and hold the mouse to hide the nested iterations.

Once again, examine the equality \(\varphi = 1+ 1/\varphi\). Substitute \(x\) for \(\varphi\), and we obtain \(x = 1 + 1/x\). Multiply this result by \(x\): \[x^{2} = x + 1\] Hence, the characteristic equation for \(\varphi\) is: \[x^{2} - x -1 = 0\] We can solve this like any other quadratic function, noticing immediately that the function will have two real roots, given that the determinant (\(b^{2}-4ac\)) is positive. Our roots will be given by: \[\frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)}\] So on one hand we have \[\frac{1+\sqrt{5}}{2} = \varphi = 1.6180339\dots\] and on the other hand we have \[\frac{1-\sqrt{5}}{2} = -\frac{1}{\varphi} = -0.6180339\dots\] Similarly, we can instead substitute \(x\) for \(1/\varphi\). In this case, our characteristic equation turns out to be slightly different. The equation \(\varphi - 1/\varphi = 1\) becomes \(1/x - x = 1\), and once we multiply by \(x\) and move terms, we obtain the equation \(x^{2} + x - 1\). The roots of this equation turn out to be \(-\varphi\) and \(1/\varphi\). (For clarification on some of these results, see ¹ ).

Now we prove that \(\varphi\) is irrational. A loose sketch of a traditional proof might go as follows:

1. It suffices to show that \(\sqrt{5}\) is irrational . . .
2. Proof by contradiction: suppose not. Suppose \(\sqrt{5}\) is rational so that by definition \(\sqrt{5} = p/q\) for some \(p,q\in\mathbb{Z}\), assuming reduced form.
3. . . .
4. . . .
5. . . .
6. . . . Perform algebraic and definitional manipulation (\(a|b\) means \(b = na\) for some \(n\)) until you induce contradiction by showing the fraction introduced in (2) was not in reduced form. The final step should yield \(5|q^{2}\).

A more interesting and intuitive approach (full proof):

1. It suffices to show that \(\sqrt{5}\) is irrational.
2. Proof by contradiction: suppose not. Suppose \(\sqrt{5}\) is rational. So by definition \(\sqrt{5} = p/q\) for some \(p,q\in\mathbb{Z}\), assuming \(p/q\) cannot be further reduced.
3. Now square both sides and we have \(5 = p^{2}/q^{2}\).
4. By the fundamental theorem of arithmetic, every integer is a product of prime numbers or is prime itself, and each integer has a unique prime factorization. Therefore, \(p\) and \(q\) each have unique prime factorizations such that without loss of generality, \(p = a_{1}a_{2}a_{3}\cdots a_{n}\) for some \(n\) number of primes \(a_{1},a_{2},\ldots,a_{n}\).
5. So \(p^{2}\) and \(q^{2}\) will both be products of pairs of primes, because WLOG \(p\cdot p = (a_{1}a_{2}\cdots a_{n}) \cdot (a_{1}a_{2}\cdots a_{n}) = (a_{1}a_{1})\cdot(a_{2}a_{2})\cdots (a_{n}a_{n})\).
6. Now we multiply the original equation by \(q^{2}\) to obtain \(5q^{2} = p^{2}\). We see that one side will be a product of pairs of primes, whereas the other side will have one misfit prime, \(5\). So this equality is false, and therefore \(\sqrt{5}\) is not rational. \(\square\)

Phi and Fibonacci Numbers

We recall that \(\varphi^{2} = \varphi + 1\). In addition to phi being the only number that differs from its reciprocal by 1, it is also the only number that differs from its square by one. Now we see, following from this, that phi cubed will be given by \[\begin{align}\varphi^{3} &= \varphi(\varphi^{2}) \\&= \varphi(\varphi + 1) \\&= \varphi^{2} + \varphi\\ &= 2\varphi + 1\end{align}\]

And we find \(\varphi^{4}\):

\[\begin{align}\varphi^{4} &= \varphi(\varphi^{3})\\ &= \varphi(2\varphi + 1)\\ &= 2\varphi^{2} + \varphi\\ &= 2(\varphi + 1) + \varphi\\ &= 3\varphi + 2\end{align}\]

Continuing on to higher powers, we will come to see that

\[\begin{align}\varphi^{2} &= 1\varphi + 1\\ \varphi^{3} &= 2\varphi + 1\\ \varphi^{4} &= 3\varphi + 2\\ \varphi^{5} &= 5\varphi + 3\\ \varphi^{6} &= 8\varphi + 5\\ \varphi^{7} &= 13\varphi + 8\\ \varphi^{8} &= 21\varphi + 13\\ \varphi^{9} &= 34\varphi + 21\\ \varphi^{10} &= 55\varphi + 34\\ \varphi^{11} &= 89\varphi + 55\\ etc.\end{align}\]

Strangely enough, the natural numbers in these expressions are precisely the numbers of the Fibonacci sequence. We recall the definition of the Fibonacci sequence:

\[F(n) = \begin{cases} 0 &\text{if } n = 0 \\ 1 & \text{if } n = 1 \\ F(n-1) + F(n-2) &\text{otherwise } \end{cases}\]

So we have 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, and so on. As well as this connection to power of phi, it can be shown that ratio of consecutive Fibonacci numbers comes increasingly close to phi as the sequence continues.

\[\begin{align}\frac{F_{2}}{F_{1}} &= \frac{1}{1} = 1\\ \frac{F_{3}}{F_{2}} &= \frac{2}{1} = 2\\ \frac{F_{4}}{F_{3}} &= \frac{3}{2} = 1.5\\ \frac{F_{5}}{F_{4}} &= \frac{5}{3} = 1.\overline{6}\\ \frac{F_{6}}{F_{5}} &= \frac{8}{5} = 1.6\\ \frac{F_{7}}{F_{6}} &= \frac{13}{8} = 1.625\\ \frac{F_{8}}{F_{7}} &= \frac{21}{13} = 1.\overline{615384}\end{align}\]

Why does this occur? The answer: any ratio of consecutive Fibonacci numbers can be given in the following way:

\[\frac{F_{n}}{F_{n-1}} = 1 + \cfrac{1}{\cfrac{F_{n-1}}{F_{n-2}}}\]

Try this for yourself to see that it is correct. Now, examine this specific case:

\[\begin{align}\frac{F_{6}}{F_{5}} = \frac{8}{5} &= 1+ \cfrac{1}{\cfrac{5}{3}}\\ &=1+\cfrac{1}{1+\cfrac{1}{\cfrac{3}{2}}}\\ &=1+\cfrac{1}{1+\cfrac{1}{1+ \cfrac{1}{\cfrac{2}{1}}}}\\ &=1+\cfrac{1}{1+\cfrac{1}{1+ \cfrac{1}{1+ \cfrac{1}{1}}}}\end{align}\]

And as we established earlier, \(\varphi\) is an infinite continuing fraction consisting of this very form. So any ratio of consecutive Fibonacci numbers will ultimately be equivalent to some part of this infinite continuing fraction, and the approximation comes closer and close to \(\varphi\) as the sequence goes on. This fact is given explicitly by

\[\lim_{n\to\infty}\frac{F_{n}}{F_{n-1}} = \varphi\]

And it can also be shown that

\[\lim_{n\to\infty}\frac{F_{n-1}}{F_{n}} = \frac{1}{\varphi}\]

• A question for the reader: is it possible that elements in any recursively defined sequence can approximate \(\varphi\)? See here.

Phi and the Fibonacci sequence are both commonly observed in natural patterns. A fascinating example of a natural pattern related to these objects is phyllotaxis: the arrangment of leaves in plants. Particularly, spiral growth patterns, such as those most clearly seen in sunflowers.

Phyllotaxis

On the left is an image of a sunflower head. On the right is an image generated from a computer program that simply draws circles with a position and size given according to a simple, fixed rule. The rule for forming these intriguing patterns is not at all complicated. The pattern in question is called parastichy, and it appears because of the tendency of the human eye to connect near objects into continuous lines. We see double spirals in some of these arrangments because the objects they comprise have multiple “neighbors”. But what accounts for the specific manner in which these relationships between neighboring growths emerge? They occur when the divergence angle of the plant’s primordia involves, in some way, phi. Precisely, the “golden angle” is \(360 (1 - 1/\varphi) \approx 137.507 \text{ degrees}\).

(in progress)

Phi and Fibonacci Numbers (in progress)

“Most irrational” number (in progress)

Phyllotaxis (in progress)