# The Golden Ratio

Since the days of old, artists as well as mathematicians have known that there is a special, aesthetically pleasing rectangle with width $1$, length $x$, and the following property:
When a square of side $1$ is removed, the rectangle that remains has the same proportions as the original rectangle. Since the new rectangle has a width $x-1$ and a length of $1$, the equivalent of the proportions means that
$$\dfrac{x-1}{1} = \dfrac{1}{x}$$

Golden Ratio is made from a rectangle to keep the same ratio by taking squares

The number $x$, which is called the golden ratio, must therefore satisfy the equation
$$x^2 – x – 1 = 0 \cdots (1)$$
This equation can be solved by the quadratic formula $x=\dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}$
\begin{align} x &=\dfrac{-(-1) \pm \sqrt{(-1)^2 – 4 \times 1 \times (-1)}}{2 \times 1} \\ &= \dfrac{1 \pm \sqrt{5}}{2} \end{align}
We take the positive root of this equation.
$$x = \dfrac{1 + \sqrt{5}}{2} \approx 1.168 \cdots$$
Now, we try to find a rational number that approximates the irrational number $x$. To do this, we note that the equation $(1)$ allows us to write that
\begin{align} \displaystyle x^2 – x – 1 &= 0 \cdots (1)\\ x – 1 – \dfrac{1}{x} &= 0 &\textit{divide by }x \\ x &= 1 + \dfrac{1}{x} &\textit{substitute itself} \\ x &= 1 + \dfrac{1}{1 + \dfrac{1}{x}} \\ x &= 1 + \dfrac{1}{1 + \dfrac{1}{1 + \dfrac{1}{x}}} \\ x &= 1 + \dfrac{1}{1 + \dfrac{1}{1 + \dfrac{1}{1 + \cdots }}} \\ \end{align}
By carrying out further substitutions of the same kind and by omitting the last $\dfrac{1}{x}$, we arrive the following serious numbers;
$$1, 1+\dfrac{1}{1}, 1+\dfrac{1}{1+\dfrac{1}{1}}, 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1}}}, 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1}}}},\cdots$$

To present this phenomenon of convergence, we can write the series of the Golden ratio.

In practice, it is tedious to continue to calculate these numbers indefinitely, but the Golden Ratio is nevertheless possible to see that they get closer and closer to the exact value of
$$\dfrac{1+\sqrt{5}}{2}$$
If the Golden Ratio were rational, then it would be the ratio of sides of a rectangle with integer sides. But it would also be a ratio of integer sides of the smaller rectangle obtained by deleting a square. The sequence of decreasing integer side lengths formed by deleting squares cannot be continued indefinitely because the integers have a lower bound, so the Golden Ratio cannot be rational.