I read this post (Adi Dani) he wrote $\phi=\dfrac{-1+\sqrt{5}}{2}$ but Wikipedia shows that there is $+1$ not $-1$ involved in Numerator, please clear this confusion of mine that who is correct or it can be written in that way too.

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    $\begingroup$ It should be +1 $\endgroup$ – Simply Beautiful Art Jan 30 '17 at 14:10
  • $\begingroup$ @SimplyBeautifulArt but he gained 5 votes for that answer, how come ! $\endgroup$ – mathlover Jan 30 '17 at 14:11
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    $\begingroup$ @mathlover This is when peer review fails! Correct him please! $\endgroup$ – Pythagoricus Jan 30 '17 at 14:14
  • $\begingroup$ @mathlover We all make mistakes, and the difference between a $+1$ and a $-1$ is small. Also, make note that he wrote the above line wrong as well, as it should be $\phi^2=1\color{red}+\phi$ $\endgroup$ – Simply Beautiful Art Jan 30 '17 at 14:20

In a sense, either is correct. $\frac{1 + \sqrt{5}}{2} \approx 1.618\ldots$, while $\frac{-1 + \sqrt{5}}{2} \approx 0.618\ldots = \frac{1}{1.618\ldots}$. Most people would say that the first one is the golden ratio, but the thing about a ratio is that it can be approached from either side - for example, the ratio between $6$ and $3$ can be thought of as either $2$ or $\frac{1}{2}$, depending on which number you think of as "first".

As far as I know, $\frac{1+\sqrt{5}}{2}$ is the one that is usually called the "Golden Ratio"; but everything interesting to say about $\frac{1+\sqrt{5}}{2}$ is also true of $\frac{-1+\sqrt{5}}{2}$, so the difference is not important.


The key to understanding the golden ratio is self-similarity. Ponder the following question.

Let $R$ be a rectangle of height $1$ and length $x>1$ such that the following holds: if we take away a square of side $1$ from $R$, then the resulting rectangle is similar to $R$ (in the sense that the ratio of the sides of the new rectangle is equal to the ratio of the sides of $R$). What is $x$?

The solution to this problem is the golden ratio $x = \frac{1+\sqrt{5}}{2}$. You can immediately see why it played a prominent role already in greek architecture.

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    $\begingroup$ I don't think you really hit on the question $\endgroup$ – Simply Beautiful Art Jan 30 '17 at 14:20

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