Technical Algebra
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Dr.Peterson
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[MATH]\phi[/MATH] is not a variable; it is a very specific number. It is defined by [MATH]\phi^2 = \phi + 1[/MATH].
So to transform the LHS to the RHS, just use that definition.
So to transform the LHS to the RHS, just use that definition.
They use [MATH]\phi[/MATH] as a common name for the golden ratio. I've also seen it described as [MATH]\phi = \frac{1}{\phi - 1}[/MATH], which is the proportion of partitioning a golden rectangle into a square and smaller golden rectangle.
[MATH]\phi = \frac{1}{\phi - 1}[/MATH]
[MATH]\phi(\phi - 1) = 1[/MATH]
[MATH]\phi^2 - \phi = 1[/MATH] (same as [MATH]\phi^2 = \phi + 1[/MATH])
[MATH]\left(\phi - \frac{1}{2}\right)^2 = 1 + \frac{1}{4}[/MATH]
[MATH]\phi - \frac{1}{2} = \sqrt{\frac{5}{4}}[/MATH]
[MATH]\phi = \frac{1}{2} + \frac{1}{2}\sqrt{5}[/MATH]
[MATH]\phi = \frac{1 + \sqrt{5}}{2} \approx 1.618[/MATH]
firemath
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Beautiful work, @Mr. Bland!
[MATH]\phi - \frac{1}{2} = \sqrt{\frac{5}{4}}[/MATH] \(\displaystyle \ \ \ \)
Starting with this line, you would need the plus or minus symbol:
[MATH]\phi - \frac{1}{2} = \pm \sqrt{\frac{5}{4}}[/MATH] \(\displaystyle \ \ \ \)
At the end of the simplification steps, it would need to be decided which branch to choose for the \(\displaystyle \ \phi \ \) expression.
Dr.Peterson
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Yes, I was a little loose in saying that [MATH]\phi[/MATH] is defined by [MATH]\phi^2 = \phi + 1[/MATH], since that defines two numbers. It can be defined as the positive solution to that equation; the negative solution is sometimes called [MATH]-\Phi = -0.618...[/MATH], where [MATH]\Phi[/MATH] is [MATH]1/\phi[/MATH].
The original definition is as a ratio (of positive numbers), which implies all this.
The original definition is as a ratio (of positive numbers), which implies all this.
There is but one golden ratio.