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Where φ is the golden ratio. $\frac{1 + \sqrt 5}{2}$

How can I use basic mathematical skills to show that:

$ψ = 1 - φ$

This is from wikipedia. I do not think I have done anything quite like this before, so I hope that someone can show me in more detailed steps how one can show these relationships. Thank you!

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You just do the arithmetic. – Andrew D Jan 8 '13 at 1:31
up vote 2 down vote accepted

$$\psi = \dfrac{1-\sqrt{5}}2 = \dfrac{2-(1+\sqrt{5})}2 = \dfrac22 - \dfrac{1+\sqrt{5}}2 = 1 - \phi$$ or $$\psi = \dfrac{1-\sqrt{5}}2 = \dfrac{1-\sqrt{5}}2 \times \dfrac{1+\sqrt{5}}{1+\sqrt{5}} = \dfrac{1^2 - 5}{2(1+\sqrt{5})} = -\dfrac{4}{2(1+\sqrt{5})} = -\dfrac2{1+\sqrt{5}} = - \dfrac1{\phi} = 1- \phi$$ since $\dfrac1{\phi} = \phi - 1$.

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Thank you, that does not make me feel dumb at all! – Leonardo Jan 8 '13 at 1:32

The best answer (imho) is already posted. But whenever $\alpha$ and $\beta$ are the two roots of a quadratic polynomial $x^2+b\,x+c$, then two things happen:

  1. $\alpha\cdot\beta=c$
  2. $\alpha+\beta=-b$

Since $\varphi$ and $\psi$ are roots of $x^2-x-1$, the second fact implies $$\varphi+\psi=1$$ which implies $$\psi=1-\varphi$$

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Simply do the arithmetic, as adviced by Andrew:

$$1-\phi=1-\frac{1+\sqrt 5}{2}=\frac{2-1-\sqrt 5}{2}=\frac{1-\sqrt 5}{2}=\psi$$

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