¿como comprobar el número áureo?

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¿como comprobar el número áureo?

en wikipedia ponen la siguiente expresión

$\varphi^2 = \frac{ ( 1 + \sqrt{5} )^2 }{2^2} = \frac{1 + 2\sqrt{5} + 5}{2^2} = \frac{6 + 2\sqrt{5}}{2^2} = \frac{3 + \sqrt{5}}{2}$

$\varphi + 1 = \frac{1 + \sqrt{5}}{2} + \frac{2}{2} = \frac{3 + \sqrt{5}}{2}$

pero no se porque $${\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{5}}}}\right)}^{{\mathtt{2}}}$$ termina siendo 1 + 2√5 + 5

Guest 24 sept 2014

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1⁺⁰ Answers

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Sucede que desarrollaron el binomio al cuadrado, es decir, tomaron la expresión

$${\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{5}}}}\right)}^{{\mathtt{2}}} = \left(\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{5}}}}\right){\mathtt{\,\times\,}}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{5}}}}\right)\right)$$ y realizaron propiedad distributiva

$$\left(\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{1}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{1}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{1}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}\right)\right)$$ que al desarrollar da $${\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}} = {\mathtt{1}}{\mathtt{\,\small\textbf+\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{5}}$$ lo cual por último termina dando

$${\mathtt{6}}{\mathtt{\,\small\textbf+\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}\right)$$

Invitado 26 sept 2014

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