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Question Number 56772 by Hassen_Timol last updated on 23/Mar/19

$$\mathrm{Given}\mathrm{that}:$$$$\frac{1}{\mathrm{\Phi }}=\frac{\mathrm{\Phi }}{1+\mathrm{\Phi }}$$$$$$$$\mathrm{Prove}\mathrm{that}:withoutusingtheexact$$$$valueof\mathrm{\Phi }...$$$$\frac{1}{\mathrm{\Phi }}=\mathrm{\Phi }-1$$$$$$$$$$$$\mathrm{Thank}\mathrm{you}$$

Answered by JDamian last updated on 23/Mar/19

$$$$$$\frac{1}{\mathrm{\Phi }}=\frac{\mathrm{\Phi }}{1+\mathrm{\Phi }}$$$$1+\mathrm{\Phi }={\mathrm{\Phi }}^2$$$$1={\mathrm{\Phi }}^2-\mathrm{\Phi }$$$$1=\mathrm{\Phi }(\mathrm{\Phi }-1)$$$$\frac{1}{\mathrm{\Phi }}=\mathrm{\Phi }-1$$

Commented by Hassen_Timol last updated on 23/Mar/19

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}$$

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