power proof

hey,guys,i'm practicing proofs .so, i'll be writing some proofs for the community to point out flaws, and many others to learn\[\] this are all about the golden ratios. often written as$ \phi \quad or\quad\varphi=\dfrac{1+\sqrt{5}}{2}$ first is the power proof, i.e \[\phi^n=F_{n}\phi+F_{n-1}\] where$F_{n}$ is the nth Fibonacci number, which is defined as \[F_n=\begin{cases} 1,& n=1\\1,&n=2\\F_{n-1}+F_{n-2},&n\geq 3\end{cases}\] \[the\quad proof\] we see that it is satisfied at n=1,2. we know that \[\phi^2=\phi+1\longrightarrow \phi^n=\phi^{n-1}+\phi^{n-2}\] since $\phi^{n-1}=F_{n-1}\phi+F_{n-2}, \phi^{n-2}=F_{n-2}\phi+F_{n-3}$, \[\begin{array} &\phi^n=\phi^{n-1}+\phi^{n-2}\\ \phi^n=F_{n-1}\phi+F_{n-2}+F_{n-2}\phi+F_{n-3}\\ \phi^n=(F_{n-1}+F_{n-2})\phi+(F_{n-2}+F_{n-3})\\ \phi^n=F_{n}\phi+F_{n-1} \end{array}\] since it satisfies all three at n,n-1,n-2.it must satisfy for all number. induction, hence proved.

#NumberTheory #GoldenRatio

Easy Math Editor

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Comments

Hello Aareyan,

Your proof is perfectly okay and completely free of any flaw.

You might also want to check out the articles we have on induction and its variants.

Good luck to you in your quest of proving things!

Mursalin Habib - 6 years, 3 months ago

thanks,sir. i was worried about 2 cases.

Aareyan Manzoor - 6 years, 3 months ago

You are using a variant called 'Strong induction' that uses more statements in the inductive hypothesis.

And 'sir' sounds really formal. You can call me Mursalin if you want.

Mursalin Habib - 6 years, 3 months ago

@Mursalin Habib – thanks,sir. learned something new, and sir, i learned English through formal speaking.,so it is an habit.

Aareyan Manzoor - 6 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$