User:Foxjwill/Proofs

Proof that $p^{1/n}$ , where $p$ is prime, is irrational

Assume that $p^{1/n}$ is rational. Then $\exists a,b \in \mathbb{Z}$ such that $a$ is coprime to $b$ and $p^{1/n}={a \over b}$ .
It follows that $p = {a^n \over b^n}$ , and that $a^n=pb^n$ .
So, by the properties of exponents along with the unique factorization theorem, $p$ divides both $a^n$ and $a$ .
Factoring out $p$ from (2), we have $a^n=p^{n-1}b'$ for some $b'\in \mathbb{Z}$ .
Therefore $p$ divides $a$ .
But this contradicts the assumption that $a$ and $b$ are coprime.
Therefore $p^{1/n}\not\in \mathbb{Q}$ .

Q.E.D.

A theorem

THEOREM. Let $C$ be a circle of radius $r$ , let $A$ be the set of chords of $C$ , for all $p\in \mathbb{R}^+$ , let $S_p\equiv \{B\subset A|B\mbox{ is finite}, \sum_{b\in B}|b|=p\}$ . Then for all $p\in \mathbb{R}^+$ , there exists an angle $\phi\in \mathbb{R}^+$ such that for all $B\in S_p$ , there exists a positive integer $k$ such that for all sets $\Theta\equiv \{\mbox{angle }\theta| \theta \in \Theta\mbox{ iff there exists a }b\in B\mbox{ that cuts },\theta\mbox{ and for all }b\in B,b\mbox{ cuts exactly one element in }\Theta\}$ , $\sum_{\theta\in \Theta}\theta = \phi$

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User:Foxjwill/Proofs

Proof that $p^{1/n}$ , where $p$ is prime, is irrational

A theorem

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User:Foxjwill/Proofs

Proof that , where is prime, is irrational

A theorem

Proof that $p^{1/n}$ , where $p$ is prime, is irrational