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

DEFINITION. Let $a$ be a chord of some circle $C$ . Then the small angle of $a$ , denoted $S(a)$ , is the smaller of the two angles cut by $a$ .

THEOREM. Let $p\in \mathbb{R}^+$ , and let $C$ be a circle. Then there exists a $\theta\in \mathbf{R}^+$ such that for every set A of chords of $C$ with lengths adding to $p$ , $\sum_{a\in A}S(a) = \theta.$

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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