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Difference between revisions of "User:Foxjwill/Proofs"

(→‎Proof that \sqrt{2} is irrational)
(→‎A theorem)
 
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# Therefore <math>p^{1/n}\not\in \mathbb{Q}</math>.
 
# Therefore <math>p^{1/n}\not\in \mathbb{Q}</math>.
 
::'''''Q.E.D.'''''
 
::'''''Q.E.D.'''''
 +
 +
==A theorem==
 +
'''DEFINITION.''' Let <math>a</math> be a chord of some circle <math>C</math>. Then the ''small angle'' of <math>a</math>, denoted <math>S(a)</math>, is the smaller of the two angles cut by <math>a</math>.
 +
 +
'''THEOREM.''' Let <math>p\in \mathbb{R}^+</math>, and let <math>C</math> be a circle. Then there exists a <math>\theta\in \mathbf{R}^+</math> such that for every set A of chords of <math>C</math> with lengths adding to <math>p</math>,
 +
<cmath>
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\sum_{a\in A}S(a) = \theta.
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</cmath>

Latest revision as of 18:13, 13 January 2009

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

  1. Assume that $p^{1/n}$ is rational. Then $\exists a,b \in \mathbb{Z}$ such that $a$ is coprime to $b$ and $p^{1/n}={a \over b}$.
  2. It follows that $p = {a^n \over b^n}$, and that $a^n=pb^n$.
  3. So, by the properties of exponents along with the unique factorization theorem, $p$ divides both $a^n$ and $a$.
  4. Factoring out $p$ from (2), we have $a^n=p^{n-1}b'$ for some $b'\in \mathbb{Z}$.
  5. Therefore $p$ divides $a$.
  6. But this contradicts the assumption that $a$ and $b$ are coprime.
  7. 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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