Difference between revisions of "User:Foxjwill/Proofs"
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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== | ||
+ | '''THEOREM.''' Let <math>C</math> be a circle of radius <math>r</math>, let <math>A</math> be the set of chords of <math>C</math>, for all <math>p\in \mathbb{R}^+</math>, let <math>S_p\equiv \{B\subset A|B\mbox{ is finite}, \sum_{b\in B}|b|=p\}</math>. <!--Additionally, let <math>\Theta\equiv \{\mbox{angle }\theta| \mbox{for all }a\in</math>--> Then for all <math>p\in \mathbb{R}^+</math>, there exists an angle <math>\phi\in \mathbb{R}^+</math> such that for all <math>B\in S_p</math>, there exists a positive integer <math>k</math> such that for all sets <math>\Theta\equiv \{\mbox{angle }\theta| \mbox{for all }b\in B,b\mbox{ cuts exactly one element in }\Theta\}</math>, | ||
+ | \[ | ||
+ | \sum_{\theta\in \Theta}\theta = \phi | ||
+ | \] |
Revision as of 14:12, 13 January 2009
Proof that
, where
is prime, is irrational
- Assume that
is rational. Then
such that
is coprime to
and
.
- It follows that
, and that
.
- So, by the properties of exponents along with the unique factorization theorem,
divides both
and
.
- Factoring out
from (2), we have
for some
.
- Therefore
divides
.
- But this contradicts the assumption that
and
are coprime.
- Therefore
.
- Q.E.D.
A theorem
THEOREM. Let be a circle of radius
, let
be the set of chords of
, for all
, let
. Then for all
, there exists an angle
such that for all
, there exists a positive integer
such that for all sets
,
\[
\sum_{\theta\in \Theta}\theta = \phi
\]