Difference between revisions of "User:Foxjwill/Proofs"
(New page: ==Proof that <math>\sqrt{2}</math> is irrational== # Assume that <math>p^{1/n}</math> is rational. Then <math>\exists a,b \in \mathbb{Z}</math> such that <math>a</math> is coprime to <math...) |
(→Proof that \sqrt{2} is irrational) |
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| − | ==Proof that <math> | + | ==Proof that <math>p^{1/n}</math>, where <math>p</math> is prime, is irrational== |
# Assume that <math>p^{1/n}</math> is rational. Then <math>\exists a,b \in \mathbb{Z}</math> such that <math>a</math> is coprime to <math>b</math> and <math>p^{1/n}={a \over b}</math>. | # Assume that <math>p^{1/n}</math> is rational. Then <math>\exists a,b \in \mathbb{Z}</math> such that <math>a</math> is coprime to <math>b</math> and <math>p^{1/n}={a \over b}</math>. | ||
# It follows that <math>p = {a^n \over b^n}</math>, and that <math>a^n=pb^n</math>. | # It follows that <math>p = {a^n \over b^n}</math>, and that <math>a^n=pb^n</math>. | ||
Revision as of 15:49, 8 July 2008
Proof that 
, where 
is prime, is irrational
, 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
.
such that 
is coprime to 
and 
.
, and that 
.
and 
for some 
.
.- Q.E.D.
