User:Wim Coenen/sandbox

A simple proof of the famous "Last" theorem of Pierre de Fermat.

Preliminary “calculation”

$s\in Z^{+}$
$t\in Z^{+}$
$s<=t$
$n\in N$
$n>2$

$q^{\prime \prime \prime \prime \prime }={\sqrt[{n}]{\frac {(s+t)^{n}}{t^{n}}}}$
$q^{\prime \prime \prime \prime \prime }={\frac {s+t}{t}}$
$q^{\prime \prime \prime \prime \prime }$ is a rational number.
${\textstyle q^{\prime \prime \prime \prime }={\sqrt[{n}]{\frac {(s+t)^{n}+1}{t^{n}}}}}$
$q^{\prime \prime \prime \prime }$ is an irrational number.
$q^{\prime \prime \prime }={\sqrt[{n}]{\frac {(s+t)^{n}-1}{t^{n}}}}$
$q^{\prime \prime \prime }$is an irrational number.
$q^{\prime \prime }={\sqrt[{n}]{\frac {(s+t)^{n}-{\binom {n}{1}}s^{n-1}t^{1}}{t^{n}}}}$
$q^{\prime \prime }={\sqrt[{n}]{\frac {\sum _{k=0}^{n}{\binom {n}{k}}s^{n-k}t^{k}-{\binom {n}{1}}s^{n-1}t^{1}}{t^{n}}}}$
$q^{\prime \prime }={\sqrt[{n}]{\frac {s^{n}+{\binom {n}{1}}s^{n-1}t^{1}+\sum _{k=2}^{n}{\binom {n}{k}}s^{n-k}t^{k}-{\binom {n}{1}}s^{n-1}t^{1}}{t^{n}}}}$
$q^{\prime \prime }={\sqrt[{n}]{\frac {s^{n}+0+\sum _{k=2}^{n}{\binom {n}{k}}s^{n-k}t^{k}}{t^{n}}}}$
$q^{\prime \prime }$ is an irrational number.

Now let's take it a few notches higher.

$q^{\prime }={\sqrt[{n}]{\frac {(s+t)^{n}-\sum _{k=1}^{n-1}{\binom {n}{k}}s^{n-k}t^{k}}{t^{n}}}}$
$q^{\prime }={\sqrt[{n}]{\frac {\sum _{k=0}^{n}{\binom {n}{k}}s^{n-k}t^{k}-\sum _{k=1}^{n-1}{\binom {n}{k}}s^{n-k}t^{k}}{t^{n}}}}$
$q^{\prime }={\sqrt[{n}]{\frac {s^{n}+\sum _{k=1}^{n-1}{\binom {n}{k}}s^{n-k}t^{k}-\sum _{k=1}^{n-1}{\binom {n}{k}}s^{n-k}t^{k}+t^{n}}{t^{n}}}}$
$q^{\prime }={\sqrt[{n}]{\frac {s^{n}+t^{n}}{t^{n}}}}={\sqrt[{n}]{{\frac {s^{n}}{t^{n}}}+{\frac {t^{n}}{t^{n}}}}}$
$q^{\prime }={\sqrt[{n}]{{\frac {t^{n}}{t^{n}}}+{\frac {s^{n}}{t^{n}}}}}={\sqrt[{n}]{1+{\frac {s^{n}}{t^{n}}}}}$
$q^{\prime }={\sqrt[{n}]{1+{\frac {s^{n}}{t^{n}}}}}$
$\beta =q^{\prime }={\sqrt[{n}]{1+{\frac {s^{n}}{t^{n}}}}}$

For every conceivable positive whole number $s$ and positive whole number $t$, the number $\beta$ is irrefutably an irrational number.
If during the creation of a proof we encounter a number $\alpha$ that resembles the number $\beta$, which we found in our preliminary "calculation", then we immediately know that the number $\alpha$ is an irrational number.

The actual proof of the "last" theorem of Pierre de Fermat.

$x\in \mathrm {Z} ^{+}$
$y\in \mathrm {Z} ^{+}$
$x\leq y$
$n\in N$
$n>2$

$z^{n}=x^{n}+y^{n}$
$z={\sqrt[{n}]{z^{n}}}={\sqrt[{n}]{x^{n}+y^{n}}}$
$z={\sqrt[{n}]{y^{n}{\frac {x^{n}+y^{n}}{y^{n}}}}}={\sqrt[{n}]{y^{n}}}{\sqrt[{n}]{\frac {x^{n}+y^{n}}{y^{n}}}}=y{\sqrt[{n}]{\frac {x^{n}+y^{n}}{y^{n}}}}$
$z=y{\sqrt[{n}]{{\frac {y^{n}}{y^{n}}}+{\frac {x^{n}}{y^{n}}}}}=y{\sqrt[{n}]{1+{\frac {x^{n}}{y^{n}}}}}$
$\alpha ={\sqrt[{n}]{1+{\frac {x^{n}}{y^{n}}}}}$
$\beta ={\sqrt[{n}]{1+{\frac {s^{n}}{t^{n}}}}}$

$\alpha$ is a number similar to the irrational number $\beta$ we found in our preliminary "calculation". We therefor conclude that $\alpha$ is an irrational number. Thereby the number $z$ can't ever be a whole number.

An irrational number $\alpha$ multiplied with a positive whole number $y$ leads to an irrational number $z$.

$z=y{\sqrt[{n}]{1+{\frac {x^{n}}{y^{n}}}}}\Longrightarrow z\in R/Q$

Analytical proof: Binomial development

$(a+b)^{n}=\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n-k}b^{k}$

$(a+b)^{n+1}=(a+b)(a+b)^{n}=(a+b)\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n-k}b^{k}$

$(a+b)^{n+1}=a\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n-k}b^{k}+b\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n-k}b^{k}$

$(a+b)^{n+1}=\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n+1-k}b^{k}+\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n-k}b^{k+1}$

$\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n-k}b^{k+1}=\textstyle \sum _{k=1}^{n+1}\displaystyle {\binom {n}{k-1}}a^{n+1-k}b^{k}$

$(a+b)^{n+1}=a^{n+1}+\textstyle \sum _{k=1}^{n}\displaystyle {\binom {n}{k}}a^{n+1-k}b^{k}+\textstyle \sum _{k=1}^{n}\displaystyle {\binom {n}{k-1}}a^{n+1-k}b^{k}+b^{n+1}$

$(a+b)^{n+1}=a^{n+1}+\textstyle \sum _{k=1}^{n}{\biggl (}\displaystyle {\binom {n}{k}}+{\binom {n}{k-1}}{\biggr )}a^{n+1-k}b^{k}+b^{n+1}$

${\binom {n}{k}}+{\binom {n}{k-1}}={\frac {n!}{(n-k)!k!}}+{\frac {n!}{(n-k+1)!(k-1)!}}$
${\binom {n}{k}}+{\binom {n}{k-1}}={\frac {n!(n-k+1)}{(n-k)!k!(n-k+1)}}+{\frac {n!k}{(n-k+1)!(k-1)!k}}$
${\binom {n}{k}}+{\binom {n}{k-1}}={\frac {{(n+1)}!}{(n+1-k)!k!}}={\binom {n+1}{k}}$

$(a+b)^{n+1}=a^{n+1}+\textstyle \sum _{k=1}^{n}\displaystyle {\binom {n+1}{k}}a^{n+1-k}b^{k}+b^{n+1}$

$(a+b)^{n+1}={\binom {n+1}{0}}a^{n+1}b^{0}+\textstyle \sum _{k=1}^{n}\displaystyle {\binom {n+1}{k}}a^{n+1-k}b^{k}+{\binom {n+1}{n+1}}a^{n+1-(n+1)}b^{n+1}$

$(a+b)^{n+1}=\textstyle \sum _{k=0}^{n+1}\displaystyle {\binom {n+1}{k}}a^{n+1-k}b^{k}$