Gebruiker:Wim Coenen/Kladblok

Gedachtenexperiment: Fermat

$A,p$ zijn gehele getallen.

$p$ is een priemgetal.

Beschouw $A=p+1$ als representant van alle gehele getallen die niet deelbaar zijn door $p$.

$\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {(p+1)^{p-1}-1}{p}}$

$\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-1-k}1^{k}-1}{p}}$

$\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {\sum _{k=0}^{p-2}{\binom {p-1}{k}}p^{p-1-k}1^{k}+1-1}{p}}$

$\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {\sum _{k=0}^{p-2}{\binom {p-1}{k}}p^{p-1-k}1^{k}}{p}}$

$\displaystyle {\frac {A^{p-1}-1}{p}}=\sum _{k=0}^{p-2}{\binom {p-1}{k}}p^{p-2-k}1^{k}$

Voor alle gehele getallen $A$ die niet deelbaar zijn door $p$ geldt: $\displaystyle A^{p-1}-1$ is deelbaar door $p$.

Fermat's little theorem unravelled.

$A,B,p,q,r$ are whole numbers.

p is a prime number.

Here is the proof that $A^{p-1}-1$ is divisible by $p$. ${\frac {A^{p-1}-1}{p}}=?$

${\frac {A^{p-1}-1^{p-1}}{p}}={\frac {(p+q)^{p-1}-\{p+(-p+1)\}^{p-1}}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-1-k}q^{k}-\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-1-k}(-p+1)^{k}}{p}}$

${\frac {A^{p-1}-1}{p}}=\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-2-k}q^{k}-\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-2-k}(-p+1)^{k}$

${\frac {A^{p-1}-1}{p}}=\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-2-k}\{q^{k}-(-p+1)^{k}\}$

The following is worth mentioning..

${\binom {p-1}{p-1}}p^{p-2-(p-1)}\{q^{p-1}-(-p+1)^{p-1}\}={\frac {{\binom {p-1}{p-1}}\{q^{p-1}-(-p+1)^{p-1}\}}{p}}$

$\{q^{p-1}-(-p+1)^{p-1}\}$ is divisible by $p$.

De kleine stelling van Fermat ontrafeld.

Hier volgt het bewijs dat $A^{p-1}-1$ deelbaar is door $p$. ${\frac {A^{p-1}-1}{p}}=?$

${\frac {A^{p-1}-1^{p-1}}{p}}={\frac {A^{p-1}-B^{p-1}}{p}}$

${\frac {A^{p-1}-B^{p-1}}{p}}={\frac {(p+q)^{p-1}-(p+r)^{p-1}}{p}}$

${\frac {A^{p-1}-1^{p-1}}{p}}={\frac {(p+q)^{p-1}-\{p+(-p+1)\}^{p-1}}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-1-k}q^{k}-\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-1-k}(-p+1)^{k}}{p}}$

${\frac {A^{p-1}-1}{p}}=\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-2-k}q^{k}-\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-2-k}(-p+1)^{k}$

${\frac {A^{p-1}-1}{p}}=\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-2-k}\{q^{k}-(-p+1)^{k}\}$

${\frac {A^{p-1}-1}{p}}=\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-2-k}\{(A-p)^{k}-(-p+1)^{k}\}$

Het volgende is het vermelden waard.

${\binom {p-1}{p-1}}p^{p-2-(p-1)}\{q^{p-1}-(-p+1)^{p-1}\}={\frac {{\binom {p-1}{p-1}}\{q^{p-1}-(-p+1)^{p-1}\}}{p}}$

$\{q^{p-1}-(-p+1)^{p-1}\}$ is deelbaar door $p$.

Is $A^{p-1}-B^{p-1}$ deelbaar door $p$?

${\frac {A^{p-1}-B^{p-1}}{p}}={\frac {A^{p-1}-1-(B^{p-1}-1)}{p}}$

$\{A^{p-1}-1-(B^{p-1}-1)\}{\bmod {p}}=(A^{p-1}-1){\bmod {p}}-(B^{p-1}-1){\bmod {p}}$

$(A^{p-1}-1){\bmod {p}}-(B^{p-1}-1){\bmod {p}}=0{\bmod {p}}-0{\bmod {p}}=0{\bmod {p}}$

Is a tot de macht p equivalent aan a mod p?

$a$ en $p$ zijn gehele getallen.

$p$ is een priemgetal.

$a=2$

$\displaystyle {\frac {a^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}1^{p-1-k}1^{k}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}$

$\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5,\ldots }^{p-2}{\binom {p}{k+1}}=\sum _{k=1,3,5,\ldots }^{p-2}{\frac {p(p-1)!}{(p-(k+1))!(k+1)!}}$

${\frac {2^{p-1}-1}{p}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5\ldots }^{p-2}{\frac {(p-1)!}{(p-(k+1))!(k+1!}}$

$2^{p-1}-1\equiv 0{\bmod {p}}$

$2^{p-1}\equiv 1{\bmod {p}}$

${\underline {2^{p}\equiv 2{\bmod {p}}}}$

Voor $a=2$ is de bewering waar.

Laten we aannemen dat de volgende bewering voor het getal $a$ waar is:${\underline {a^{p}\equiv a{\bmod {p}}}}$

Dan moet deze bewering ook waar zijn voor het getal $a+1$.

$(a+1)^{p}\equiv (a+1){\bmod {p}}$

$\textstyle \sum _{k=0}^{p}{\binom {p}{k}}a^{p-k}1^{k}\equiv {(a+1){\bmod {p}}}\displaystyle$

$\textstyle \sum _{k=0}^{p-1}{\binom {p}{k}}a^{p-k}+1\equiv {(a+1){\bmod {p}}}\displaystyle$

$\textstyle \sum _{k=0}^{p-1}{\binom {p}{k}}a^{p-k}\equiv {a{\bmod {p}}}\displaystyle$

$\textstyle \{\sum _{k=1}^{p-1}{\binom {p}{k}}a^{p-k}+a^{p}\}{\bmod {p}}\equiv {a{\bmod {p}}}\displaystyle$

$p\textstyle \sum _{k=1}^{p-1}{\frac {(p-1!)}{(p-k)!k!}}a^{p-k}{\bmod {p}}+{a^{p}}{\bmod {p}}\equiv a{\bmod {p}}\displaystyle$

$0{\bmod {p}}+a^{p}{\bmod {p}}\equiv a{\bmod {p}}\displaystyle$

$a^{p}\equiv {a{\bmod {p}}}$

De bewering ${\underline {a^{p}\equiv {a{\bmod {p}}}}}$ i s waar.

Bewijs van de kleine stelling van Fermat.

De bewering is dat $a^{p-1}-1\equiv 0{\bmod {p}}$

$(a+1)^{p}\equiv (a+1){\bmod {p}}$

$\textstyle \sum _{k=0}^{p}{\binom {p}{k}}a^{p-k}1^{k}\equiv {(a+1){\bmod {p}}}\displaystyle$

$\textstyle \sum _{k=0}^{p-1}{\binom {p}{k}}a^{p-k}k+1\equiv {(a+1){\bmod {p}}}\displaystyle$

$\textstyle \sum _{k=0}^{p-1}{\binom {p}{k}}a^{p-k}\equiv {a{\bmod {p}}}\displaystyle$

$\{\textstyle \sum _{k=1}^{p-1}{\binom {p}{k}}a^{p-k}+a^{p}\}{\bmod {p}}\equiv {a{\bmod {p}}}\displaystyle$

$p\textstyle \sum _{k=1}^{p-1}{\frac {(p-1!)}{(p-k)!k!}}a^{p-k}{\bmod {p}}+{a^{p}}{\bmod {p}}\equiv a{\bmod {p}}\displaystyle$

$0{\bmod {p}}+a^{p}{\bmod {p}}\equiv a{\bmod {p}}\displaystyle$

$a^{p}\equiv {a{\bmod {p}}}$

${\frac {a^{p}}{a}}\equiv {{\frac {a}{a}}{\bmod {p}}}$

$a^{p-1}\equiv {1{\bmod {p}}}$

${a^{p-1}-1}\equiv {0{\bmod {p}}}$

De bewering is $a^{p-1}-1\equiv 0{\bmod {p}}$ is waar.

De kleine stelling van Fermat oud bewijs.

$A,B,p,q,r$ zijn gehele getallen..

$p$ is een priemgetal.

Is $A^{p-1}-1$ deelbaar door $p$. ${\frac {A^{p-1}-1}{p}}=?$

${\frac {A^{p-1}-1^{p-1}}{p}}={\frac {A^{p-1}-B^{p-1}}{p}}$

${\frac {A^{p-1}-B^{p-1}}{p}}={\frac {(p+q)^{p-1}-(p+r)^{p-1}}{p}}$

${\frac {A^{p-1}-1^{p-1}}{p}}={\frac {(p+q)^{p-1}-\{p+(-p+1)\}^{p-1}}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-1-k}q^{k}-\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-1-k}(-p+1)^{k}}{p}}$

${\frac {A^{p-1}-1}{p}}=\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-2-k}q^{k}-\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-2-k}(-p+1)^{k}$

${\frac {A^{p-1}-1}{p}}=\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-2-k}\{q^{k}-(-p+1)^{k}\}$

Het volgende is het vermelden waard.

${\binom {p-1}{p-1}}p^{p-2-(p-1)}\{q^{p-1}-(-p+1)^{p-1}\}={\frac {{\binom {p-1}{p-1}}\{q^{p-1}-(-p+1)^{p-1}\}}{p}}$

$\{q^{p-1}-(-p+1)^{p-1}\}$ is deelbaar door p.

Is A^(p-1)-B^(p-1) divisible by p?

$A,B,p,q$ are whole numbers.

$p$ is a prime number.

$A\neq qp$

You choose a random number $B$.

Then there is always a number $A=B+p$

$A=B+p$

${\frac {A^{p-1}-B^{p-1}}{p}}={\frac {(B+p)^{p-1}-B^{p-1}}{p}}$

${\frac {A^{p-1}-B^{p-1}}{p}}={\frac {\textstyle \sum _{k=0}^{p-1}\displaystyle {\tbinom {p-1}{k}}{B^{p-1-k}p^{k}-B^{p-1}}}{p}}$

${\frac {A^{p-1}-B^{p-1}}{p}}={\frac {\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{B^{p-1-k}p^{k}+B^{p1}-B^{p-1}}}{p}}$

${\frac {A^{p-1}-B^{p-1}}{p}}={\frac {\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{B^{p-1-k}p^{k}}}{p}}$

${\frac {A^{p-1}-B^{p-1}}{p}}=\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{B^{p-1-k}p^{k-1}}$

$A^{P-1}-B^{p-1}$ is divisible by $p$.

Is A^(p-1)-B^(p-1) divisible by p? A different path

$A,B,p,q$ are whole numbers.

$p$ is a prime number.

$A\neq qp$

There is a number $A=p+q$ and there is a number $B=p-q$.

$A=p+q$

$B=p-q$

${\frac {A^{p-1}-B^{p-1}}{p}}={\frac {(p+q)^{p-1}-(p-q)^{p-1}}{p}}$

${\frac {A^{p-1}-B^{p-1}}{p}}={\frac {\textstyle \sum _{k=0}^{p-1}\displaystyle {\tbinom {p-1}{k}}{p^{k}q^{p-1-k}}-{\textstyle \sum _{k=0}^{p-1}\displaystyle {\tbinom {p-1}{k}}{p^{k}(-q)^{p-1-k}}}}{p}}$

${\frac {A^{p-1}-B^{p-1}}{p}}={\frac {\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{p^{k}q^{p-1-k}}-{\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{p^{k}(-q)^{p-1-k}+q^{p-1}-(-q)^{p-1}}}}{p}}$

${\frac {A^{p-1}-B^{p-1}}{p}}=\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{p^{k-1}q^{p-1-k}}-{\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{p^{k-1}(-q)^{p-1-k}}}$

$A^{P-1}-B^{p-1}$ is divisible by $p$.

Quod erat demonstrandum

Oud materiaaal

Inductie stap een.

$C=2$

$C^{p}=2^{p}$

$2^{p}\equiv 2{\bmod {p}}$

${\frac {2^{p}}{2}}\equiv {\frac {2}{2}}{\bmod {p}}$

$2^{p-1}\equiv 1{\bmod {p}}$

$2^{p-1}-1\equiv 0{\bmod {p}}$

$\displaystyle {\frac {C^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(2-1)^{k}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}$

De som van de binomiaalcoëfficiënten. 

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

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

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

$\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5,\ldots }^{p-2}{\binom {p}{k+1}}=\sum _{k=1,3,5,\ldots }^{p-2}{\frac {p(p-1)!}{(p-(k+1))!(k+1)!}}$

${\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5\ldots }^{p-2}{\frac {(p-1)!}{(p-(k+1))!(k+1!}}$

Voor $C=2$ geldt $2^{p}\equiv \mod p$

Inductie stap twee.

Voor een zeker getal $B$ geldt:

${\frac {B^{p-1}-1}{p}}=Q$

$Q\in \mathbb {N}$

Inductie stap drie.

Het getal $A$ is gelijk aan: $A=p+q$

Het getal $B$ is gelijk aan: $B=p-q$

Dan volgt daaruit: $A+B=2p\Longrightarrow p={\frac {1}{2}}(A+B)$

${\frac {A^{p-1}-1}{p}}-{\frac {B^{p-1}-1}{p}}=2{\frac {A^{2k}-B^{2k}}{A+B}}=2\textstyle \sum _{n=1}^{2k}\displaystyle {(-1)^{n-1}}A^{2k-n}B^{n-1}=Z$

${\frac {A^{p-1}-1}{p}}={\frac {(p+q)^{p-1}}{p}}$

${\frac {B^{p-1}-1}{p}}={\frac {(p-q)^{p-1}}{p}}$

${\frac {A^{p-1}-1}{p}}-{\frac {B^{p-1}-1}{p}}={\frac {(p+q)^{p-1}}{p}}-{\frac {(p-q)^{p-1}}{p}}$

$p=3$

$q=5$

${\frac {(p+q)^{p-1}}{p}}-{\frac {(p-q)^{p-1}}{p}}={\frac {(3+5)^{3-1}}{3}}-{\frac {(3-5)^{3-1}}{3}}={\frac {64-4}{3}}=20$

${\frac {A^{p-1}-1}{p}}=Z+Q=R\Longrightarrow R\in \mathbb {N}$

Een andere weg.

Elk getal $A$ kunnen we schrijven als $A=ap+1$

${\frac {A^{p-1}-1}{p}}={\frac {(ap+1)^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=0}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(ap)}^{k}1^{p-1-k}-1}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {1+\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}a^{k}p^{k}1^{p-1-k}-1}{p}}={\frac {\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}a^{k}p^{k}}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}a^{k}p^{k}}{p}}=\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}a^{k}p^{k-1}$

${\frac {A^{p-1}-1}{p}}=\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{{\frac {(A-1)^{k}}{p^{k}}}p^{k-1}}$

${\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {1}{p}}\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}}$

$\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(A-1)}^{k}$ moet een p-voud zijn.

$\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(2-1)^{k}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}$

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

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

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

$\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=^{2}\uparrow \sum _{k=1}^{p-1}{\binom {p}{k+1}}=p(^{2}\uparrow \sum _{k=1}^{p-1}{\frac {(p-1)!}{(p-(k+1))!(k+1!}})=pq$

${\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=^{2}\uparrow \sum _{k=1}^{p-1}{\frac {(p-1)!}{(p-(k+1))!(k+1!}}=q$

${\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {1}{p}}\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}}$

Een kortere weg.

$A\neq ap$

Elk getal $A$ kan worden geschreven als $A=(A-1)+1$

${\frac {A^{p-1}-1}{p}}={\frac {((A-1)+1)^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=0}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(A-1)}^{k}1^{p-1-k}-1}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {1+\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}(A-1)^{k}1^{p-1-k}-1}{p}}={\frac {\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}(A-1)^{k}}{p}}$

${\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {1}{p}}\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}}$

$\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(A-1)}^{k}$ moet een p-voud zijn.

Inductiestap een.

$\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(2-1)^{k}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}$

De som van de binomiaalcoëfficiënten. 

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

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

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

$\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5,\ldots }^{p-2}{\binom {p}{k+1}}=\sum _{k=1,3,5,\ldots }^{p-2}{\frac {p(p-1)!}{(p-(k+1))!(k+1)!}}=pq$

${\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5\ldots }^{p-2}{\frac {(p-1)!}{(p-(k+1))!(k+1)!}}=q$

${\frac {A^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}=q$

$q\in \mathbb {N}$

Voor $A=2$ is de stelling waar.

Inductiestap twee

We nemen de waarheid van de stelling voor een zeker getal $A$ aan.

Dus:

${\frac {A^{p-1}-1}{p}}=R$

$R\in \mathbb {N}$

Inductiestap drie.

${\frac {(A+p)^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=0}^{p-1}{\binom {p-1}{k}}A^{p-1-k}p^{k}-1\displaystyle }{p}}$

${\frac {(A+p)^{p-1}-1}{p}}={\frac {\textstyle A^{p-1}+\sum _{k=1}^{p-1}{\binom {p-1}{k}}A^{p-1-k}p^{k}-1\displaystyle }{p}}$

${\frac {(A+p)^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=1}^{p-1}{\binom {p-1}{k}}A^{p-1-k}p^{k}\displaystyle }{p}}+{\frac {A^{p-1}-1}{p}}$

${\frac {(A+p)^{p-1}-1}{p}}=\textstyle \sum _{k=1}^{p-1}{\binom {p-1}{k}}A^{p-1-k}p^{k-1}\displaystyle +R=Q$

$Q\in \mathbb {N}$

De stelling is dus juist.

Een alternatief.

$p=A-B$

${\frac {A^{p-1}-1}{p}}-{\frac {B^{p-1}-1}{p}}={\frac {A^{2k}-B^{2k}}{A-B}}=\textstyle \sum _{n=1}^{2k}A^{2k-n}B^{n-1}\displaystyle$

${\frac {B^{p-1}-1}{p}}={\frac {A^{p-1}-1}{p}}-\textstyle \sum _{n=1}^{2k}A^{2k-n}B^{n-1}\displaystyle$

${\frac {B^{p-1}-1}{p}}=R-\textstyle \sum _{n=1}^{2k}A^{2k-n}B^{n-1}\displaystyle =Z$

$Z\in \mathbb {N}$

De stelling is dus juist.

De definitieve versie

$A\neq ap$

Elk getal $A$ kan worden geschreven als $A=(A-1)+1$

${\frac {A^{p-1}-1}{p}}={\frac {((A-1)+1)^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=0}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(A-1)}^{k}1^{p-1-k}-1}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {1+\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}(A-1)^{k}1^{p-1-k}-1}{p}}={\frac {\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}(A-1)^{k}}{p}}$

${\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {1}{p}}\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}}$

$\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(A-1)}^{k}$ moet een p-voud zijn.

Stap een.

$\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(2-1)^{k}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}$

De som van de binomiaalcoëfficiënten. 

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

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

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

$\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5,\ldots }^{p-2}{\binom {p}{k+1}}=\sum _{k=1,3,5,\ldots }^{p-2}{\frac {p(p-1)!}{(p-(k+1))!(k+1)!}}=pq$

${\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5\ldots }^{p-2}{\frac {(p-1)!}{(p-(k+1))!(k+1)!}}=q$

${\frac {A^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}=q$

$q\in \mathbb {N}$

Voor $A=2$ is de stelling waar.

Stap twee

${\frac {b^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}={\frac {1}{p}}\sum _{k=1,3,5,\ldots }^{p-2}{\binom {p}{k+1}}=\sum _{k=1,3,5,\ldots }^{p-2}{\frac {(p-1)!}{(p-(k+1))!(k+1)!}}=q$

$A=a+p\Longrightarrow p=A-a$

${\frac {A^{p-1}-1}{p}}-{\frac {a^{p-1}-1}{p}}={\frac {A^{2k}-a^{2k}}{A-a}}=\textstyle \sum _{n=1}^{2k}A^{2k-n}a^{n-1}\displaystyle$

${\frac {A^{p-1}-1}{p}}={\frac {a^{p-1}-1}{p}}+\textstyle \sum _{n=1}^{2k}A^{2k-n}a^{n-1}\displaystyle$

Stel dat: $a=2$

${\frac {A^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}+\textstyle \sum _{n=1}^{2k}A^{2k-n}2^{n-1}\displaystyle$

${\frac {A^{p-1}-1}{p}}=\sum _{k=1,3,5,\ldots }^{p-2}{\frac {(p-1)!}{(p-(k+1))!(k+1)!}}+\textstyle \sum _{n=1}^{2k}A^{2k-n}2^{n-1}=Q\displaystyle$

$Q\in \mathbb {N}$

Een alternatieve route.

$B=A+1$

$A+1=(A+p)-(p-1)=(A+p)-c$

${\displaystyle {\frac {((A+p)-c)^{p-1}-1}{p}}={\frac {1}{p}}\textstyle (\sum _{k=0}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(-1)^{k}(A+p)^{p-1-k}c^{k}}-1)$

${\displaystyle {\frac {((A+p)-(p-1))^{p-1}-1}{p}}={\frac {1}{p}}\textstyle \sum _{k=0}^{p-2}\displaystyle {\tbinom {p-1}{k}}}{(-1)^{k}(A+p)^{p-1-k}c^{k}}+{\frac {{\binom {p-1}{0}}(-1)^{p-1}(A+p)^{p-1}c^{0}}{p}}-{\frac {1}{p}}$

${\displaystyle {\frac {((A+p)-c)^{p-1}-1}{p}}={\frac {1}{p}}\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(-1)^{k}(A+p)^{p-1-k}c^{k}}+{\frac {(A+p)^{p-1}}{p}}-{\frac {1}{p}}$

$p=A-B$

${\frac {A^{p-1}-1}{p}}-{\frac {B^{p-1}-1}{p}}={\frac {A^{2k}-B^{2k}}{A-B}}=\textstyle \sum _{n=1}^{2k}A^{2k-n}B^{n-1}\displaystyle$

${\frac {B^{p-1}-1}{p}}={\frac {A^{p-1}-1}{p}}-\textstyle \sum _{n=1}^{2k}A^{2k-n}B^{n-1}\displaystyle$

De kleine stelling van Fermat 1

$A\neq cp$

$A=a+p\Longrightarrow p=A-a$

${\frac {A^{p-1}-1}{p}}-{\frac {a^{p-1}-1}{p}}={\frac {A^{p-1}-a^{p-1}}{A-a}}=\textstyle \sum _{n=1}^{2k}A^{2k-n}a^{n-1}\displaystyle$

We gaan er vanuit dat $A^{p-1}-1$ en $a^{p-1}-1$ beiden niet deelbaar zijn door $p$.

$B$ en $b$ zijn gehele getallen. $r_{1}$ en $r_{2}$ zijn resten na deling door $p$.

${\frac {A^{p-1}-1}{p}}-{\frac {a^{p-1}-1}{p}}=B+{\frac {r_{1}}{p}}-\left(b+{\frac {r_{2}}{p}}\right)=B-b+{\frac {r_{1}-r_{2}}{p}}=\textstyle \sum _{n=1}^{2k}A^{2k-n}a^{n-1}\displaystyle \Longrightarrow$

${\frac {r_{1}-r_{2}}{p}}=0\Longrightarrow r_{1}=r_{2}=r$

${\frac {A^{p-1}-1}{p}}-{\frac {a^{p-1}-1}{p}}=B+{\frac {r}{p}}-\left(b+{\frac {r}{p}}\right)=B-b=\textstyle \sum _{n=1}^{2k}A^{2k-n}a^{n-1}\displaystyle$

Een stap verder.

${\frac {A^{p}-A}{p}}-{\frac {a^{p}-a}{p}}={\frac {A^{p}-a^{p}}{A-a}}-{\frac {A-a}{A-a}}={\frac {A^{2k+1}-a^{2k+1}}{A-a}}-1=\textstyle \sum _{n=1}^{2k+1}A^{2k+1-n}a^{n-1}-1\displaystyle$

$A{\frac {A^{p-1}-1}{p}}-a{\frac {a^{p-1}-1}{p}}=\textstyle \sum _{n=1}^{2k+1}A^{2k+1-n}a^{n-1}-1\displaystyle$

Wederom gaan we er vanuit dat $A^{p-1}-1$ en $a^{p-1}-1$ beiden niet deelbaar zijn door $p$.

$B$ en $b$ zijn gehele getallen. $r$ is de rest na deling door $p$.

$A{\frac {A^{p-1}-1}{p}}-a{\frac {a^{p-1}-1}{p}}=A\left(B+{\frac {r}{p}}\right)-a\left(b+{\frac {r}{p}}\right)=AB-ab+(A-a){\frac {r}{A-a}}$

$A{\frac {A^{p-1}-1}{p}}-a{\frac {a^{p-1}-1}{p}}=AB-ab+r$

$AB-ab+r=\textstyle \sum _{n=1}^{2k+1}A^{2k+1-n}a^{n-1}-1\displaystyle \Longrightarrow r=0$

$r$ is nul, dus zijn $A^{p-1}-1$ en $a^{p-1}-1$ beiden deelbaar door $p$.

De kleine stelling van Fermat gevorderd

$\sum _{k=0}^{p-1}{\binom {p-1}{k}}=2^{p-1}={\frac {1}{2}}2^{p}={\frac {1}{2}}\sum _{k=0}^{p}{\binom {p}{k}}$

$2^{p-1}-1=\sum _{k=0}^{p-1}{\binom {p-1}{k}}-1={\frac {1}{2}}\sum _{k=0}^{p}{\binom {p}{k}}-1$

$\sum _{k=0}^{p-1}{\binom {p-1}{k}}-1={\frac {1}{2}}\{\sum _{k=0}^{p}{\binom {p}{k}}-2\}={\frac {1}{2}}\sum _{k=1}^{p-1}{\binom {p}{k}}$

$\sum _{k=0}^{p-1}{\binom {p-1}{k}}-1={\frac {1}{2}}\sum _{k=1}^{p-1}{\binom {p}{k}}$

${\frac {2^{p-1}-1}{p}}={\frac {\sum _{k=0}^{p-1}{\binom {p-1}{k}}-1}{p}}={\frac {1}{2}}{\frac {\sum _{k=1}^{p-1}{\binom {p}{k}}}{p}}$

${\frac {\sum _{k=0}^{p-1}{\binom {p-1}{k}}-1}{p}}={\frac {1}{2}}{\frac {\sum _{k=1}^{p-1}{\binom {p}{k}}}{p}}$

${\frac {\sum _{k=0}^{p-1}{\binom {p-1}{k}}-1}{p}}={\frac {1}{2}}{\frac {p\sum _{k=1}^{p-1}{\frac {(p-1)!}{(p-k)!k!}}}{p}}$

${\frac {\sum _{k=0}^{p-1}{\binom {p-1}{k}}-1}{p}}={\frac {1}{2}}\sum _{k=1}^{p-1}{\frac {(p-1)!}{(p-k)!k!}}$

Enkele berekeningen

$A=(A-1)+1$

$A^{p-1}-1=((A-1)+1)^{p-1}-1=\textstyle \sum _{k=0}^{p-1}\displaystyle {\binom {p-1}{k}}(A-1)^{p-1-k}1^{k}-1$

$\textstyle \sum _{k=0}^{p-1}\displaystyle {\binom {p-1}{k}}(A-1)^{p-1-k}-1=\textstyle \sum _{k=0}^{p-2}\displaystyle {\binom {p-1}{k}}(A-1)^{p-1-k}$

$p=7$

$\textstyle \sum _{k=0}^{7-2}\displaystyle {\binom {7-1}{k}}(A-1)^{7-1-k}=\textstyle \sum _{k=0}^{5}\displaystyle {\binom {6}{k}}(A-1)^{6-k}$

$A=2\Longrightarrow$$1*1^{6}+6*1^{5}+15*1^{4}+20*1^{3}+15*1^{2}+6*1^{1}=7*9$

$A=3\Longrightarrow$$1*2^{6}+6*2^{5}+15*2^{4}+20*2^{3}+15*2^{2}+6*2^{1}=7*104$

$A=4\Longrightarrow$$1*3^{6}+6*3^{5}+15*3^{4}+20*3^{3}+15*3^{2}+6*3^{1}=7*585$

$A=5\Longrightarrow$$1*4^{6}+6*4^{5}+15*4^{4}+20*4^{3}+15*4^{2}+6*4^{1}=7*2232$

$A=6\Longrightarrow$$1*5^{6}+6*5^{5}+15*5^{4}+20*5^{3}+15*5^{2}+6*5^{1}=7*6665$

$A=8\Longrightarrow$$1*7^{6}+6*7^{5}+15*7^{4}+20*7^{3}+15*7^{2}+6*7^{1}=7*37449$

$t$ is een positief geheel getal.

Op basis van bovenstaande berekeningen komen we voor $A$ tot de volgende formule: $\textstyle \sum _{k=0}^{5}\displaystyle {\binom {6}{k}}(A-1)^{6-k}=7t$

Het volgende is niet meer dan een vermoeden: $A^{p-1}-1=\textstyle \sum _{k=0}^{p-2}\displaystyle {\binom {p-1}{k}}(A-1)^{p-1-k}=pq$

$A=2$

${\frac {A^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=0}^{p-2}\displaystyle {\binom {p-1}{k}}(2-1)^{p-k}}{p}}$

${\frac {\textstyle \sum _{k=0}^{p-2}\displaystyle {\binom {p-1}{k}}1^{p-1-k}}{p}}={\frac {\textstyle \sum _{k=0}^{p-2}\displaystyle {\binom {p-1}{k}}}{p}}$

De som van de binomiaalcoëfficiënten. 

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

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

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

$\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5,\ldots }^{p-2}{\binom {p}{k+1}}=\sum _{k=1,3,5,\ldots }^{p-2}{\frac {p(p-1)!}{(p-(k+1))!(k+1)!}}=pq$

${\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5\ldots }^{p-2}{\frac {(p-1)!}{(p-(k+1))!(k+1)!}}=q$

${\frac {A^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}=q$

$q\in \mathbb {N}$

Voor $A=2$ is de stelling waar.

$A^{p-1}-1=\textstyle \sum _{k=0}^{p-2}\displaystyle {\binom {p-1}{k}}(A-1)^{p-1-k}=pq$

$p=5$

$A=3$

$3^{5-1}-1=\textstyle \sum _{k=0}^{5-2}\displaystyle {\binom {5-1}{k}}(3-1)^{5-1-k}$

$3^{5-1}-1=1.2^{4}+4.2^{3}+6.2^{2}+4.2^{1}=16+32+24+8=80=5.16$

$(3+1)^{5-1}-1=1.3^{4}+4.3^{3}+6.3^{2}+4.3^{1}=81+108+54+12=255=5.51$

$(3+1)^{5-1}-1=1.(2+1)^{4}+4.(2+1)^{3}+6.(2+1)^{2}+4.(2+1)^{1}=81+108+54+12=255$

$(A+2)^{5-1}-1=1.(A+1)^{4}+4.(A+1)^{3}+6.(A+1)^{2}+4.(A+1)$

$(A+2)^{5-1}-1=1.(A^{4}+4A^{3}+6A^{2}+4A^{1}+1)+4.(A^{3}+3A^{2}+3A^{1}+1)+6.(A^{2}+2A^{1}+1)^{2}+4.(A+1)$

$1.(A^{4}+4A^{3}+6A^{2}+4A^{1}+1)+4.(A^{3}+3A^{2}+3A^{1}+1)+6.(A^{2}+2A^{1}+1)^{2}+4.(A+1)=A^{4}+8A^{3}+24A^{2}+32A^{1}+15$$(A+2)^{5-1}-1=A^{4}+8A^{3}+24A^{2}+32A^{1}+15$

$I=1$

$I^{4}+8I^{3}+24I^{2}+32I^{1}+15=1^{4}+8.1^{3}+24.1^{2}+32.1^{1}+15=80=5.16$

$I=2$

$I^{4}+8I^{3}+24I^{2}+32I^{1}+15=2^{4}+8.2^{3}+24.2^{2}+32.2^{1}+15=255=5.51$

Conclusie: $I^{4}+8I^{3}+24I^{2}+32I^{1}+15=5v$

Hiermee is tevens bewezen dat: $A^{p-1}-1=\textstyle \sum _{k=0}^{p-2}\displaystyle {\binom {p-1}{k}}(A-1)^{p-1-k}=pq$

${\displaystyle (A+3)^{5-1}-1=1.(A+2)^{4}+4.(A+2)^{3}+6.(A+2)^{2}+4.(A+2)}$

$I=-1$.

${\displaystyle {\displaystyle (I+3)^{5-1}-1=1.(I+2)^{4}+4.(I+2)^{3}+6.(I+2)^{2}+4.(I+2)}}=15$

$I=0$

${\displaystyle {\displaystyle {\displaystyle (I+3)^{5-1}-1=1.(I+2)^{4}+4.(I+2)^{3}+6.(I+2)^{2}+4.(I+2)}}=80}$

$I=1$

${\displaystyle {\displaystyle {\displaystyle {\displaystyle (I+3)^{5-1}-1=1.(I+2)^{4}+4.(I+2)^{3}+6.(I+2)^{2}+4.(I+2)}}=255}}$

$\displaystyle (A+3)^{5-1}-1=A^{4}+12A^{3}+54A^{2}+72A^{1}+80$

$2^{5-1}-1=\textstyle \sum _{k=0}^{5-2}\displaystyle {\binom {5-1}{k}}(2-1)^{5-1-k}=1^{4}+4.1^{3}+6.1^{2}+4.1=1+4+6+4=15=5.3$

$3^{5-1}-1=\textstyle \sum _{k=0}^{5-2}\displaystyle {\binom {5-1}{k}}(3-1)^{5-1-k}=2^{4}+4.2^{3}+6.2^{2}+4.2=16+32+24+8=80=5.16$

${\underline {A^{5-1}-1=\textstyle \sum _{k=0}^{5-2}\displaystyle {\binom {5-1}{k}}(A-1)^{5-1-k}=5i}}$

$2^{3-1}-1=\textstyle \sum _{k=0}^{1}\displaystyle {\binom {2}{k}}1^{2-k}=1.1+2.1=3.1$

$4^{3-1}-1=\textstyle \sum _{k=0}^{1}\displaystyle {\binom {2}{k}}3^{2-k}=1.9+2.3=3.5$

${\underline {A^{3-1}-1=\textstyle \sum _{k=0}^{3-2}\displaystyle {\binom {3-1}{k}}(A-1)^{3-1-k}=3u}}$

$A^{p-1}-1=\textstyle \sum _{k=0}^{p-2}\displaystyle {\binom {p-1}{k}}(A-1)^{p-1-k}=pq$

$(A+2)^{p-1}-1=\textstyle \sum _{k=0}^{p-1}\displaystyle {\binom {p-1}{k}}A^{p-1-k}2^{k}-1$

$(A+2)^{p-1}=\textstyle \sum _{k=0}^{p-1}\displaystyle {\binom {p-1}{k}}A^{p-1-k}2^{k}$

$\displaystyle {\binom {p-1}{0}}(A-1)^{p-1}+\displaystyle {\binom {p-1}{1}}(A-1)^{p-2}+\displaystyle {\binom {p-1}{2}}(A-1)^{p-3}+\displaystyle {\binom {p-1}{0}}(A-1)^{p-4}+\displaystyle {\binom {p-1}{0}}(A-1)^{p-5}+\displaystyle {\binom {p-1}{0}}(A-1)^{p-6}-1$

$p=5$

$\displaystyle {\binom {4}{0}}(A-1)^{4}+\displaystyle {\binom {4}{1}}(A-1)^{3}+\displaystyle {\binom {4}{2}}(A-1)^{2}+\displaystyle {\binom {4}{3}}(A-1)^{1}+\displaystyle {\binom {4}{4}}(A-1)^{0}-1$

$1(A-1)^{4}+4(A-1)^{3}+6(A-1)^{2}+4(A-1)^{1}+1(A-1)^{0}-1$

$A=2$

$1.1^{4}+4.1^{3}+6.1^{2}+4.1^{1}+1.1^{0}-1=15$

$A=3$

$1.2^{4}+4.2^{3}+6.2^{2}+4.2^{1}+1.2^{0}-1=72$

$p=3$

$A=4$

$\displaystyle {\binom {2}{0}}(4-1)^{3-1}+\displaystyle {\binom {2}{1}}(4-1)^{3-2}+\displaystyle {\binom {2}{2}}(4-1)^{3-3}-1=1.(4-1)^{3-1}+2.(4-1)^{3-2}+1.(4-1)^{3-3}-1=16-1=15$

$A=5$

$\displaystyle {\binom {2}{0}}(5-1)^{3-1}+\displaystyle {\binom {2}{1}}(5-1)^{3-2}+\displaystyle {\binom {2}{2}}(5-1)^{3-3}-1=1.(5-1)^{3-1}+2.(5-1)^{3-2}+1.(5-1)^{3-3}-1=24$

De kleine stelling van Fermat

$A\neq ap$

$q$ is een geheel getal.

Is $A^{p-1}-1$ deelbaar door $p$ , ${\frac {A^{p-1}-1}{p}}=q?$

Inductiestap 1.

$\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(2-1)^{k}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}$

De som van de binomiaalcoëfficiënten. 

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

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

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

$\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5,\ldots }^{p-2}{\binom {p}{k+1}}=\sum _{k=1,3,5,\ldots }^{p-2}{\frac {p(p-1)!}{(p-(k+1))!(k+1)!}}=pq$

${\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5\ldots }^{p-2}{\frac {(p-1)!}{(p-(k+1))!(k+1)!}}=q$

${\frac {A^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}=q$

$q\in \mathbb {N}$

Voor $A=2$ is de stelling waar.

Inductiestap 2.

We nemen de waarheid van de stelling aan.

${\frac {A^{p-1}-1}{p}}=q\Longrightarrow {\frac {A^{p}-A}{p}}={\frac {pqA}{p}}$

$q$ is een geheel getal. Bij gevolg is ${\frac {pqA}{p}}$ ook een geheel getal.

Inductiestap 3.

${\frac {(A+1)^{p}-(A+1)}{p}}={\frac {\textstyle \sum _{k=0}^{p}\displaystyle {\tbinom {p}{k}}{A}^{p-k}1^{k}-A-1}{p}}$

${\frac {(A+1)^{p}-(A+1)}{p}}={\frac {{A}^{p}-A+\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p}{k}}{A}^{p-k}1^{k}+1-1}{p}}$

${\frac {(A+1)^{p}-(A+1)}{p}}={\frac {{A}^{p}-A+\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p}{k}}{A}^{p-k}1^{k}}{p}}$

${\frac {(A+1)^{p}-(A+1)}{p}}={\frac {A^{p}-A}{p}}+{\frac {\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p}{k}}{A}^{p-k}1^{k}}{p}}$

${\frac {(A+1)^{p}-(A+1)}{p}}={\frac {pqA}{p}}+{\frac {p\textstyle \sum _{k=1}^{p-1}{\frac {(p-1)!}{(p-k)!k!}}{A}^{p-k}}{p}}$

${\frac {(A+1)^{p}-(A+1)}{p}}=qA+\textstyle \sum _{k=1}^{p-1}{\frac {(p-1)!}{(p-k)!k!}}{A}^{p-k}$

$(A+1){\frac {(A+1)^{p-1}-1}{p}}=qA+\textstyle \sum _{k=1}^{p-1}{\frac {(p-1)!}{(p-k)!k!}}{A}^{p-k}$

$qA$ is een geheel getal en $\textstyle \sum _{k=1}^{p-1}{\frac {(p-1)!}{(p-k)!k!}}{A}^{p-k}$ is een geheel getal.

Dus is $(A+1){\frac {(A+1)^{p-1}-1}{p}}$ ook een geheel getal.

De factor ${\frac {(A+1)^{p-1}-1}{p}}$ is een geheel getal, dus is $(A+1)^{p-1}-1$ deelbaar door $p$.

Proof of Fermat's little theorem unravelled.

$A,B,p,q,r$ are whole numbers.

p is a prime number.

Here is the proof that $A^{p-1}-1$ is divisible by $p$. ${\frac {A^{p-1}-1}{p}}=?$

${\frac {A^{p-1}-1^{p-1}}{p}}={\frac {A^{p-1}-B^{p-1}}{p}}$

${\frac {A^{p-1}-B^{p-1}}{p}}={\frac {(p+q)^{p-1}-(p+r)^{p-1}}{p}}$

${\frac {A^{p-1}-1^{p-1}}{p}}={\frac {(p+q)^{p-1}-\{p+(-p+1)\}^{p-1}}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-1-k}q^{k}-\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-1-k}(-p+1)^{k}}{p}}$

${\frac {A^{p-1}-1}{p}}=\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-2-k}q^{k}-\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-2-k}(-p+1)^{k}$

${\frac {A^{p-1}-1}{p}}=\sum _{k=0}^{p-1}{\binom {p-1}{k}}p^{p-2-k}\{q^{k}-(-p+1)^{k}\}$

The following is worth mentioning..

${\binom {p-1}{p-1}}p^{p-2-(p-1)}\{q^{p-1}-(-p+1)^{p-1}\}={\frac {{\binom {p-1}{p-1}}\{q^{p-1}-(-p+1)^{p-1}\}}{p}}$

$\{q^{p-1}-(-p+1)^{p-1}\}$ is divisible by $p$.

Qud erat demonstrandum

Functions I used to check my findings in Excel.

Public Function Nbk(y As Integer, z As Integer) As Double

Nbk = Application.WorksheetFunction.Combin(y, z)

End Function

Public Function Nadine(A As Integer, p As Integer) As Double

Dim k As Integer

Dim q As Integer

q = A - p

For k = 0 To p - 1

Nadine = Nadine + Nbk(p - 1, k) * p ^ (p - 2 - k) * (q ^ k - (-p + 1) ^ k)

Next k

End Function

Proof of Fermat's little theorem oud

The numbers $A,B,a,p,k$ are positive whole numbers.

Is $A^{p-1}-1$ divisible by $p$?

${\frac {A^{p}-1}{p}}=?$

For $A$ we may write down $ap+1$.

$B=ap$ is a whole number.

${\frac {A^{p-1}-1}{p}}={\frac {(ap+1)^{p-1}-1}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {\sum _{k=0}^{p-1}{\binom {p-1}{k}}(ap)^{p-1-k}1^{k}-1}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {\sum _{k=0}^{p-2}{\binom {p-1}{k}}(ap)^{p-1-k}1^{k}+1-1}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {\sum _{k=0}^{p-2}{\binom {p-1}{k}}(ap)^{p-1-k}}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {\sum _{k=0}^{p-2}{\binom {p-1}{k}}a^{p-1-k}p^{p-1-k}}{p}}$

${\frac {A^{p-1}-1}{p}}=\sum _{k=0}^{p-2}{\binom {p-1}{k}}a^{p-1-k}p^{p-2-k}$

Modulo rekenen met positieve c.q. negatieve getallen

$\lfloor q\rfloor$ wil zeggen de hele waarde van $q$, de entier van $q$.

Als $a>0$ dan:

$a{\pmod {b}}\equiv a-b\lfloor {\frac {a}{b}}\rfloor$

Als $a<0$ dan:

$a{\bmod {b}}\equiv a+b\lfloor {\frac {-a+b}{b}}\rfloor$

Twee rekenvoorbeelden.

$a=161$

$b=20$

$a{\pmod {b}}\equiv 161-20\lfloor {\frac {161}{20}}\rfloor \equiv 161-160\equiv 1$

$a=-135$

$b=12$

$a{\pmod {b}}\equiv -135+12\lfloor {\frac {135+12}{12}}\rfloor \equiv -135+144\equiv 9$

Binomiale ontwikkeling

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

Inductiestap een:

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

Voor $n=0$ is de bewering waar.

Inductiestap twee:

We nemen aan dat de bewering voor $n$ waar is:

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

Inductiestap drie:

We onderzoeken of de bewering voor $n+1$ waar is.

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

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

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

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

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

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

De som van de binomiaalcoëfficiënten. 

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

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

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

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

De bewering is waar voor $n+1$.

Dus de bewering $(a+b)^{n}=\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n-k}b^{k}$ is dus waar.

QED.

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Pythagorese drietallen

Voor het genereren van Pythagorese drietallen zijn er een zestal formules.

Voor de even waarden $2p$:

$(2p)^{2}=4p^{2}=2p^{2}-(-2p^{2})$

$4p^{2}=p^{4}+2p^{2}+1-(p^{4}-2p^{2}+1)$

$(2p)^{2}=(p^{2}+1)^{2}-(p^{2}-1)^{2}$

$x=2p$ $y=p^{2}-1$ $z=p^{2}+1$

Vult men de waarde $p=2$ in, dan vindt men $x=4$, $y=3$, $z=5$

Voor de oneven waarden $2p+1$:

$(2p+1)^{2}=4p^{2}+4p+1^{2}=4p(p+1)+1^{2}$

$(2p+1)^{2}=(2p(p+1))^{2}+2.2p(p+1)+1^{2}-(2p(p+1))^{2}$

$(2p+1)^{2}=(2p(p+1)+1)^{2}-(2p(p+1))^{2}$

$x=2p+1$ $y=2p(p+1)$ $z=2p(p+1)+1$

Vult men de waarde $p=1$ in, dan vindt men $x=3$, $y=4$, $z=5$

Fermat

$p$ is geen deler van $A$.

$A=p{\frac {A-1}{p}}+1=pq+1$

${\frac {A^{p-1}-1}{p}}={\frac {(pq+1)^{p-1}-1}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {(pq+1)^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=0}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(pq)}^{p-1-k}1^{k}-1}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=0}^{p-2}\displaystyle {\tbinom {p-1}{k}}{(pq)}^{p-1-k}1^{k}+1-1}{p}}$

${\frac {A^{p-1}-1}{p}}=\textstyle \sum _{k=0}^{p-2}\displaystyle {\tbinom {p-1}{k}}p^{p-2-k}q^{p-1-k}$

${\frac {A^{p-1}-1}{p}}=\textstyle \sum _{k=0}^{p-2}\displaystyle {\tbinom {p-1}{k}}p^{p-2-k}{\biggl (}{\frac {A-1}{p}}{\biggr )}^{p-1-k}$

De kleine stelling van Fermat 2

$A\in {Z}$

$p=2$

$A=2k+1$

${\frac {A^{p-1}-1}{p}}={\frac {(2k+1)^{2-1}-1}{2}}={\frac {2k}{2}}=k$

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.$\{A,a,b,D,E,r_{1},r_{2},q_{1},q_{2}\}\subset \mathrm {Z}$

$p$ is een priemgetal.

$p\neq 2$

$A\neq bp$

$A=a+p$

${\frac {A^{p-1}-1}{p}}={\frac {(a+p)^{p-1}-1}{p}}$

${\frac {A^{p-1}-1}{p}}-{\frac {a^{p-1}-1}{p}}={\frac {(a+p)^{p-1}-1}{p}}-{\frac {a^{p-1}-1}{p}}={\frac {(a+p)^{p-1}-a^{p-1}}{p}}$

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

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

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

${\frac {A^{p-1}-1}{p}}-{\frac {a^{p-1}-1}{p}}=\textstyle \sum _{k=1}^{p-1}{\binom {p-1}{k}}a^{p-1-k}p^{k-1}\displaystyle$

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Het onderstaande getallenvoorbeeld is gekozen om aan te tonen dat bij deling door een getal anders dan p=3

er een rest $q_{1}$ en een rest $q_{2}$ ontstaan. Deze resten zullen blijken ongelijk te zijn.

$A=8$

$p=3$

$A=a+p\Longrightarrow a=A-p=8-3=5$

$a=5$

$p+4=3+4=7$

${\frac {(a+p)^{p-1}-a^{p-1}}{p+4}}={\frac {(5+3)^{3-1}-1}{7}}-{\frac {5^{3-1}-1}{7}}={\frac {63}{7}}-{\frac {24}{7}}=9+{\frac {0}{7}}-(3+{\frac {3}{7}})$

${\frac {q_{1}}{7}}={\frac {0}{7}}\land {\frac {q_{2}}{7}}={\frac {3}{7}}\Longrightarrow q_{1}\neq q_{2}$

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$A^{p-1}-1$ en $a^{p-1}-1$ zijn ten opzichte van elkaar isomorf.

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Als $(a+p)^{p-1}-1$ niet deelbaar is door $p$ dan ontstaat er een rest $r_{1}\neq 0$.

$a^{p-1}-1$ is dan automatisch niet deelbaar door $p$ en geeft dan als rest $r_{2}\neq 0$.

Omdat $r_{1}$en $r_{2}$ dan altijd ongelijk zijn geldt : ${\frac {r_{1}-r_{2}}{p}}\neq 0$

Dus $r_{1}=r_{2}\Longrightarrow {\frac {r_{1}-r_{2}}{p}}={\frac {r_{1}-r_{1}}{p}}={\frac {0}{p}}=0$ kan nooit als $r_{1}\neq 0$

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Als $(a+p)^{p-1}-1$ echter wel deelbaar is door $p$ dan is de rest $r_{1}=0$

Automatisch is $a^{p-1}-1$ dan ook deelbaar door $p$, rest $r_{2}=0$ .

$r_{1}=r_{2}=0$

In dat geval is: ${\frac {r_{1}-r_{2}}{p}}={\frac {0-0}{p}}=0$

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${\frac {(a+p)^{p-1}-1}{p}}=D+{\frac {r_{1}}{p}}$

${\frac {a^{p-1}-1}{p}}=E+{\frac {r_{2}}{p}}$

Als $(r_{1}\neq 0\land r_{1}\neq r_{2})\Longrightarrow {\frac {r_{1}}{p}}\neq {\frac {r_{2}}{p}}$

${\frac {(a+p)^{p-1}-a^{p-1}}{p}}=(D-E)+{\frac {r1-r2}{p}}$

1) ${\frac {(a+p)^{p-1}-a^{p-1}}{p}}=\textstyle \sum _{k=1}^{p-1}{\binom {p-1}{k}}a^{p-1-k}p^{k-1}\displaystyle \Longrightarrow {\frac {r_{1}-r_{2}}{p}}=0$

2) Als: $(r_{1}\neq 0\Longrightarrow r_{2}\neq 0)\land (r_{1}\neq r_{2})\Longrightarrow {\frac {r_{1}-r_{2}}{p}}\neq 0$

Als $(r_{1}=0\Longrightarrow r_{2}=0)\Longrightarrow ({\frac {r_{1}-r_{2}}{p}}={\frac {0-0}{p}}=0)$

Als $r_{1}\neq 0$ dan kan $r_{2}$ in dat geval nooit gelijk zijn aan $r_{1}$.

We hebben gevonden dat ${\frac {r_{1}-r_{2}}{p}}=0$, (zie 1) Dit kan alleen maar zo zijn als.$r_{1}=r_{2}=0$ (zie 2)

Zowel $A^{p-1}-1$ als $a^{p-1}-1$ zijn dus deelbaar door $p$.

De abc-formule

$ax^{2}+bx+c=0$

$a(x^{2}+2{\frac {b}{2a}}x)+c=0$

$a(x^{2}+2{\frac {b}{2a}}x+{\bigl (}{\frac {b}{2a}})^{2}-{\bigl (}{\frac {b}{2a}})^{2})+c=0$

$a(x+{\frac {b}{2a}})^{2}-a{\frac {b^{2}}{4a^{2}}}+c=0$

$a(x+{\frac {b}{2a}})^{2}-{\frac {b^{2}}{4a}}+{\frac {4ac}{4a}}=0$

$(x+{\frac {b}{2a}})^{2}={\frac {b^{2}-4ac}{4a^{2}}}$

$x+{\frac {b}{2a}}={\frac {\pm {\sqrt {(b^{2}-4ac)}}}{2a}}$

$x={\frac {-b\pm {\sqrt {(b^{2}-4ac)}}}{2a}}$

Goniometrie

In een rechthoekige driehoek met een schuine zijde $c=1$ geldt:

$\alpha +\beta +\gamma =\pi$

$\gamma ={\frac {1}{2}}\pi$

$\alpha +\beta +{\frac {1}{2}}\pi =\pi$

$\alpha +\beta ={\frac {1}{2}}\pi$

$c=1$

$\sin \alpha ={\frac {a}{c}}={\frac {a}{1}}=a\Longrightarrow a=\sin \alpha$

$\cos \alpha ={\frac {b}{c}}={\frac {b}{1}}=b\Longrightarrow b=\cos \alpha$

$a=sin\alpha$

$b=cos\alpha$

$\sin \beta ={\frac {b}{c}}={\frac {\cos \alpha }{1}}=\cos \alpha$

$\cos \beta ={\frac {a}{c}}={\frac {\sin \alpha }{1}}=\sin \alpha$

${\underline {\sin \beta =\cos \alpha }}$

${\underline {\cos \beta =\sin \alpha }}$

De hypotenusa $c$ is: $c=1$

$a^{2}+b^{2}=\sin ^{2}\alpha +cos^{2}\alpha =c^{2}=1^{2}=1$

${\underline {\sin ^{2}\alpha +cos^{2}\alpha =1}}$ geldt voor elke willekeurige hoek $\alpha$.

$\sin ^{2}\alpha +cos^{2}\alpha =1=\sin({\frac {1}{2}}\pi )=\sin(\alpha +\beta )$

$\sin(\alpha +\beta )=\sin ^{2}\alpha +cos^{2}\alpha$

$\sin \alpha =\cos \beta$

$\cos \alpha =\sin \beta$

$\sin(\alpha +\beta )=\sin ^{2}\alpha +cos^{2}\alpha =\sin \alpha \cos \beta +\cos \alpha \sin \beta$

${\underline {\sin(\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta }}$

Bovenstaande formule geldt voor willekeurige hoeken $\alpha$ en $\beta$.

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Een getallenvoorbeeld:

$\alpha =4$

$\beta =5$

$\sin(\alpha +\beta )=\sin(4+5)=\sin 9=0,4121184$

$\sin(4+5)=\sin 4\cos 5+\cos 4\sin 5=0,4121184$

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$\sin(\alpha -\beta )=\sin(\alpha +(-\beta ))$

$\sin(\alpha +(-\beta ))=\sin \alpha \cos(-\beta )+\cos \alpha \sin(-\beta )$

$\cos(-\beta )=\cos \beta$

$\sin(-\beta )=-\sin \beta$

$\sin(\alpha -\beta )=\sin \alpha \cos \beta +\cos \alpha \sin(-\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta$

${\underline {\sin(\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta }}$

Bovenstaande formule geldt voor willekeurige hoeken $\alpha$ en $\beta$.

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Een getallenvoorbeeld:

$\alpha =4$

$\beta =5$

$\sin(\alpha -\beta )=\sin(4-5)=\sin(-1)=-0,8414709$

$\sin(4-5)=\sin 4\cos 5-\cos 4\sin 5=-0,8414709$

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Voor elke rechthoekige driehoek (hypotenusa c=1) geldt:

$\cos(\alpha +\beta )=\cos({\frac {1}{2}}\pi )=0$

$\cos(\alpha +\beta )=0=\sin \alpha \cos \alpha -\sin \alpha \cos \alpha$

$\sin \alpha =\cos \beta$

$\cos \alpha =\sin \beta$

$\cos(\alpha +\beta )=\cos \alpha \sin \alpha -\sin \alpha \cos \alpha =\cos \alpha \cos \beta -\sin \alpha \sin \beta$

${\underline {\cos(\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta }}$

Bovenstaande formule geldt voor willekeurige hoeken $\alpha$ en $\beta$.

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Een getallenvoorbeeld:

$\alpha =4$

$\beta =5$

$\cos(\alpha +\beta )=\cos(4+5)=\cos 9=-0,9111302$

$\cos(4+5)=\cos 4\cos 5-\sin 4\sin 5=-0,9111302$

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$\cos(\alpha -\beta )=\cos(\alpha +(-\beta ))$

$\cos(\alpha +(-\beta ))=\cos \alpha \cos(-\beta )-\sin \alpha \sin(-\beta )$

$\cos(-\beta )=\cos \beta$

$\sin(-\beta )=-\sin \beta$

${\underline {\cos(\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta }}$

Bovenstaande formule geldt voor willekeurige hoeken $\alpha$ en $\beta$.

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Een getallenvoorbeeld:

$\alpha =4$

$\beta =5$

$\cos(\alpha -\beta )=\cos(4-5)=\cos(-1)=0,5403023$

$\cos(4-5)=\cos 4\cos 5+\sin 4\sin 5=0,5403023$

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Somregel

Uit nevenstaande afbeelding "lezen" we het volgende af:

$\sin(\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta$

$\cos(\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta$