Zatiki jarrai

Zenbaki errealen zatiki jarraien bidezko garapena.

Sarrera: Zenbaki errealak sistema hamartar posizionala erabiliz adieraz daitezkeen bezala, zatiki jarraien bidezko garapena eginez ere adieraz daitezke.

Atal honetan zenbaki erreal baten zatiki jarraien bidezko garapena definituko da. Lehenik zenbaki arrazionalen gain eta ondoren zenbaki irrazionalengain definituz.

Zenbaki baten zatiki jarraien bidezko garapena definitzeko beharrezkoak dira oinarrizko zenbait kontzeptu, eta horiek aztertuko dira lehen atalean.

Zenbaki irrazionalen zatiki jarraien bidezko garapena definitu ondoren, garapen honetako ${\displaystyle n}$. hondarrak, zenbaki irrazionalaren hurbilketa egokia direla ikusiko da zentzu honetan: Zenbaki irrazionaletik ''hurbil'' dauden zenbaki arrazional guztiek, ${\displaystyle n}$.hondarra berdina dute. (Sistema hamartar posizionalarekin gertatzen den gisara).

Oinarrizko kontzeptuak

Notazioa

${\displaystyle a_{0},a_{1},...,a_{n}\in \mathbb {R} }$, eta ${\displaystyle a_{1},...,a_{n}>0}$, izanik.

${\displaystyle [a_{0},a_{1},...,a_{n}]=a_{0}+{\frac {1}{a_{1}+{\frac {1}{a_{2}+...{\frac {1}{a_{n-1}+{\frac {1}{a_{n}}}}}}}}}}$ adieraziko da. Notazio honen baliokidea dugu: ondorengo adierazpena. ${\displaystyle \forall x>0;\forall n\in \mathbb {N} \cup \{0\}\Rightarrow [a_{0},a_{1},...,a_{n},x]=[a_{0},a_{1},...,a_{n}+{\frac {1}{x}}]}$.

Oharra: ${\displaystyle a_{0}}$zenbaki erreala emanik, bere garapena: ${\displaystyle [a_{0}]=a_{0}}$ adierazten da ( ez da beraz parte osoa), eta parte osoa adierazteko ere ${\displaystyle [x]=x}$ notazioa erabiliko denez, testuinguruaren arabera bereiziko da.

Proposizioa1

${\displaystyle a_{0}\in \mathbb {Z} }$ eta ${\displaystyle a_{1},a_{2},...,a_{n}\in \mathbb {N} }$ bada, ${\displaystyle [a_{0},a_{1},...,a_{n}]\in \mathbb {Q} }$

Froga:

${\displaystyle [a_{n}]=a_{n}\in \mathbb {N} \Rightarrow [a_{n-1},a_{n}]=a_{n-1}+{\frac {1}{[a_{n}]}}\in \mathbb {Q} -\{0\}}$

${\displaystyle \Rightarrow [a_{n-2},a_{n-1},a_{n}]=a_{n-2}+{\frac {1}{[a_{n-1},a_{n}]}}\in \mathbb {Q} -\{0\}}$

...

${\displaystyle \Rightarrow [a_{1},a_{2},...,a_{n}]=a_{1}+{\frac {1}{[a_{2},...,a_{n}]}}\in \mathbb {Q} -\{0\}}$

${\displaystyle \Rightarrow [a_{0},a_{1},...,a_{n}]=a_{0}+{\frac {1}{[a_{1},...,a_{n}]}}\in \mathbb {Q} -\{0\}}$

Definizioa1.

${\displaystyle a_{0}\in \mathbb {Z} }$ eta ${\displaystyle a_{1},a_{2},...,a_{n}\in \mathbb {N} }$ bada, ${\displaystyle [a_{0},a_{1},...,a_{n}]={\frac {p}{q}}\in \mathbb {Q} }$, idazten bada, orduan ${\displaystyle {\frac {p}{q}}}$ zenbakiaren zatiki jarraien bidezko garapena: ${\displaystyle [a_{0},a_{1},...,a_{n}]}$, dela esango dugu.

• Zatiki jarraien bidezko definizio hau soilik zenbait zenbaki arrazionalentzako da baliogarria. Zenbaki arrazional guztientzako baliogarria den definizio bat lortu nahi da. Ondorengo leman zatiki hauen ezaugarriak zehaztuko dira.

Lema1

Biz ondorengo ezaugarriak dituzten zenbakiak: ${\displaystyle a_{0}\in \mathbb {Z} }$ eta ${\displaystyle a_{1},a_{2},...,a_{n},...\in \mathbb {N} }$. Orduan:

${\displaystyle [a_{0},a_{1},...,a_{n}]={\frac {p_{n}}{q_{n}}}}$betetzen da. Zeinetan: ${\displaystyle p_{0}=a_{0},q_{0}=1,p_{1}=a_{1}a_{0}+1,q_{1}=a_{1}}$, eta

${\displaystyle {\begin{cases}p_{n+2}=a_{n+2}p_{n+1}+p_{n}\\q_{n+2}=a_{n+2}q_{n+1}+q_{n}\end{cases}}\forall n\geq 0}$, gisara definitzen diren.

Froga:

Ondorengo emaitza erabiliko da frogapenean: ${\displaystyle \forall x>0}$, rentzat ${\displaystyle [a_{0},a_{1},...,a_{n},x]=[a_{0},a_{1},...,a_{n}+{\frac {1}{x}}]}$

${\displaystyle [a_{0}]=a_{0}={\frac {a_{0}}{1}}\Rightarrow p_{0}=a_{0},q_{0}=1}$

${\displaystyle [a_{0},x_{1}]=[a_{0}+{\frac {1}{x_{1}}}]={\frac {a_{0}x_{1}+1}{x_{1}}}}$

${\displaystyle x_{1}=a_{1}\Rightarrow [a_{0},a_{1}]=[a_{0}+{\frac {1}{a_{1}}}]={\frac {a_{0}a_{1}+1}{a_{1}}}\Rightarrow p_{1}=a_{0}a_{1}+1,q_{1}=a_{1}}$

${\displaystyle x_{1}=a_{1}+{\frac {1}{x_{2}}}\Rightarrow [a_{0},a_{1},x_{2}]=[a_{0},a_{1}+{\frac {1}{x_{2}}}]={\frac {a_{0}(a_{1}+{\frac {1}{x_{2}}})+1}{a_{1}+{\frac {1}{x_{2}}}}}={\frac {p_{1}x_{2}+p_{0}}{q_{1}x_{2}+q_{0}}}}$

${\displaystyle x_{2}=a_{2}\Rightarrow [a_{0},a_{1},a_{2}]={\frac {p_{1}a_{2}+p_{0}}{q_{1}a_{2}+q_{0}}}={\frac {p_{2}}{q_{2}}}}$

${\displaystyle x_{2}=a_{2}+{\frac {1}{x_{3}}}\Rightarrow [a_{0},a_{1},a_{2},x_{3}]={\frac {p_{1}(a_{2}+{\frac {1}{x_{3}}})+p_{0}}{q_{1}(a_{2}+{\frac {1}{x_{3}}})+q_{0}}}={\frac {p_{2}x_{3}+p_{1}}{q_{2}x_{3}+q_{1}}}}$

Eta formula ${\displaystyle n}$. terminuraino egia dela suposatzen bada hots:

${\displaystyle [a_{0},a_{1},...,a_{n},x_{n+1}]={\frac {p_{n}x_{n+1}+p_{n-1}}{q_{n}x_{n+1}+q_{n-1}}}}$. Orduan:

${\displaystyle x_{n+1}=a_{n+1}\Rightarrow [a_{0},a_{1},...,a_{n},a_{n+1}]={\frac {p_{n}a_{n+1}+p_{n-1}}{q_{n}a_{n+1}+q_{n-1}}}={\frac {p_{n+1}}{q_{n+1}}}}$

${\displaystyle x_{n+1}=a_{n+1}+{\frac {1}{x_{n+2}}}\Rightarrow [a_{0},a_{1},...,a_{n},a_{n+1},x_{n+2}]={\frac {p_{n}(a_{n+1}+{\frac {1}{x_{n+2}}})+p_{n-1}}{q_{n}(a_{n+1}+{\frac {1}{x_{n+2}}})+q_{n-1}}}=}$

${\displaystyle ={\frac {(p_{n}a_{n+1}+p_{n-1})x_{n+2}+p_{n}}{(q_{n}a_{n+1}+q_{n-1})x_{n+2}+q_{n}}}={\frac {p_{n+1}x_{n+2}+p_{n}}{q_{n+1}x_{n+2}+q_{n}}}}$

Formula ${\displaystyle (n+1)}$.terminuraino egiazkoa dela ondorioztatzen da. Honela indukzio printzipioa aplikatuz:

${\displaystyle \forall n\geq 0,\forall t>0\Rightarrow [a_{0},a_{1},...,a_{n},a_{n+1},t]={\frac {p_{n+1}t+p_{n}}{q_{n+1}t+q_{n}}}}$

Eta ${\displaystyle t=a_{n+2}}$, aukeratuz:

${\displaystyle \forall n\geq 0\Rightarrow [a_{0},a_{1},...,a_{n},a_{n+1},a_{n+2}]={\frac {p_{n+1}a_{n+2}+p_{n}}{q_{n+1}a_{n+2}+q_{n}}}}$

Lema2

${\displaystyle a_{0}\in \mathbb {Z} }$ eta ${\displaystyle a_{1},a_{2},...,a_{n}\in \mathbb {N} }$, eta ${\displaystyle [a_{0},a_{1},...,a_{n}]={\frac {p_{n}}{q_{n}}}}$, adierazten bada, lema1-en bezala.

Orduan ondorengo ezaugarriak betezen dira:

1. ${\displaystyle \forall n\geq 0}$, ${\displaystyle q_{n}}$Segida gorakorra da eta: ${\displaystyle \phi ^{n-1}\leq q_{n}}$, zeinetan ${\displaystyle \phi ={\frac {1+{\sqrt {5}}}{2}}}$
2. ${\displaystyle \forall n\geq 0}$, ${\displaystyle q_{n+1}p_{n}-p_{n+1}q_{n}=(-1)^{n+1}}$, eta ${\displaystyle q_{n+2}p_{n}-p_{n+2}q_{n}=(-1)^{n+1}a_{n+2}}$.
3. ${\displaystyle \forall n\geq 0,\forall t>0}$,orduan: ${\displaystyle [a_{0},a_{1},...,a_{n},a_{n+1},t]={\frac {p_{n+1}t+p_{n}}{q_{n+1}t+q_{n}}}}$
4. ${\displaystyle \forall n\geq 0}$: ${\displaystyle zkh(p_{n},q_{n})=1}$.

Froga:

Lehen atala.

${\displaystyle q_{0}=1\leqslant q_{1}=a_{1}\in N;\forall n\geq 0\Rightarrow q_{n+2}=a_{n+2}q_{n+1}+q_{n}\geq q_{n+1}}$.

Ondorioz ${\displaystyle \forall n\geq 0}$, ${\displaystyle q_{n}}$ segida gorakorra da.

${\displaystyle \phi ^{-1}={\frac {{\sqrt {5}}-1}{2}}\leq 1=q_{0};\phi ^{0}=1\leq q_{1}\in \mathbb {N} }$.

${\displaystyle \forall n\geq 0}$ rentzat: ${\displaystyle q_{n+2}=a_{n+2}q_{n+1}+q_{n}\geq q_{n+1}+q_{n}\geq \phi ^{n}+\phi ^{n-1}=\phi ^{n-1}(1+\phi )=\phi ^{n+1}}$.

Ondorioz: ${\displaystyle \forall n\geq 0}$-rentzat, ${\displaystyle \phi ^{n-1}\leq q_{n}}$.

Bigarren atala.

Lehen azpiatala:${\displaystyle \forall n\geq 0}$, ${\displaystyle q_{n+1}p_{n}-p_{n+1}q_{n}=(-1)^{n+1}}$, frogatuko da.

${\displaystyle {\begin{cases}p_{n+2}=a_{n+2}p_{n+1}+p_{n}\\q_{n+2}=a_{n+2}q_{n+1}+q_{n}\end{cases}}\Leftrightarrow {\begin{cases}p_{n+2}q_{n+1}=a_{n+2}p_{n+1}q_{n+1}+p_{n}q_{n+1}\\q_{n+2}p_{n+1}=a_{n+2}q_{n+1}p_{n+1}+q_{n}p_{n+1}\end{cases}}}$adierazpenen kenketa eginez:

${\displaystyle q_{n+2}p_{n+1}-p_{n+2}q_{n+1}=-(q_{n+1}p_{n}-q_{n}p_{n+1})}$.

Honela ${\displaystyle f(n)=(q_{n+1}p_{n}-p_{n+1}q_{n})}$adieraziz:

${\displaystyle \forall n\geq 0,f(n)=-f(n-1)\Rightarrow f(n)=(-1)^{n}f(0)=(-1)^{n}(q_{1}p_{0}-p_{1}q_{0})}$.

${\displaystyle f(n)=(-1)^{n}(q_{1}p_{0}-p_{1}q_{0})=(-1)^{n}(a_{1}a_{0}-a_{0}a_{1}-1)=(-1)^{n+1}}$

lehen azpiatala ondorioztatuz.

Bigarren azpiatala:${\displaystyle \forall n\geq 0}$, ${\displaystyle p_{n+2}q_{n}-q_{n+2}p_{n}=a_{n+2}(p_{n+1}q_{n}-q_{n+1}p_{n})=(-1)^{n}a_{n+2}}$, frogatuko da.

${\displaystyle {\begin{cases}p_{n+2}=a_{n+2}p_{n+1}+p_{n}\\q_{n+2}=a_{n+2}q_{n+1}+q_{n}\end{cases}}\Leftrightarrow {\begin{cases}p_{n+2}q_{n}=a_{n+2}p_{n+1}q_{n}+p_{n}q_{n}\\q_{n+2}p_{n}=a_{n+2}q_{n+1}p_{n}+q_{n}p_{n}\end{cases}}}$, eta kenketa eginez:

${\displaystyle p_{n+2}q_{n}-q_{n+2}p_{n}=a_{n+2}(p_{n+1}q_{n}-q_{n+1}p_{n})=(-1)^{n}a_{n+2}}$, bigarren azpiatala ondorioztatuz.

Hirugarren atala:

Lema1-en frogatu da.

Laugarren atala

Bigarren ataleko lehen azpiatala erabiliz:

${\displaystyle \forall n\geq 0,q_{n+1}p_{n}-p_{n+1}q_{n}=(-1)^{n+1}\Rightarrow zkh(p_{n},q_{n})=1.}$

Zenbaki arrazional baten zatiki jarraien bidezko garapena

Definizioa2:

${\displaystyle x\in \mathbb {R} }$, emanik, bi segida eraikitzen dira errekurrentziz: ${\displaystyle (x_{n},a_{n})}$, zeinetan ${\displaystyle a_{n}=[x_{n}]}$ (${\displaystyle x_{n}}$-ren parte osoa testuinguruan). ${\displaystyle x_{0}=x}$, baliotik hasiko gara eta prozesua eten egingo da:${\displaystyle x_{0}\in \mathbb {Z} }$, edo ${\displaystyle x_{n}\in \mathbb {N} }$ ematen bada, eta horrela ez bada: ${\displaystyle x_{n+1}={\frac {1}{x_{n}-a_{n}}}}$, adieraziz prozesua jarraituko da. Honela:

${\displaystyle R_{n}=[a_{0},a_{1},...,a_{n}]}$, zenbaki arrazionala, ${\displaystyle x}$ -en ${\displaystyle n}$. hondarra izendatzen da.

${\displaystyle x_{n}}$, ${\displaystyle x}$-en, ${\displaystyle n}$.zatiki osatua (cociente completo) izendatzen da.

${\displaystyle a_{0},a_{1},...,a_{n}}$, zenbakiak ${\displaystyle R_{n}}$-ren koefizienteak izendatzen dira.

• Ondorengo lemari esker, zenbaki arrazionalen zatiki jarraien bidezko garapena definituko da.

Lema3:

${\displaystyle x\in \mathbb {R} }$, emanik.

1. ${\displaystyle n\in \mathbb {N} \cup \{0\}}$-rentzat ${\displaystyle x_{n}}$ existitzen bada: ${\displaystyle x=[a_{0},a_{1},...,a_{n-1},x_{n}]}$
2. ${\displaystyle (x_{n})_{n\geq 0}}$segida finitua da, baldin eta soilik baldin ${\displaystyle x\in \mathbb {Q} }$.

Froga.

Lehena:

${\displaystyle n=0\Rightarrow x=[x_{0}]=x_{0}}$( Garapena kasu honetan, ez parte osoa ).

${\displaystyle n\in \mathbb {N} }$ zeinentzat existitzen den ${\displaystyle x_{n}}$, orduan: ${\displaystyle x_{1},...,x_{n-1}\not \in \mathbb {N} }$, eta ${\displaystyle x_{0}\not \in \mathbb {Z} }$, bestela prozesua eten egin bai da.

${\displaystyle x=x_{0}=[x_{0}]+x_{0}-[x_{0}]=a_{0}+{\frac {1}{\frac {1}{x_{0}-a_{0}}}}=[a_{0},{\frac {1}{x_{0}-a_{0}}}]=[a_{0},x_{1}]}$

${\displaystyle x=[a_{0},x_{1}]=[a_{0},a_{1}+{\frac {1}{\frac {1}{x_{1}-a_{1}}}}]=[a_{0},a_{1},{\frac {1}{x_{1}-a_{1}}}]=[a_{0},a_{1},x_{2}]}$.

${\displaystyle x_{1},...,x_{n-1}\not \in \mathbb {N} }$ betetzen denez, prozesua jarrai dezakegu: ${\displaystyle x_{n}}$ koefizienterarte.

${\displaystyle x=[a_{0},a_{1},...,a_{n-1},x_{n}]}$

Bigarrena

${\displaystyle \Rightarrow }$

${\displaystyle (x_{n})_{n\geq 0}}$ Segida finitua da.

Soilik ${\displaystyle x_{0}}$, terminua badu: ${\displaystyle x=x_{0}\in \mathbb {Z} }$, eta prozesua eten egin da.

${\displaystyle x\not \in \mathbb {Z} }$ ${\displaystyle \Rightarrow \exists n\in \mathbb {N} :x_{n}\in \mathbb {N} }$, eta prozesua eten egin da.

Ondorioz ${\displaystyle x=[a_{0},a_{1},...,a_{n-1},x_{n}]}$.

Proposizioa1-engatik: ${\displaystyle a_{0}\in \mathbb {Z} ;a_{1},...,a_{n-1},x_{n}=a_{n}\in \mathbb {N} \Rightarrow x\in \mathbb {Q} }$.

${\displaystyle \Leftarrow }$

${\displaystyle x\in \mathbb {Q} \Rightarrow x={\frac {a}{b}}:a\in \mathbb {Z} ;b\in \mathbb {N} }$, orduan ${\displaystyle a}$ eta ${\displaystyle b}$-ren arteko zatidura euklidearra eginez:

${\displaystyle a=ba_{0}+r_{1},0\leq r_{1}<b\Rightarrow {\frac {a}{b}}=[a_{0},{\frac {b}{r_{1}}}]}$

${\displaystyle b=r_{1}a_{1}+r_{2},0\leq r_{2}<r_{1}\Rightarrow {\frac {a}{b}}=[a_{0},a_{1},{\frac {r_{1}}{r_{2}}}]}$

${\displaystyle r_{1}=r_{2}a_{2}+r_{3},0\leq r_{3}<r_{2}\Rightarrow {\frac {a}{b}}=[a_{0},a_{1},a_{2},{\frac {r_{2}}{r_{3}}}]}$

...

Prozesua errepikatuz, ondorengo segida eraiki daieteke:

${\displaystyle 0\leq r_{n}<...<r_{2}<r_{1}<b\Rightarrow \exists n\in \mathbb {N} :r_{n+1}=0\Rightarrow r_{n-1}=a_{n}r_{n}}$

${\displaystyle {\frac {a}{b}}=[a_{0},a_{1},...,a_{n-1},a_{n}]}$

• Ondorioz: Zenbaki arrazionalentzat, zatiki jarraien bidezko garapena defini daiteke. Gainera ${\displaystyle x}$ zenbakia arrazionala ez bada, bere koefizienteen garapena ez da finitua ondorioztatzen da azken lematik. Kasu horretan zatiki jarraien bidezko bere garapena, limite moduan definituko da.

Definizioa4:

${\displaystyle x\in \mathbb {Q} }$ bada aurreko lemagatik:${\displaystyle (x_{k})_{k\geq 0}}$ segida finitua da, eta ${\displaystyle x_{n}}$bada azken terminua:

${\displaystyle x=[a_{0},a_{1},...,a_{n-1},a_{n}]}$, adierazpena, ${\displaystyle x}$-en zatiki jarraien bidezko garapena izendatuko da.

Zenbaki irrazional baten zatiki jarraien bidezko garapena

• ${\displaystyle \alpha }$ Zenbaki irrazionala bada eta: ${\displaystyle a_{0},a_{1},...,a_{n-1},a_{n},...}$, bere koefizienteen segida, ${\displaystyle \alpha =[a_{0},a_{1},...,a_{n},...]}$dela frogatzea da ondorengo proposizioaren helburua. Modu honetan, definizioa osatu egingo da.

Proposizioa2:

Izan bedi ${\displaystyle \alpha }$, zenbaki irrazionala, eta ${\displaystyle (R_{n})_{n\geq 0}}$, ${\displaystyle n}$. hondarren segida, orduan:

1. ${\displaystyle (R_{2n})_{n\geq 0}}$, segida hertsiki gorakorra da eta ${\displaystyle \forall n\geq 0,R_{2n}<\alpha }$.
2. ${\displaystyle (R_{2n+1})_{n\geq 0}}$, segida hertsiki beherakorra da eta ${\displaystyle \forall n\geq 0,R_{2n+1}>\alpha }$.
3. ${\displaystyle \forall n\geq 0,\left\vert R_{2n+1}-R_{2n}\right\vert \leq {\frac {1}{\phi ^{4n-1}}}}$
4. ${\displaystyle \alpha =\lim _{n\to \infty }R_{n}}$

Froga

Frogapenean erabiliko diren ondorengo funtzioak definituko dira:

${\displaystyle f_{0}(x)=x}$, zeinetan : ${\displaystyle x\in \mathbb {R} }$ eta ${\displaystyle f_{n}(x)=[a_{0},a_{1},...,a_{n-1},x]}$, zeinetan ${\displaystyle x>0}$.

${\displaystyle n\in \mathbb {N} }$-rentzat funtzioen arteko ondorengo berdintza betetzen da:

${\displaystyle f_{n}(x)=[a_{0},a_{1},...,a_{n-1},x]=[a_{0},a_{1},...,a_{n-1}+{\frac {1}{x}}]=f_{n-1}(a_{n-1}+{\frac {1}{x}})}$.

${\displaystyle \forall n\in \mathbb {N} \Rightarrow f_{n}(x)=f_{n-1}(a_{n-1}+{\frac {1}{x}})}$. Ondorioz:

${\displaystyle f_{n}(x)}$gorakorra baldin eta soilik baldin ${\displaystyle f_{n-1}(x)}$ beherakorra.

${\displaystyle f_{0}(x)}$ gorakorra denez: ${\displaystyle f_{2n}(x)}$ gorakorra da, eta ${\displaystyle f_{2n+1}(x)}$ beherakorra ${\displaystyle \forall n\in \mathbb {N} \cup \{0\}}$.

Honetaz gain: ${\displaystyle f_{n}(a_{n})=R_{n}}$, ${\displaystyle \alpha }$-ren ${\displaystyle n}$. hondarra da, ${\displaystyle \alpha =f_{n}(\alpha _{n})}$, 3.Lemagatik, eta ${\displaystyle a_{n}=[\alpha _{n}]<\alpha _{n}}$.

Lehen atala:

${\displaystyle R_{2n+2}=f_{2n+2}(a_{2n+2})=f_{2n+1}(a_{2n+1}+{\frac {1}{a_{2n+2}}})=f_{2n}(a_{2n}+{\frac {1}{a_{2n+1}+{\frac {1}{a_{2n+2}}}}})>}$

${\displaystyle >f_{2n}(a_{2n})=R_{2n}\Rightarrow (R_{2n})_{n\geq 0}\uparrow }$, hertsiki gorakorra.

Gainera: ${\displaystyle R_{2n}=f_{2n}(a_{2n})=f_{2n}([\alpha _{2n}])<f_{2n}(\alpha _{2n})=\alpha }$.

Bigarren atala:

${\displaystyle R_{2n+3}=f_{2n+3}(a_{2n+3})=f_{2n+2}(a_{2n+2}+{\frac {1}{a_{2n+3}}})=f_{2n+1}(a_{2n+1}+{\frac {1}{a_{2n+2}+{\frac {1}{a_{2n+3}}}}})<}$

${\displaystyle <f_{2n+1}(a_{2n+1})=R_{2n+1}\Rightarrow (R_{2n+1})_{n\geq 0}\downarrow }$, hertsiki beherakorra.

Gainera: ${\displaystyle R_{2n+1}=f_{2n+1}(a_{2n+1})=f_{2n+1}([\alpha _{2n+1}])>f_{2n+1}(\alpha _{2n+1})=\alpha }$.

Hirugarren atala:

${\displaystyle n}$.hondarrak zatiki moduan adieraziko dira: ${\displaystyle R_{n}={\frac {p_{n}}{q_{n}}}}$. Lema2-ko: 1 eta 2 emaitzak erabiliz:${\displaystyle |R_{2n+1}-R_{2n}|=|{\frac {p_{2n+1}}{q_{2n+1}}}-{\frac {p_{2n}}{q_{2n}}}|={\frac {|p_{2n+1}q_{2n}-q_{2n+1}p_{2n}|}{q_{2n+1}q_{2n}}}={\frac {1}{q_{2n+1}q_{2n}}}<{\frac {1}{\phi ^{4n-1}}}}$

Laugarren atala:

${\displaystyle |R_{n}-\alpha |={\begin{cases}\alpha -R_{2k}<|R_{2k+1}-R_{2k}|<{\frac {1}{\phi ^{4k-1}}},&{\text{n=2k,bada}}\\R_{2k+1}-\alpha <|R_{2k+1}-R_{2k}|<{\frac {1}{\phi ^{4k-1}}},&{\text{n=2k+1,bada}}\end{cases}}\Rightarrow }$

${\displaystyle \lim _{n\to \infty }|R_{n}-\alpha |\leq \lim _{n\to \infty }{\frac {1}{\phi ^{2n-3}}}=0\Rightarrow \lim _{n\to \infty }R_{n}=\alpha }$

Definizioa5:

${\displaystyle \alpha }$ zenbaki irrazionala bada, eta ${\displaystyle a_{0},a_{1},...,a_{n}.n}$.hondarraren koefizienteak:${\displaystyle R_{n}=[a_{0},a_{1},...,a_{n-1},a_{n}]}$.

${\displaystyle \alpha }$-ren zatiki jarraien bidezko garapena honela definitzen da.

${\displaystyle \alpha =\lim _{n\to \infty }R_{n}=\lim _{n\to \infty }[a_{0},a_{1},...,a_{n-1},a_{n}]=[a_{0},a_{1},...,a_{n-1},a_{n},...]}$. Zeinetan azken berdintza notazioa den.

Hurbilketa egokiak

Proposizioa3:

Izan bedi: ${\displaystyle \alpha }$ zenbaki irrazionala eta ${\displaystyle R_{n}={\frac {p_{n}}{q_{n}}}}$, zatiki jarraien bidezko n.hondarra, zatiki laburtezin gisara adierazia. Izan bedi: ${\displaystyle r={\frac {p}{q}}}$, zatiki laburtezin bat. Orduan:

1. ${\displaystyle \forall n\in \mathbb {N} \cup \{0\}}$, ${\displaystyle {\frac {1}{q_{n}(q_{n}+q_{n+1})}}<\left\vert \alpha -R_{n}\right\vert <{\frac {1}{q_{n}q_{n+1}}}}$
2. Baldin eta ${\displaystyle \exists m\in \mathbb {N} \cup \{0\}:q<q_{m+1}\Rightarrow |\alpha -R_{n}|\leq {\frac {q}{q_{m}}}|\alpha -r|}$eta ${\displaystyle |\alpha -R_{n}|={\frac {q}{q_{m}}}|\alpha -r|\Leftrightarrow r=R_{m}}$

Froga:

Lehena:

${\displaystyle \forall k\in \mathbb {N} \cup \{0\}\Rightarrow R_{2k}<R_{2k+2}<\alpha <R_{2k+3}<R_{2k+1}}$

Batetik:

${\displaystyle \Rightarrow {\begin{cases}R_{n}<\alpha <R_{n+1}&n=2k{\text{, Bikoitia}}\\R_{n+1}<\alpha <R_{n}&n=2k+1{\text{, Bakoitia}}\end{cases}}}$${\displaystyle \Rightarrow |\alpha -R_{n}|<|R_{n+1}-R_{n}|}$

Bestetik:

${\displaystyle \Rightarrow {\begin{cases}R_{n}<R_{n+2}<\alpha &n=2k{\text{, Bikoitia}}\\\alpha <R_{n+2}<R_{n}&n=2k+1{\text{, Bakoitia}}\end{cases}}}$ ${\displaystyle \Rightarrow |\alpha -R_{n}|>|R_{n+2}-R_{n}|}$

Bi emaitzak bateratuz:

${\displaystyle |R_{n+2}-R_{n}|<|\alpha -R_{n}|<|R_{n+1}-R_{n}|}$

Lema2-ko 2.ataleko emaitzak erabiliz, eta ${\displaystyle R_{n}={\frac {p_{n}}{q_{n}}}}$ dela jakinik:

${\displaystyle |R_{n+2}-R_{n}|=|{\frac {p_{n+2}}{q_{n+2}}}-{\frac {p_{n}}{q_{n}}}|=|{\frac {p_{n+2}q_{n}-q_{n+2}p_{n}}{q_{n}q_{n+2}}}|={\frac {a_{n+2}}{q_{n}q_{n+2}}}}$

${\displaystyle |R_{n+1}-R_{n}|=|{\frac {p_{n+1}}{q_{n+1}}}-{\frac {p_{n}}{q_{n}}}|=|{\frac {p_{n+1}q_{n}-q_{n+1}p_{n}}{q_{n}q_{n+1}}}|={\frac {1}{q_{n}q_{n+1}}}}$

Eta ondorioz: ${\displaystyle {\frac {a_{n+2}}{q_{n}q_{n+2}}}<|\alpha -{\frac {p_{n}}{q_{n}}}|<{\frac {1}{q_{n}q_{n+1}}}}$

Ondorengo funtzioa kontsideratzen da: ${\displaystyle f(x)={\frac {x}{q_{n}(xq_{n+2}+q_{n})}},x>0}$

${\displaystyle f'(x)={\frac {q_{n}(xq_{n+2}+q_{n})-xq_{n}q_{n+2}}{q_{n}^{2}(xq_{n+2}+q_{n})^{2}}}={\frac {1}{(xq_{n+2}+q_{n})^{2}}}>0}$, eta beraz ${\displaystyle f(x)}$gorakorra.

${\displaystyle 1\leq a_{n+2}\Rightarrow f(1)\leq f(a_{n+2})\Leftrightarrow {\frac {1}{q_{n}(q_{n}+q_{n+1})}}\leq {\frac {a_{n+2}}{q_{n}(a_{n+2}q_{n+1}+q_{n})}}={\frac {a_{n+2}}{q_{n}q_{n+2}}}}$, eta lehen atala ondorioztatzen da:${\displaystyle {\frac {1}{q_{n}(q_{n}+q_{n+1})}}<|\alpha -{\frac {p_{n}}{q_{n}}}|<{\frac {1}{q_{n}q_{n+1}}}}$.

Bigarren atala.

Suposa bedi ${\displaystyle \exists m\in \mathbb {N} \cup \{0\}:q<q_{m+1}}$. Ondorengo ekuazio sistema kontsideratzen da:

${\displaystyle {\begin{cases}p_{m+1}u+p_{m}v=p\\q_{m+1}u+q_{m}v=q\end{cases}}}$, Ekuazio sistemaren determinantea: ${\displaystyle p_{m+1}q_{m}-q_{m+1}p_{m}=(-1)^{m}\neq 0}$

Ondorioz koefizienteen matrizeen heina, matrize zabalduaren heina da, eta ezezagun kopuruaren berdina: Sitema bateragarri determinatua beraz.

Bere ebazpena:

${\displaystyle {\begin{cases}u=(-1)^{m}(pq_{m}-qp_{m})\\v=(-1)^{m}(qp_{m+1}-pq_{m+1})\end{cases}}:(u,v)\in \mathbb {Z} \times \mathbb {Z} }$

1. ${\displaystyle v\neq 0}$ da. Absurdura bideraztuz: ${\displaystyle v=0\Leftrightarrow {\frac {p_{m+1}}{q_{m+1}}}={\frac {p}{q}}}$, eta ${\displaystyle zkh(p_{m+1},q_{m+1})=zkh(p,q)=1\Rightarrow p=p_{m+1};q=q_{m+1}}$, hipotesiagatik: ${\displaystyle q<q_{m+1}}$, ezinezkoa.
2. ${\displaystyle u=0\Leftrightarrow |\alpha -R_{n}|={\frac {q}{q_{m}}}|\alpha -r|\Leftrightarrow r=R_{m}}$ frogatuko da. ${\displaystyle u=0\Leftrightarrow {\frac {p_{m}}{q_{m}}}={\frac {p}{q}}}$, eta ${\displaystyle zkh(p_{m},q_{m})=zkh(p,q)=1\Rightarrow p=p_{m};q=q_{m}}$, ondorioz: ${\displaystyle \Rightarrow |\alpha -R_{m}|=|\alpha -r|={\frac {q}{q_{m}}}|\alpha -r|.}$ Eta alderantziz ,azken berdintza betetzen bada: ${\displaystyle |\alpha -R_{m}|=|\alpha -r|={\frac {q}{q_{m}}}|\alpha -r|\Rightarrow }$i) ${\displaystyle \alpha -R_{m}={\frac {q}{q_{m}}}(\alpha -r)}$ edo ii) ${\displaystyle \alpha -R_{m}=-{\frac {q}{q_{m}}}(\alpha -r)}$, eta ii) kasua ematea ezinezkoa da zeren kasu horretan: ${\displaystyle \alpha ={\frac {p+p_{m}}{q+q_{m}}}\in \mathbb {Q} }$, eta ${\displaystyle \alpha }$irrazionala. Ondorioz, ${\displaystyle \alpha -R_{m}={\frac {q}{q_{m}}}(\alpha -r)}$ betetzen da ${\displaystyle \Rightarrow \alpha (q_{m}-q)=p_{m}-p}$, eta ${\displaystyle \alpha }$ irrazionala: ${\displaystyle \Rightarrow q_{m}=q;p_{m}=p\Rightarrow R_{m}=r}$.
3. ${\displaystyle u\neq 0\Leftrightarrow |\alpha -R_{n}|<{\frac {q}{q_{m}}}|\alpha -r|}$ frogatuko da. ${\displaystyle 0<q=q_{m+1}u+q_{m}v<q_{m+1}}$, betetzen da hipotesiagatik. ${\displaystyle u<0;v<0\Rightarrow 0<q<0}$, ezinezkoa da eta ${\displaystyle u>0;v>0\Rightarrow q_{m+1}<q_{m+1}u+q_{m}v<q_{m+1}}$, ere ezinezkoa, ondorioz ${\displaystyle uv<0}$. ${\displaystyle {\begin{cases}R_{2k}<\alpha <R_{2k+1}\\R_{2k+2}<\alpha <R_{2k+1}\end{cases}}\Rightarrow (\alpha -R_{m})(\alpha -R_{m+1})<0}$ ${\displaystyle \Rightarrow v(\alpha -R_{m})u(\alpha -R_{m+1})>0\Rightarrow v(q_{m}\alpha -p_{m})u(q_{m+1}\alpha -p_{m+1})>0}$, honela: ${\displaystyle |q\alpha -p|=|(q_{m+1}u+q_{m}v)\alpha -(p_{m+1}u+p_{m}v)|=|u(q_{m+1}\alpha -p_{m+1})+v(q_{m}\alpha -p_{m})|}$Eta azken desberdintzagatik: ${\displaystyle |q\alpha -p|=|u(q_{m+1}\alpha -p_{m+1})|+|v(q_{m}\alpha -p_{m})|}$, betetzen da. Honela:${\displaystyle |q\alpha -p|=|u(q_{m+1}\alpha -p_{m+1})|+|v(q_{m}\alpha -p_{m})|>|v(q_{m}\alpha -p_{m})|\geq |q_{m}\alpha -p_{m}|}$eta beraz: ${\displaystyle \Rightarrow |\alpha -R_{n}|<{\frac {q}{q_{m}}}|\alpha -r|}$. Eta alderantziz ${\displaystyle |\alpha -R_{n}|<{\frac {q}{q_{m}}}|\alpha -r|}$, betetzen bada, eta ${\displaystyle u=0}$, orduan 2.atalagatik: ${\displaystyle |\alpha -R_{n}|={\frac {q}{q_{m}}}|\alpha -r|}$, ezinezkoa.

Proposizioa4

Izan bedi: ${\displaystyle \alpha }$ zenbaki irrazionala eta ${\displaystyle R_{n}={\frac {p_{n}}{q_{n}}}}$, zatiki jarraien bidezko n.hondarra, zatiki laburtezin gisara adierazia. Orduan:

1. ${\displaystyle \forall n\in \mathbb {N} \cup \{0\}}$ ${\displaystyle \Rightarrow \left\vert \alpha -R_{n}\right\vert <{\frac {1}{2q_{n}^{2}}}}$ edo ${\displaystyle \left\vert \alpha -R_{n+1}\right\vert <{\frac {1}{2q_{n+1}^{2}}}}$
2. ${\displaystyle r={\frac {p}{q}}\in \mathbb {Q} ,zkh(p,q)=1;|\alpha -r|<{\frac {1}{2q^{2}}}\Rightarrow r=R_{n}}$non ${\displaystyle q\in [q_{n},q_{n+1}{\bigr )}\cap \mathbb {Z} }$

Froga:

Lehena:${\displaystyle |\alpha -R_{n}|+|\alpha -R_{n+1}|=|R_{n+1}-R_{n}|={\frac {|p_{n+1}q_{n}-q_{n}p_{n+1}|}{q_{n}q_{n+1}}}={\frac {1}{q_{n}q_{n+1}}}}$

Bestalde: ${\displaystyle a\neq 0,b\neq 0\Rightarrow ab<{\frac {a^{2}+b^{2}}{2}}\Rightarrow {\frac {1}{q_{n}q_{n+1}}}<{\frac {1}{2q_{n}^{2}}}+{\frac {1}{2q_{n+1}^{2}}}}$, eta beraz ${\displaystyle \left\vert \alpha -R_{n}\right\vert <{\frac {1}{2q_{n}^{2}}}}$edo ${\displaystyle \left\vert \alpha -R_{n+1}\right\vert <{\frac {1}{2q_{n+1}^{2}}}}$.

Bigarrena: 2.Lemagatik: ${\displaystyle q_{n}\uparrow \infty }$, eta ${\displaystyle q_{0}=1\leq q\Rightarrow \exists !n\in \mathbb {N} \cup \left\{0\right\}}$, non ${\displaystyle q\in [q_{n},q_{n+1}{\bigr )}\cap \mathbb {Z} }$.

${\displaystyle q<q_{n+1}\Rightarrow |\alpha -R_{n}|\leq {\frac {q}{q_{n}}}|\alpha -r|}$, 2.proposizioagaitik.

${\displaystyle |r-R_{n}|\leq |\alpha -R_{n}|+|\alpha -r|\leq (1+{\frac {q}{q_{n}}})|\alpha -r|<(1+{\frac {q}{q_{n}}}){\frac {1}{2q^{2}}}}$ ${\displaystyle |r-R_{n}|<{\frac {q_{n}+q}{2q_{n}q^{2}}}\leq {\frac {2q}{2q_{n}q^{2}}}={\frac {1}{qq_{n}}}\Rightarrow |{\frac {p}{q}}-{\frac {p_{n}}{q_{n}}}|<{\frac {1}{qq_{n}}}}$, eta ondorioz: ${\displaystyle |pq_{n}-qp_{n}|<1;pq_{n}-qp_{n}\in \mathbb {Z} \Rightarrow pq_{n}-qp_{n}=0\Rightarrow r=R_{n}}$.

Kanpo estekak