Zenbaki errealen zatiki jarraien bidezko garapena.
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**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
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### **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.

### **Definizioa**4:

{\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
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* {\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
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### **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}}.