Ulwazi olusisiseko malunga namaqhezu: Ukwenza lula izinto ezifana nezi kuvumelekile. Ukwenza lula amanani afana nezi akuvumelekile. Sekunjalo, kukho amaqhezu apho oku kulula okungavumelekanga kubonakala kusebenza khona. Kufanelekile ukuphonononga usapho olulula kakhulu lwamaqhezu: amaqhezu apho idijithi efanayo ibonakala ekupheleni kwenani kunye nasekuqaleni kwenani eliphantsi.
I-numerator ngoku kufuneka ibe nenani \(a\) kunye nenani elifakelweyo \(x\) , i-denominator yenombolo efanayo \(x\) kunye nenani elifakelweyo \(c\):
$$\frac{\overline{a\,x}}{\overline{x\,c}}$$
Masithi inani lilonke lamanani kwinani kunye nenani eliphantsi libe \(n \ge 2\) .
Emva koko \(a\) kunye \(c\) nganye ineendawo zedesimali ze \(k=n-1\) . Ngaphezu koko, vumela \(x \in \{1,\dots,9\}\) .
Kwi-decimal notation eqhelekileyo, oku kuthetha:
$$\overline{a\,x}\,=10a+x$$
kwaye
$$\overline{x\,c}\,=x10^k+c.$$
Ukususwa okungavumelekanga, okuphononongwa apha, kuya kuba:
$$\frac{\overline{a\,x}}{\overline{x\,c}} \longmapsto \frac{a}{c}$$
Sikhangela ngqo ezo meko apho ixabiso lingatshintshiyo.:
$$\frac{10a+x}{x10^k+c}=\frac{a}{c}$$
Lonke ulwakhiwo ngoku lunokufumaneka ekuguquleni okulula. Ukuphindaphinda ngobuninzi kuvelisa...:
$$c(10a+x)=a(x10^k+c)$$
Xa iphindaphindwe,:
$$10ac+cx=ax10^k+ac$$
Ukuba udibanisa la magama ngokufanelekileyo, kulandela ukuba:
$$9ac=x(a10^k-c)$$
Ekubeni \(a\) kunye \(c\) zizonke \(k\) iinombolo ezipheleleyo, \(a10^kc>0\) . Oku kungenxa yokuba \(a10^k\) isoloko inkulu kunayo nayiphi na inombolo yenombolo \ \(k\) \(c\) . Ke ngoko, singahlulahlula size sifumane imeko ephakathi.:
$$\boxed{x=\frac{9ac}{a10^k-c}}$$
Le fomula ichaza kanye xa ukwenza lula okubonakalayo kusebenza kule fomu ikhethekileyo. Akufuneki nje kuphela kodwa kwanele: Ukuba le equation iyasebenza, zonke iinguqu zinokuguqulwa kwaye zibuyele kwisiphumo sokuqala.:
$$\frac{10a+x}{x10^k+c}=\frac{a}{c}$$
Ingongoma ebalulekileyo kukuba ibinzana elithi \(\frac{9ac}{a10^kc}\) ekugqibeleni kufuneka livelise idijithi enye yedesimali ukusuka ku \(\{1,\dots,9\}\) . Kulapho kuphela apho kuvela khona iqhezu elinjalo. Kwiqhezu elifanelekileyo \(a<c\) kufuneka kwakhona. Emva koko \(\frac{a}{c}<1\) nalo liyinyani, kwaye ngenxa yokulingana kwamaxabiso, iqhezu lokuqala nalo lifanelekile.
Ifomula \(\displaystyle x=\frac{9ac}{a10^kc}\) ilula kakhulu ekuqinisekiseni ubungqina. Nangona kunjalo, ifom ehlelwe ngokutsha kancinci iluncedo ngakumbi ekufumaneni imizekelo enjalo. Kwi-equation \(\frac{10a+x}{x10^k+c}=\frac{a}{c}\) sele sifumene \(\displaystyle 9ac=x(a10^kc)\) . Ngokufanayo, sine \(\displaystyle c(9a+x)=xa10^k\) .
Ngoku sijonga umahluko oqhelekileyo we \(a\) kunye \(x\) . Masivumele \(\displaystyle g=\gcd(a,x)\) . Emva koko kukho amanani \(b\) kunye \(y\) anjalo \(\displaystyle a=gb\) , \(\displaystyle x=gy\) , kunye \(\displaystyle \gcd(b,y)=1\) . Sifaka oku kwi \(\displaystyle c(9a+x)=xa10^k\) sifumana \(\displaystyle c(9b+y)=x b10^k\) . Ekubeni \(\displaystyle \gcd(9b+y,b)=\gcd(y,b)=1\) , i-factor \(9b+y\) kufuneka yahlule ngokupheleleyo ibinzana \(\displaystyle x10^k\) . Ukuba simisela \(\displaystyle d=9b+y\) , emva koko \(\displaystyle d\mid x10^k\) kunye ne \(\displaystyle d\equiv y \pmod 9\) . Ngokwahlukileyo koko, kwii-divisors ezinjalo singazifumana ngokuthe ngqo
$$\displaystyle a=g\frac{d-y}{9}$$
kwaye
$$\displaystyle c=\frac{x10^k(d-y)}{9d}.$$
Oku kuthetha ukuba akusekho mfuneko yokuba ube yimfama. \(a\) kwaye \(c\) Zizame. Ngenombolo nganye. \(x\in\{1,\dots,9\}\), isahluli ngasinye \(g\mid x\) kwaye yonke into eyahlulahlulayo ifanelekile \(d\mid x10^k\) Umntu ufumana abaviwa. Okuseleyo kukujonga ukuba ngaba \(a\) kwaye \(c\) ngokwenene \(k\)Zii-digits kwaye, ukuba ufuna amaqhezu okwenyani, nokuba \(a\) kwaye \(c\) ngokwenene \(k\)Zii-digits kwaye, ukuba ufuna amaqhezu okwenyani, nokuba \(a<c\) iyasebenza.
Imizekelo emibini:
$$\frac{16}{64}=\frac{1}{4}$$
Apha, idijithi \(6\) iyacinywa.
Umzekelo omde kakhulu oneendawo \(42\) nganye kunye nokurhoxiswa okuphindaphindwayo ngu:
$$\frac{166666666666666666666666666666666666666666}{666666666666666666666666666666666666666664}=\frac{1}{4}$$
Nalapha, idijithi efanayo iyasuswa: kwinani eliyinombolo yokugqibela \(6\) , kwinani eliyinombolo yokuqala \(6\) .