Ulwazi oluyisisekelo mayelana nezingxenyana: Ukwenza izinto zibe lula njengezinto ezifanayo kuyavunyelwa. Ukwenza izinto zibe lula njengezinombolo akuvumelekile. Noma kunjalo, kunezingxenyana lapho lokhu kwenza izinto zibe lula okwenqatshelwe kubonakala kusebenza khona. Kuhle ukuhlola umndeni olula kakhulu wezingxenyana: izinxenyana lapho kuvela khona idijithi efanayo ekugcineni kwenombolo kanye nasekuqaleni kwe-denominator.
I-numerator kufanele manje ibe nenombolo \(a\) kanye nenombolo engeziwe \(x\) , i-denominator yenombolo efanayo \(x\) kanye nenombolo engeziwe \(c\):
$$\frac{\overline{a\,x}}{\overline{x\,c}}$$
Vumela inani eliphelele lezinombolo ku-numerator kanye ne-denominator libe yi- \(n \ge 2\) .
Bese kuthi \(a\) kanye \(c\) ngayinye ibe nezindawo zedesimali ze- \(k=n-1\) . Ngaphezu kwalokho, vumela \(x \in \{1,\dots,9\}\) .
Ku-notation evamile yedesimali, lokhu kusho ukuthi:
$$\overline{a\,x}\,=10a+x$$
futhi
$$\overline{x\,c}\,=x10^k+c.$$
Ngakho-ke ukususwa okwenqatshelwe, okuhlolwa lapha, kungaba:
$$\frac{\overline{a\,x}}{\overline{x\,c}} \longmapsto \frac{a}{c}$$
Sifuna ngqo lawo macala lapho inani lingashintshi khona.:
$$\frac{10a+x}{x10^k+c}=\frac{a}{c}$$
Isakhiwo sonke manje singatholakala ekuphatheni okulula. Ukuphindaphinda okuhlanganisiwe kuveza...:
$$c(10a+x)=a(x10^k+c)$$
Uma kuphindaphindwa,:
$$10ac+cx=ax10^k+ac$$
Uma uhlanganisa imigomo ngendlela efanele, kulandela lokho:
$$9ac=x(a10^k-c)$$
Njengoba \(a\) kanye \(c\) kuyizinombolo eziphelele zedijithi ngayinye \(k\) eziqondile, \(a10^kc>0\) . Lokhu kungenxa yokuthi \(a10^k\) ihlala inkulu kunanoma iyiphi inombolo yedijithi \(k\) \(c\) . Ngakho-ke, singahlukanisa futhi sithole isimo esimaphakathi.:
$$\boxed{x=\frac{9ac}{a10^k-c}}$$
Le fomula ichaza ngqo ukuthi ukwenza lula okubonakalayo kusebenza nini kule fomu ekhethekile. Akudingeki nje kuphela kodwa futhi kwanele: Uma lesi sibalo sisebenza, khona-ke zonke izinguquko zingabuyiselwa emuva futhi ziholele emuva emphumeleni wokuqala.:
$$\frac{10a+x}{x10^k+c}=\frac{a}{c}$$
Iphuzu elibalulekile ukuthi inkulumo ethi \(\frac{9ac}{a10^kc}\) kumele ekugcineni iphumele enanini elilodwa ledesimali elivela ku \(\{1,\dots,9\}\) . Kulapho kuphela lapho kuvela khona ingxenye enjalo. Kuma-fraction afanele \(a<c\) kudingeka futhi. Khona-ke \(\frac{a}{c}<1\) nayo iyiqiniso, futhi ngenxa yokulingana kwamanani, ingxenye yokuqala nayo ifanelekile.
Ifomula \(\displaystyle x=\frac{9ac}{a10^kc}\) ilula kakhulu ebufakazini. Kodwa-ke, ifomu elihlelwe kabusha kancane liwusizo kakhulu ekutholeni izibonelo ezinjalo. Kusukela ku-equation \(\frac{10a+x}{x10^k+c}=\frac{a}{c}\) sesivele sithole \(\displaystyle 9ac=x(a10^kc)\) . Ngokufanayo, sine \(\displaystyle c(9a+x)=xa10^k\) .
Manje sesibheka isihlukanisi esivamile se- \(a\) kanye ne- \(x\) . Ake \(\displaystyle g=\gcd(a,x)\) . Bese kuba nezinombolo \(b\) kanye ne- \(y\) kangangokuthi \(\displaystyle a=gb\) , \(\displaystyle x=gy\) , kanye ne- \(\displaystyle \gcd(b,y)=1\) . Sifaka lokhu ku- \(\displaystyle c(9a+x)=xa10^k\) sithola \(\displaystyle c(9b+y)=x b10^k\) . Njengoba \(\displaystyle \gcd(9b+y,b)=\gcd(y,b)=1\) , isici \(9b+y\) kumele sihlukanise ngokuphelele inkulumo \(\displaystyle x10^k\) . Uma sibeka \(\displaystyle d=9b+y\) , khona-ke \(\displaystyle d\mid x10^k\) kanye no \(\displaystyle d\equiv y \pmod 9\) . Ngokuphambene nalokho, kusukela kulezo zihlukanisi singathola ngqo
$$\displaystyle a=g\frac{d-y}{9}$$
futhi
$$\displaystyle c=\frac{x10^k(d-y)}{9d}.$$
Lokhu kususa isidingo sokuzama ngokungacabangi \(a\) kanye \(c\) . Ngenombolo ngayinye \(x\in\{1,\dots,9\}\) , i-divisor ngayinye \(g\mid x\) , kanye ne-divisor ngayinye efanelekile \(d\mid x10^k\) kutholakala ama-candidate. Okusele nje ukuhlola ukuthi \(a\) kanye \(c\) ziyizinombolo \(k\) ngempela futhi, uma kudingeka izingxenyana ezifanele, ukuthi \(a<c\) . Lokhu kushesha kakhulu ngoba \(\displaystyle 10^k=2^k5^k\) inezihlukanisi ezimbalwa kakhulu uma kuqhathaniswa nenani eliphelele lezinombolo \(k\) ezingaba khona.
Izibonelo ezimbili:
$$\frac{16}{64}=\frac{1}{4}$$
Lapha, inombolo \(6\) iyasuswa.
Isibonelo eside kakhulu esinezindawo \(42\) ngasinye kanye nokukhanselwa okuphindaphindiwe yilesi:
$$\frac{166666666666666666666666666666666666666666}{666666666666666666666666666666666666666664}=\frac{1}{4}$$
Nalapha futhi, idijithi efanayo iyasuswa: ku-numerator okokugcina \(6\) , ku-denominator owokuqala \(6\) .