Qoldiqli bo`lish haqida teorema

Teorema. Har qanday f(x) va g(x) 0 ko`phadlar uchun shunday yagona h(x) va r(x) ko`phadlar mavjudki , ular uchun deg r(x) < degg(x) va deg h(x)f(x)=g(x) h(x)+ r(x) (2.2)
Isboti Agar f(x) ko`phaddan a <sub>m</sub>x<sup>m-n</sup>g(x) ko`phaddi ayirsak , f(x)- a _mx^m-ng(x) = r₁(x) ko`phadda a <sub>m</sub>x<sup>m</sup> had bo`lmaydi . Bu yerda quyidagi ikki hol bo`lishi mumkin :
a) r<sub>1</sub>(x) x ning darajasi g(x) ning darajasidan kichik ;
b) r<sub>1</sub>(x) ning rarajasi g(x) darajasidan katta yoki unga teng .
Agar a) hol yuz bersa , h(x) =a <sub>m</sub>x<sup>m-n</sup> ; r(x)= r<sub>1</sub>(x) bo`lib , teorema isbotlangan bo`ladi . Biz b) hol ustida to`xtalib o`tamiz . faraz qilaylik , dar r<sub>1</sub>(x) ≥ deg g(x) bo`lib , r₁(x) = c₀+c₁x +c₂x²+… +c_kx^k ko`rinishga ega bo`lsin .
Endi g(x) ko`phadni c<sub>k</sub>x<sup>k-n</sup> ga ko`paytirib , natijasini r₁(x) dan ayiramiz . U holda r₁(x) - c_kx^k-ng(x) = r₂(x) bo`lib , r<sub>2</sub>(x) ko`phadga c_kx^k had bo`lmaydi .
r<sub>2</sub>(x)=d<sub>0</sub>+d<sub>1</sub>x+d<sub>2</sub>x<sup>2</sup>+…+d<sub>l</sub>x<sup>l</sup> bo`lsin . Bu yerda yana yuqoridagi ikki holdan biri yuz berishi mumkin :
1) Agar l≥n bo`lsa , ushbu ayrmani tuzamiz :
r<sub>2</sub>(x)- d<sub>1</sub>x<sup>l-n</sup> g(x)= r<sub>3</sub>(x) ,
jarayonni davom ettirib , biror v qadamdan so`ng dar r_v(x)< dar g(x) ga erishamiz .Boshqacha aytganda , r_v-1(x)- g(x)=r_v(x) tenglikda dar r_v(x)< dar g(x) bo`ladi .
Endi ushbu tengliklarni hadlab qo`shamiz :
f(x)- a _mx^m-ng(x)= r₁(x)
r₁(x)- c _kx^k-n g(x)= r₂(x),
r₂(x)- d _lx^l-n g(x)= r₃(x),
……………………….
r_v-1(x)- g(x)= r_v(x).
unda
f(x)-( a _mx^m-n+ c _kx^k-n + d _lx^l-n+…+ )g(x)= r_v(x)
hosil bo`ladi . bu yerda
a <sub>m</sub>x<sup>m-n</sup>+ c <sub>k</sub>x<sup>k-n</sup> +…+ =h(x) va r<sub>v</sub>(x) =r(x)
desak, f(x)=g(x)·h(x)+r(x) tenglik hosil bo`ladi .
f(x)=g(x)·h(x)+r(x) tenglikdagi f(x) bo`linuvchi , g(x) bo`luvchi , h(x) chala bo`linma , r(x) esa qoldiq ko`phadlar deyiladi .
Endi (1) tenglikning yagonaligini isbotlaymiz .
Aytaylik , (1) shartni qanoatlantiruvchi yana bir juft va
Ko`phad mavjud , yani
f(x)=g(x) · + (2.3)
tenglik o`rinli bo`lsin . (2.2) va (2.3) tengliklarni hadlab ayirib
o=g(x)(h(x)- )+(r(x)- )
yoki
g(x)·(h(x)- )= - r(x) (2.4)
ni hosil qilamiz . Bu yerda r(x) va ning aniqlanishiga asosan dar
( -r(x))< dar g(x) bo`ladi . Agar chap tomonda h(x)- 0 bo`lsa , ( -r(x)) ning darajasi (2.4) ga asosan g(x) ning darajasidan kichik emas. Bu esa r(x) va ning aniqlanishiga ziddir . Shuning uchun h(x) = bo`ladi . bunga ko`ra (3) dan r(x)= kelib chiqadi .
Bu teoremani bazan f(x) ko`phadni g(x) ko`phadga qoldiqli bo`lish teoremasi deb yuritiladi .