Question

Prove the following version of the division algorithm, which holds for both positive and negative divisors. Extended Division Algorithm: Let a and b be integers with b = 0. Then there exist unique integers q and r such that a = bq + r and 0 sr<|bl| [ Hint: Apply Theorem 1.1 when a divided by [b]. Then consider two cases (b >0 and b < 0) Explain the answer and visible for read

Answer 1

Theeran: Extended divison algorithm theorem: Given integer a and b with bto. there exist unique integers q and r such that a = bq to and ocr<lbl. Cres Proof Case - I: when bio. Let us consider the subset of integers Saya ba : 467, a-ba 0} First we show that S is non-empty since byt lal bylal Therefore at lalby at lalyo This shown that a-bf-lalles and therefore Sit e non-empty. Since I is a non-empty set of non-negative integers, either 6 s contains O as its least element or a l ① s contains a smallest possi positive integer as its least element by the well ordering property of the set in In either cash we call it n. Therefore I an integer a such that a-bq aro 18yo ... we assert that rabi Beacuse if ryb. then a-(a +i) 6 = (a-an) -6 = 1-670 This shows that a- (2+1) bts. and also a- (a +1) b en-b in. This leads to a Contsradiation to the fact that r is the least element in S. Hence rab and consequently a= bqtr where os rab. In order to establish uniqueness of a andr, let us suppose that a fas two representation : a = bq tr, a= bq, tr, where oar, zh our <b.

ant Then b(q-a)=nur.or. blana, 1 = 10,-r1 But osrab and _bcorso yield - b<"-rcb ie In-rlebo Consente Consequently lq-alt f Since q and or, are integers, the only possibility is q=%, and therefore rar, Case-It i'mit Case-It wtien b<0. a ti I Then Iblyo. By the previous theorem 3 unique integers q, and p such that a = 1619, +r o<n<6. = - bą, trist with Therefore a = bq tr where a = -9, This completes the proof all