Without loss of generality, let us suppose that r2≥r1. Substituting for b, we haveq1a+r1=q2a+r2But thenq1a−q2ar2−r1==r2−r1and thus,a(q1−q2)Consequently, r2−r1 is some multiple of a.However, since 0≤r1,r2r1, it must be the case that 0≤r2−r1Further, as seen before,q2=b−r2aandq1=b−r1aSince r1 and r2 agree in value, this forces q1 and q2 to agree in value as well.Thus, having shown that if b=q1a+r1 and b=q2a+r2, then it can only be the case that both q1=q2 and r1=r2, we can conclude that if there exist integers q and r such that b=qa+r with 0≤rSince the first part of our argument established the existence of these integers q and r, we are done. It is indeed the case that given integers a and b, with a>0, there exist unique integers q and r such that b=qa+r and 0≤rQED.
การแปล กรุณารอสักครู่..