# 整数的整除性 - PowerPoint PPT Presentation

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，则

，则必有

（或2）或

（或1）

a=bq+r 0<r<b

①a和b的任一公因数也是b和r的公因数；

②b和r的任一公因数也是a和b的公因数；

③(a,b)=(b,r);

④若(a,b)=d, 则 (a|d, b|d)=1.

①若d|a且b|d, 则d|(a-bq)或d|r. 这即表明: d是a和b的公因数, d必是b和r的公因数.

②若d|b且d|r, 则: d|(bq＋r)或d|a. 这即是说d是b和 r的公因数, d也必是a和b的公因数.

③由①和②知, a和b的公因数集合等于b和r的公因数集合，故两个集合的最大整数相同, 即(a,b)=(b,r).

(a,b)=(b,r)的证明也可不基于①和②, 另有证法. 因为(a,b)|a, (a,b)|b, 故(a,b)|(a-bq)或(a,b)|r. 因而(a,b)≤(b,r). 同法可证(b,r) ≤(a,b).于是得到(a,b)=(b,c).

④假设c=(a/d,b/d). 显然c≥1, 只要再证明c≤1即得c=1. 因为c|(a/d), c|(b/d), 于是a/d=cu, b/d=cv或(cd)u=a, (cd)v=b, 即知d是a和b的公因数, 所以ad≤(a,b), 即cd≤d. 由于d≥1, 可得c≤1, 因此, c=1.

a=bq1+r1, 0＜rl＜b

b=r1q2+r2, 0＜r2＜r1

r1=r2q3+r3, 0＜r3＜r2

…….

rn-2=rn-1qn+rn, 0＜rn＜rn-1

rn-1=rnqn+1+0

(ⅰ) 由bac及(a, b)=1可以推出bc；

(ⅱ) 由bc，ac及(a, b)=1可以推abc。

ax  by =1。

acx  bcy = c。 (2)

(ⅱ) 若(a, b)=1，则存在整数x，y使得式(2)成立。由bc与ac得到abac，abbc，再由式(2)得到abc。结论(ⅱ)得证。