1.1 Introduction
The principle of induction 归纳法则
意思应该是:证明了1成立,如果n成立意味着n+1成立,那么对于所有大于等于1的数都成立
Well-ordering principle 良序原理
在学习本书过程中作为公理用
1.3 Greatest common divisor
定理1.2用归纳法证明其实原理还是a-b b和a b有共同的最大公约数
定理1.4(b)证明
d = (a,(b,c)),则d|a,d|(b,c),那么d|a,d|b,d|c
那么d|(a,b),那么d|((a,b),c)
同理,如果e=((a,b),c),则e|(a,(b,c))
所以d=e
1.4 Prime numbers
定理1.9书中证明不清楚,查看了wiki百科证明如下
px+ay=1
xpb+yab=b
p|ab
所以p|(xpb+yab)=b
1.6 The series of reciprocals of the primes
这一章最难懂的证明就是下面这步:
为什么说不等式右边项包含全部左边项
对于任意
为什么
因为
Exercise
- 1=ax+by, a=cp, b=dq, so 1=cpx+dqy
- 1=ax+by, 1=ax+cz, so 1=(ax+by)(ax+cz)=x(aax+by+cz)+bcyz
- (a, b)=1, so a and b have no commom prime factor, so do
an andbk - 1=ax+by, let m=a+b, n=a-b, then 2=(m+n)x+(m-n)y=m(x+y)+n(x-y)
if 2|x+y and 2|x-y then (a+b, a-b)=1 else (a+b, a-b)=2 - as exercise 4, but more complex:
1=ax+by, let m=a+b, n=a2−ab+b2 , then a/b=3m±12n−3m2√6
so(12n−3m2)(x−y)2=(6−3m(x+y))2
so3=n(x−y)2+m(3(x+y)−m(x2+xy+y2))=ni+mk
so the result is 1 or 3 depend on whether 3|(i,k) - 1=ax+by, cd=a+b then 1=(cd-b)x+by=dcx+b(y-x) so (d,b)=1
- (a/b)+(c/d)=n, so ad+bc=nbd, so (a-bn)d+bc=0, because (a,b)=1, b|d, in the same way d|b so |d|=|b|
- if n is squarefree, choose the max integer c such that
c2|n , then a=c,b=n/c2 , if n is not squarefree, then a=1, b=n - (a) 4|36 and 9|36 and 4<9 but
2∤3
(b)b2 is the largest, soa2|b2 , so a|b - m=ax+by,n=cx+dy, so (a,b)|m and (a,b)|n. so (a,b)|(m,n), the prove of (m,n)|(a,b) is similar
- the lowest digit must be odd:1 3 5 7, 5|
14+4or34+4or74+4 , only leave numbers of n=10k+5, how to prove this?n4+4=(n2+2n+2)(n2−2n+2)) from newsmth.net - (a) true let n=1 we get a|b, even n>1, we have m =
(b/a)n , m is not integer unless b/a is integer, so a|b
(b) false44|1010 , however4∤10
(c) true the difference with (a) is only factor 2, leta=2ka b=2kb , we getnka<nkb+1 for all n, soka<kb - (a)
am=nbm , sobm|am , so b=1 because 12(a)
(b) letc=n1/m=p/q so(p/q)m=n , so q=1 andpm=n , contradict with the assumption of n is not the mth power of n - assume a=x^m and m
Chapter2 Arithmetical Functions and Dirichlet Multiplication
定理2.2的证明中说所有A(d)的并是S,因为S中任意数与n的gcd肯定是n的因子,某个d;而且对于某个确定的数a,gcd(a,n)是固定的,因此a只属于某个A(d),而且不同的d,A(d)没有交集
A(d)与
3
来源:CSDN
作者:buck
链接:https://blog.csdn.net/buck84/article/details/47732443