University-数学讨论区-Abstract Algebra-Set Theories, Infinities, Number Theory

~HeBe~_@

Set Theories, Infinities, Number Theory

看到你们也这样开了STPM，SPM的数学讨论区

希望有任何难题可以在此帖讨论哦！

[ 本帖最后由 ~HeBe~_@ 于 27-4-2007 07:07 PM 编辑 ]

dunwan2tellu

(i) Prove that there are infinity many prime of the form 4k+1 (without Dirichlet Theorem).

(ii)Prove that there are infinity many prime of the form 4k+3 (without Dirichlet Theorem).

[ 本帖最后由 dunwan2tellu 于 30-4-2007 12:00 PM 编辑 ]

Take the primes of the form 4n+3 (n >= 1)
Assume there are only finitely many.
Suppose they are p1, p2, ..., pk.
Consider N = 4*p1*p2*...*pk + 3.
Clearly N must have a prime factor of the form 4n+3.
But this prime was not in our original list; dividing N by any of the primes in our list leaves a remainder of 3.
Contradiction and there are infinitely many primes of the form 4n+3.

Take positive integer N.
Let M = (N!)^2 + 1.
Consider smallest prime factor of M, called P.
Note P > N.
We have M = 0 (mod P), or (N!)^2 = -1 (mod P).
Now we raise both sides to the (P-1)/2-th power: (N!)^(P-1) = (-1)^((P-1)/2) (mod P). But by Fermat's Little Theorem, (N!)^(P-1) = 1 (mod P).
So (-1)^((P-1)/2) = 1 mod P. Thus (P-1)/2 is even, say (P-1)/2 = 2n.
Then P = 4n +1. So given any integer N, we can find a prime P > N of the form 4n+1. There must be infinitely many primes of the form 4n+1.

不好意思，习惯以英文回答，好久没用华文做数学题了，好多名词都不会用。

cutiepipie

不好意思噢。。这些问题必须要用英文来写噢。。请大家帮帮忙。。谢谢。。越快越好。。
Abstract algebra..
1. Let G be a group with |G|= pq, where p>q are distinct primes.
  (i) Show that G has normal Sylow p-subgroup.   
  (ii) If G is non- abelian, then show that q | (p-1) and calculate the number of Sylow q-subgroups in G.
2.Let G be a group with |G|=p²q, where p and q are distinct primes,show that G has a normal Sylow p-subgroups or a normal Sylow q- subgroup.
3.Let G be a group with |G|=p³q,where p and q are distinct primes. Show that either G has a normal Sylow p- or q- subgroup or |G|=24.
4.Let G be a group with |G|=p²q² where p>q are primes. If G does not have a unique normal Sylow p-subgroup, then determine the possible value(s) of |G|.
5.If |G|=pqr where p,q,r are primes, show whether G is simple.
6.If |G|<100 and G is non-abelian and simple, then show that|G|=60.

晨天

我有几个问题想问大家

关于set relation
reflexive 　　　　- for all x, xRx always TRUE
irreflexive　　　 - for all x, xRx always FALSE
not reflexive　　 - mixed
not irreflexive   - mixed


symmetric　　　　　　- 100% symmetric
asymmetric　　　　　 - must anti-sym and irreflexive
anti asymmetric　　　- 100% opposite symmetric
not symmetric　　　　- mixed
not asymmetric　　　 - mixed
not anti asymmetric　- mixed

transitive 　　　- TRUE if any one of element meet xRy, yRz => xRz

这是我听科回来的， 请问我理解的对吗？？？
关于 antisymmetric， 是不是 一定要
xRy must always different with yRx????

transitive....
一个element meet就成立吗？？？ 不是要全体互相meet 吗？？

dunwan2tellu

我学 Abstract Algebra 只学 equivalence relations 里需要用的的
reflexive
symmetry
transitive

关于 Antisymmetry 我找 wikipedia 的 definition 是
a set X is antisymmetric if, for all a,b in X
if aRb and bRa then a=b

asymmetry 是只要他不能symmetry 就是了，所以除了 antisymmetry 我们还需要 irreflexive. 不然的话 就会有 aRa 而这就表示他 symmetry 因为
aRa ==> aRa 符合 symmetry 的 definition.

一个 relation 是 antisymmetry的话，他可以是 reflexive 的.

Transitive 的definition 应该是 a set X is transitive if whenever aRb AND bRc , THEN aRc for any a,b,c in X

reflexive 　　　　- for all x, xRx always TRUE
irreflexive　　　 - for all x, xRx always FALSE
not reflexive　　 - mixed
not irreflexive   - mixed

解释不够清楚......

Let R be a relation on set X
Reflexive:  for any a in X, aRa
Not Reflexive: there exist an a in X such that a is not related to itself
Irreflexive: for any a in X, a is not related to itself
Not Irreflexive:  there exist an a in X such that aRa

差别就在 “for all" 和 ”there exist"



[ 本帖最后由 dunwan2tellu 于 12-4-2009 04:58 PM 编辑 ]