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 Socratic
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Mar
20
comment Pull back image of maximal ideal under surjective ring homomorphism is maximal
But $J \subseteq f^{-1}(M)$ is a contradiction because I assumed $f^{-1}(M)$ is a "proper subset" of $J$ ...
Mar
20
asked Pull back image of maximal ideal under surjective ring homomorphism is maximal
Feb
23
comment If $G$ is an uncountable group and $H$ is a subgroup then $G$ \ $H$ is uncountable ?
@FrankScience: But how ?? a countable union of uncountable sets is also uncountable ...
Feb
23
comment If $G$ is an uncountable group and $H$ is a subgroup then $G$ \ $H$ is uncountable ?
@FrankScience: Yes $G-H$ is indeed the set difference and $xH$ do have same cardinality as that of $H$
Feb
23
asked If $G$ is an uncountable group and $H$ is a subgroup then $G$ \ $H$ is uncountable ?
Feb
22
accepted How to prove $(\space|\sin n|\space)$ does not converge?
Feb
8
answered Does $G\cong G/H$ imply that $H$ is trivial?
Jan
24
comment A simple way to obtain $\prod_{p\in\mathbb{P}}\frac{1}{1-p^{-s}}=\sum_{n=1}^{\infty}\frac{1}{n^s}$
@nullUser: You have shown $E_p$ 's are independent , but in the last line of the proof aren't you assuming $E^c_p$'s are independent ?
Jan
16
awarded  Socratic
Jan
5
comment Finding partitions $\{A,B\}$ of the set $\mathbb N_n:=\{1,…,n\}$ such that $\Big|\prod_{i \in A}p_i-\prod_{i\in B}p_i\Big|=1$
@GerryMyerson : :) yes , I have seen your answer also upvoted it
Jan
5
comment Any ring of prime order commutative ?
How can I get $x$ from $\bar x$ ?
Jan
3
comment Proving that $f(n)=n$ if $f(n+1)>f(f(n))$
@MikeSpivey: In the proof of claim 3 : from $f(f(k-1))<n$ , how does it follow $f(f(k-1))=f(k-1)$ ?
Jan
3
comment Annoying Polynomial Inequality
@23rd: From $h$ decreasing and $\lim _{x \to \infty} h(x)=0$ ,how do you conclude either $h \equiv 0$ or $h>0$ ? I seem to only conclude $h \ge 0$ ....
Dec
29
comment On integer $n>1$ and prime $p$ such that $p<n$ , $p$ does not divide $n$ and $n-p$ is a prime
@MattSamuel: hmm, you are right , if $n$ is odd and $p \ne 2$ then $n-p$ is even so is either $2$ or not a prime ; then we are reduced to $n$ is even and then if $p$ and $n-p$ are primes then their sum is $n$ satisfying Goldbach. I don't know about that "twin" prime conclusion of yours though
Dec
29
asked On integer $n>1$ and prime $p$ such that $p<n$ , $p$ does not divide $n$ and $n-p$ is a prime
Dec
29
accepted For any integers $m,n>1$ , does there exist a group $G$ with elements $a,b \in G$ such that $o(a)=m , o(b)=n$ but $ab$ has infinite order ?
Dec
28
comment For any integers $m,n>1$ , does there exist a group $G$ with elements $a,b \in G$ such that $o(a)=m , o(b)=n$ but $ab$ has infinite order ?
@mesel: but this doesn't assure we can always find a group with required elements for arbitrary integers $m,n >1$ ; there is a result stating "For any integers $m,n,r $, all greater than $1$ , there is a finite group $G$ with elements $a,b\in G$ such that $o(a)=m,o(b)=n,o(ab)=r$ perhaps you can see my motivation ...
Dec
28
comment For any integers $m,n>1$ , does there exist a group $G$ with elements $a,b \in G$ such that $o(a)=m , o(b)=n$ but $ab$ has infinite order ?
@yes , but I said any integer $m,n>1$ ...
Dec
28
asked For any integers $m,n>1$ , does there exist a group $G$ with elements $a,b \in G$ such that $o(a)=m , o(b)=n$ but $ab$ has infinite order ?
Dec
28
accepted For any integers $m,n,r>1$ , does there exist an infinite group $G$,with elements $a,b\in G$ such that $o(a)=m , o(b)=n , o(ab)=r $?