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" The moving power of mathematical invention is not reasoning but imagination. "

— Augustus De Morgan


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
comment Finite non-abelian group $G$ with non-identity elements with co-prime order and $a,b \in G $ , g.c.d.$\big(o(a),o(b)\big)=1 \implies o(ab)=o(a).o(b)$
you are very right ; I upvote ; but that is not what I wanted ; I have edited now
Dec
29
revised Finite non-abelian group $G$ with non-identity elements with co-prime order and $a,b \in G $ , g.c.d.$\big(o(a),o(b)\big)=1 \implies o(ab)=o(a).o(b)$
added 82 characters in body; edited title
Dec
29
asked Finite non-abelian group $G$ with non-identity elements with co-prime order and $a,b \in G $ , g.c.d.$\big(o(a),o(b)\big)=1 \implies o(ab)=o(a).o(b)$
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 $?
Dec
28
comment 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 $?
true , missed it by that much ...
Dec
28
asked 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 $?
Dec
28
revised Finding partitions $\{A,B\}$ of the set $\mathbb N_n:=\{1,…,n\}$ such that $\Big|\prod_{i \in A}p_i-\prod_{i\in B}p_i\Big|=1$
added 261 characters in body
Dec
28
asked Finding partitions $\{A,B\}$ of the set $\mathbb N_n:=\{1,…,n\}$ such that $\Big|\prod_{i \in A}p_i-\prod_{i\in B}p_i\Big|=1$