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visits member for 1 year, 11 months
seen Aug 25 at 16:46

UCR Fourth Year student.


Aug
7
awarded  Critic
Aug
7
accepted Showing that $\lambda(n)|\phi(n)$ where $n$ is a positive integer.
Aug
6
comment A consequence of Wilson's Theorem
Let me make sure I understand everything. First we have $(p-1)!=(p-1)(p-2)\cdots(p-k)(p-(k+1))!=\pm k!(p-(k+1))!.$ I'm not sure how this is equivalent to $(-1)^{k+1}$ modulo $p$.
Aug
6
comment A consequence of Wilson's Theorem
This is really interesting.
Aug
6
comment A consequence of Wilson's Theorem
-k!(p-(k+1))! I believe
Aug
6
comment A consequence of Wilson's Theorem
The first part reduces to $-1$, $-2$, etc correct?
Aug
6
asked A consequence of Wilson's Theorem
Jul
2
awarded  Curious
Jun
25
asked Constructing a field with the roots of the polynomial $x^2-3$ and $F=\mathbb{Z}_5$
Apr
11
comment Showing that $\lambda(n)|\phi(n)$ where $n$ is a positive integer.
In the course I'm in we don't have the machinery of Groups. That would have made the proof nicer.
Apr
11
asked Showing that $\lambda(n)|\phi(n)$ where $n$ is a positive integer.
Mar
28
accepted $Ord_n(ab)$ when $(a,n)=(b,n)=1$ but $(Ord_n(a), Ord_n(b))\neq 1$
Mar
28
comment $Ord_n(ab)$ when $(a,n)=(b,n)=1$ but $(Ord_n(a), Ord_n(b))\neq 1$
Because $t=Ord_n(ab)$.
Mar
28
comment $Ord_n(ab)$ when $(a,n)=(b,n)=1$ but $(Ord_n(a), Ord_n(b))\neq 1$
So the definition of order, for relatively prime $b$ and $n$, is the least positive integer $x$ such that $b^{x} \equiv 1 \mod n$. Multiples of $x$ will still make it one, but they won't be the least such.
Mar
28
comment $Ord_n(ab)$ when $(a,n)=(b,n)=1$ but $(Ord_n(a), Ord_n(b))\neq 1$
My reasoning is that $b^t \equiv 1$. Hence $t$ must be some multiple of $s$.
Mar
28
asked $Ord_n(ab)$ when $(a,n)=(b,n)=1$ but $(Ord_n(a), Ord_n(b))\neq 1$
Mar
18
accepted Showing that $\sum\limits_{n=1}^{\infty}f_n-f_{n+1}$ converges normally if $f_n$ converges normally
Mar
18
asked Showing that the sequence $z^n$ is normally but not uniformly convergent
Mar
18
accepted Property of a normally convergence squence of functions
Mar
18
comment Showing that $\sum\limits_{n=1}^{\infty}f_n-f_{n+1}$ converges normally if $f_n$ converges normally
Would it be correct to say that because $\lim\limits_{m\to \infty}s_m=f_1-f$. where $f_{m+1} \to f$ normally we know that $\sum\limits_{n=1}^{\infty}(f_n-f_{n+1})$ converges normally?