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Oct
28
comment What are the generators of $\mathbb{Z}_9^*$?
@JyrkiLahtonen But that seems contrary to what I was taught... but again I may have misunderstood
Oct
28
comment What are the generators of $\mathbb{Z}_9^*$?
@coffeemath notice the headers in my table. Are the headers wrong? I listed all the elements $n$ such that $(n,9)=1$
Oct
28
comment What are the generators of $\mathbb{Z}_9^*$?
@coffeemath but I thought I had to take $2^4=16\equiv 7\mod{9}$
Oct
28
comment What are the generators of $\mathbb{Z}_9^*$?
Where are my errors? I've gone over this twice and I got the same thing each time.
Oct
28
asked What are the generators of $\mathbb{Z}_9^*$?
Oct
27
asked Find the natural numbers $n$ for which $\varphi(n)$ is not divisible by $4$.
Oct
27
revised Solve for n: $\varphi(2n)=\varphi(3n)$
deleted 5 characters in body
Oct
25
comment Solve for n: $\varphi(2n)=\varphi(3n)$
@lhf Updated the math
Oct
25
revised Solve for n: $\varphi(2n)=\varphi(3n)$
added 32 characters in body
Oct
25
comment Solve for n: $\varphi(2n)=\varphi(3n)$
@lhf I think you may be right, but I don't know how to show it... My math above used the assumption that if $(3,n)=3$, then $\varphi(3n)=\varphi(3^2n)=\varphi(3^2)\varphi(n)=3(3-1)\varphi(n)$... but now I see that I didn't factor the $3$ out of the $n$, so I should probably have $\varphi(n/3)$ instead...
Oct
25
answered Solve for n: $\varphi(2n)=\varphi(3n)$
Oct
24
comment Solve for n: $\varphi(2n)=\varphi(3n)$
All I was taught was that it's multiplicative.
Oct
24
asked Solve for n: $\varphi(2n)=\varphi(3n)$
Oct
23
accepted What is an “incongruent” solution?
Oct
23
comment What is an “incongruent” solution?
@labbhattacharjee Why is it not called "congruence" then, not "incongruence"?
Oct
23
comment What is an “incongruent” solution?
Does this mean that there are 11 different equivalence classes?
Oct
23
asked What is an “incongruent” solution?
Oct
22
answered Let $ABC$ be a well-defined product of matrices. Suppose that $A,C$ are both invertible. Prove that $rank(ABC)=rank(B)$.
Oct
22
comment Let $ABC$ be a well-defined product of matrices. Suppose that $A,C$ are both invertible. Prove that $rank(ABC)=rank(B)$.
@AgustíRoig Ahh.. I think I see how I might do this, then... If $A$ and $C$ are invertible, they can be reduced to $I$... and $I$ has the same rank as $A$ and $C$. So if I have the product $I_mBI_n$, then I get $B$...
Oct
22
comment Let $ABC$ be a well-defined product of matrices. Suppose that $A,C$ are both invertible. Prove that $rank(ABC)=rank(B)$.
@AgustíRoig What do you mean? "rank" is the dimension of the column space... so if $A$ and $C$ are invertible, and $A$ is an $m\times m$ matrix, then the columns of $A$ are linearly independent, meaning the dimensions of the column space is $m$, thus the rank of $A$ is $m$... simmilarly for $C$.