agent154
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 Oct28 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 Oct28 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 $(n,9)=1$ Oct28 comment What are the generators of $\mathbb{Z}_9^*$? @coffeemath but I thought I had to take $2^4=16\equiv 7\mod{9}$ Oct28 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. Oct28 asked What are the generators of $\mathbb{Z}_9^*$? Oct27 asked Find the natural numbers $n$ for which $\varphi(n)$ is not divisible by $4$. Oct27 revised Solve for n: $\varphi(2n)=\varphi(3n)$ deleted 5 characters in body Oct25 comment Solve for n: $\varphi(2n)=\varphi(3n)$ @lhf Updated the math Oct25 revised Solve for n: $\varphi(2n)=\varphi(3n)$ added 32 characters in body Oct25 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... Oct25 answered Solve for n: $\varphi(2n)=\varphi(3n)$ Oct24 comment Solve for n: $\varphi(2n)=\varphi(3n)$ All I was taught was that it's multiplicative. Oct24 asked Solve for n: $\varphi(2n)=\varphi(3n)$ Oct23 accepted What is an “incongruent” solution? Oct23 comment What is an “incongruent” solution? @labbhattacharjee Why is it not called "congruence" then, not "incongruence"? Oct23 comment What is an “incongruent” solution? Does this mean that there are 11 different equivalence classes? Oct23 asked What is an “incongruent” solution? Oct22 answered Let $ABC$ be a well-defined product of matrices. Suppose that $A,C$ are both invertible. Prove that $rank(ABC)=rank(B)$. Oct22 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 $B$... Oct22 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 $C$.