It is currently Tue Sep 25, 2018 11:37 am

 All times are UTC [ DST ]

 Page 1 of 1 [ 6 posts ]
 Print view Previous topic | Next topic
Author Message
 Post subject: Division RingPosted: Sat Jun 25, 2016 11:45 am
 Team Member

Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Prove that the set $\displaystyle{\mathbb{H}=\left\{\begin{pmatrix} z & w\\ -\bar{w} & \bar{z} \end{pmatrix} : z,w\in\mathbb{C}\right\}\subseteq \mathbb{M}_{2}\,\left(\mathbb{C}\right)}\,$ equipped with the usual operations of addition and multiplication of matrices, is a division ring, known as Tetranion division ring of $\displaystyle{\rm{Hamilton}}$ .

Top

 Post subject: Re: Division RingPosted: Sat Jun 25, 2016 11:46 am
 Team Member

Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
1st part : $\displaystyle{\mathbb{H}=\left\{\begin{pmatrix} \,\,\,\,z && w\\ -\overline{w} && \overline{z} \end{pmatrix} : z\,,w\in\mathbb{C}\right\}\subseteq \mathbb{M_2}\,(\mathbb{C})}$

Obviously, $\displaystyle{\mathbb{H}\neq \varnothing}$ cause

$\displaystyle{\mathbb{O}= \begin{pmatrix} 0 && 0\\ 0 && 0 \end{pmatrix}\in \mathbb{H}\,\,,I_2=\begin{pmatrix} 1 && 0\\ 0 && 1 \end{pmatrix}=1_{\mathbb{M_2}\,(\mathbb{C})}\in\mathbb{H}}$

Consider now

$\displaystyle{A=\begin{pmatrix} \,\,\,\,z_1 && w_1\\ -\overline{w_1} && \overline{z_1} \end{pmatrix}\in\mathbb{H}\,\,,B=\begin{pmatrix} \,\,\,\,z_2 && w_2\\ -\overline{w_2} && \overline{z_2} \end{pmatrix}\in \mathbb{H}}$

where $\displaystyle{z_i\,,w_i\in\mathbb{C}\,,i\in\left\{1,2\right\}}$ .

$\displaystyle{A+B=\begin{pmatrix} z_1+z_2 && w_1+w_2\\ -\overline{w_1}-\overline{w_2} && \overline{z_1}+\overline{z_2} \end{pmatrix}=\begin{pmatrix} \,\,z_1+z_2 && w_1+w_2\\ -(\overline{w_1+w_2}) && \overline{z_1+z_2} \end{pmatrix}\in\mathbb{H}}$

and $\displaystyle{-A=\begin{pmatrix} -z_1 && -w_1\\ -(-\overline{w_1}) && -\overline{z_1} \end{pmatrix}=\begin{pmatrix} -z_1 && -w_1\\ -(\overline{-w_1}) && \overline{-z_1} \end{pmatrix}\in\mathbb{H}}$

Also,

$\displaystyle{A\cdot B=\begin{pmatrix} \,\,z_1\,z_2-w_1\,\overline{w_2} && z_1\,w_2+w_1\,\overline{z_2}\\ -z_2\,\overline{w_1}-\overline{z_1\,w_2} && -w_2\,\overline{w_1}+\overline{z_1\,z_2}\end{pmatrix}}$

where :

$\displaystyle{ -z_2\,\overline{w_1}-\overline{z_1\,w_2}=-(z_2\,\overline{w_1}+\overline{z_1\,w_2})=-\overline{z_1\,w_2+w_1\,z_2}}$

and

$\displaystyle{-w_2\,\overline{w_1}+\overline{z_1\,z_2}=\overline{z_1\,z_2-w_1\,\overline{w_2}}}$

so, $\displaystyle{A\cdot B\in\mathbb{H}}$ .

Therefore, the triplet $\displaystyle{\left(\mathbb{H},+,\cdot\right)}$ is a ring with $\displaystyle{\mathbb{O}= \begin{pmatrix} 0 && 0\\ 0 && 0 \end{pmatrix}}$

as the zero-element and

$\displaystyle{I_2=\begin{pmatrix} 1 && 0\\ 0 && 1 \end{pmatrix}}$

as its unity.

Let $\displaystyle{A=\begin{pmatrix} \,\,\,\,z && w\\ -\overline{w} && \overline{z} \end{pmatrix}\in\mathbb{H}-\left\{\mathbb{O}\right\}}$. We have that :

$\displaystyle{z\neq 0}$ or $\displaystyle{w\neq 0}$ and

$\displaystyle{\det\,(A)=z\,\overline{z}+w\,\overline{w}=\left|z\right|^2+\left|w\right|^2>0}$, so the matrix $\displaystyle{A}$

is invertible. There is $\displaystyle{A^{-1}\in\mathbb{M}_{2}\,(\mathbb{C})}$ such that $\displaystyle{A\cdot A^{-1}=I_{2}=A^{-1}\cdot A}$ .

We''ll prove that $\displaystyle{A^{-1}\in\mathbb{H}}$ and then we get the desired. It's known that

$\displaystyle{A\cdot \mathrm{adj}(A)=\det\,(A)\,I_{2}=\mathrm{adj}(A)\cdot A}$ , so

$\displaystyle{A^{-1}=\dfrac{1}{\det\,(A)}\,\mathrm{adj}(A)}$

where:

$\displaystyle{\mathrm{adj}(A)=\begin{pmatrix} \overline{z} && -w\\ \overline{w} && z \end{pmatrix}}$

so, $\displaystyle{A^{-1}=\begin{pmatrix} \overline{z/a} && -w/a\\ -(\overline{-w/a}) && \overline{\overline{z/a}} \end{pmatrix}\in\mathbb{H}\,\,,a=\det\,(A)}$

Finally, $\displaystyle{\left(\mathbb{H},+,\cdot\right)}$ is a division ring but not a field because

$\displaystyle{\begin{pmatrix} i && 0\\ 0 && -i \end{pmatrix}\in\mathbb{H}\,\,,\begin{pmatrix} 0 && 1\\ -1 && 0 \end{pmatrix}\in\mathbb{H}}$ and

$\displaystyle{\begin{pmatrix} i && 0\\ 0 && -i \end{pmatrix}\cdot \begin{pmatrix} 0 && 1\\ -1 && 0 \end{pmatrix}=\begin{pmatrix} 0 && i\\ i && 0 \end{pmatrix} }$

but

$\displaystyle{ \begin{pmatrix} 0 && 1\\ -1 && 0 \end{pmatrix}\cdot \begin{pmatrix} i && 0\\ 0 && -i \end{pmatrix}= \begin{pmatrix} 0 && -i\\ -i && 0 \end{pmatrix}\neq \begin{pmatrix} 0 && i\\ i && 0 \end{pmatrix}}$

Top

 Post subject: Re: Division RingPosted: Sat Jun 25, 2016 11:48 am
 Team Member

Joined: Tue Nov 10, 2015 8:25 pm
Posts: 314
After requesting permission from Vaggelis, i am adding a similar question to the above (for those interested), instead of opening a new topic. Here it is:

Show that the set
$\displaystyle S = \left\{ m + ni\sqrt{3} \in \mathbb{C} | \text{ either both m,n } \in \mathbb{Z} \text{ or both } m,n \in \mathbb{Q} \text{ such that } 2m,2m \text{ are odd integers } \right\}$
is a (commutative) ring ( where by ring we mean an associative ring with multiplicative identity ). Find, if possible, $\displaystyle U(R)$, that is the group of invertible elements in $\displaystyle R$.

Top

 Post subject: Re: Division RingPosted: Sat Jun 25, 2016 11:49 am
 Team Member

Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Some thoughts :

We define $\displaystyle{f:\mathbb{H}\longrightarrow \mathbb{R}^{4}\,\,,f\,\left(\begin{pmatrix} \,\,\,\,z && w\\ -\overline{w} && \overline{z} \end{pmatrix}\right)=\left(Re(z),Im(z),Re(w),Im(w)\right)}$

The function $\displaystyle{f}$ is well defined . Let

$\displaystyle{A=\begin{pmatrix} \,\,\,\,z_1 && w_1\\ -\overline{w_1} && \overline{z_1} \end{pmatrix}\,\,,B=\begin{pmatrix} \,\,\,\,z_2 && w_2\\ -\overline{w_2} && \overline{z_2} \end{pmatrix}\in\mathbb{H}}$ .

Then,

\displaystyle{\begin{aligned}f\,(A+B)&=\left(Re(z_1+z_2),Im(z_1+z_2),Re(w_1+w_2),Im(w_1+w_2)\right)\\&=\left(Re(z_1)+Re(z_2),Im(z_1)+Im(z_2),Re(w_1)+Re(w_2),Im(w_1)+Im(w_2)\right)\\&=\left(Re(z_1),Im(z_1),Re(z_2),Im(z_2)\right)+\left(Re(z_2),Im(z_2),Re(w_2),Im(w_2)\right)\\&=f\,(A)+f\,(B)\end{aligned}}

Now, let $\displaystyle{A=\begin{pmatrix} \,\,\,\,z_1 && w_1\\ -\overline{w_1} && \overline{z_1} \end{pmatrix}\,\,,B=\begin{pmatrix} \,\,\,\,z_2 && w_2\\ -\overline{w_2} && \overline{z_2} \end{pmatrix}\in\mathbb{H}}$

with $\displaystyle{f\,(A)=f\,(B)}$. Then,

$\displaystyle{\left(Re(z_1),Im(z_1),Re(w_1),Im(w_1)\right)=\left(Re(z_2),Im(z_2),Re(w_2),Im(w_2)\right)}$,

so :

$\displaystyle{Re(z_1)=Re(z_2)\,\land Im(z_1)=Im(z_2)\,\land Re(w_1)=Re(w_2)\,\land Im(w_1)=Im(w_2)}$

and thus :

$\displaystyle{z_1=z_2\,\land w_1=w_2\implies A=B}$ , which means that $\displaystyle{f}$ is $\displaystyle{1-1}$ at $\displaystyle{\mathbb{H}}$

Consider $\displaystyle{\left(a,b,x,y\right)\in\mathbb{R}^4}$.

Setting $\displaystyle{z=a+i\,b\,,w=x+i\,y\in\mathbb{C}}$, we have that

$\displaystyle{\begin{pmatrix} \,\,\,\,z && w\\ -\overline{w} && \overline{z} \end{pmatrix}\in\mathbb{H}}$ and

$\displaystyle{f\,\left(\begin{pmatrix} \,\,\,\,z && w\\ -\overline{w} && \overline{z} \end{pmatrix}\right)=\left(Re(z),Im(z),Re(w),Im(w)\right)=\left(a,b,x,y\right)}$

which means that $\displaystyle{f}$ is onto.

Therefore, $\displaystyle{\left(\mathbb{H},+\right)\simeq \left(\mathbb{R}^4,+\right)}$ .

Also,

$\displaystyle{f\,(I_2)=f\,(1_{\mathbb{M_{2}}\,(\mathbb{C}}))=\left(Re(1),Im(1),Re(0),Im(0)\right)=\left(1,0,0,0\right)}$

If $\displaystyle{A=\begin{pmatrix} \,\,\,\,z_1 && w_1\\ -\overline{w_1} && \overline{z_1} \end{pmatrix}\,\,,B=\begin{pmatrix} \,\,\,\,z_2 && w_2\\ -\overline{w_2} && \overline{z_2} \end{pmatrix}\in\mathbb{H}}$ ,

then :

$\displaystyle{A\,B=A\cdot B=\begin{pmatrix} \,\,z_1\,z_2-w_1\,\overline{w_2} && z_1\,w_2+w_1\,\overline{z_2}\\ -z_2\,\overline{w_1}-\overline{z_1\,w_2} && -w_2\,\overline{w_1}+\overline{z_1\,z_2}\end{pmatrix}}$

and if $\displaystyle{z_1=a_1+i\,b_1\,,w_1=c_1+i\,d_1\,,z_2=a_2+i\,b_2\,,w_2=c_2+i\,d_2}$, then :

\displaystyle{\begin{aligned}z_1\,z_2-w_1\,\overline{w_2}&=\left(a_1+i\,b_1 \right )\,\left(a_2+i\,b_2 \right )-\left(c_1+i\,d_1 \right )\,\left(c_2-i\,d_2 \right )\\&=\left[(a_1\,a_2-b_1\,b_2)+i\,(a_2\,b_1+a_1\,b_2) \right ]-\left[(c_1\,c_2+d_1\,d_2)+i\,(d_1\,c_2-c_1\,d_2) \right ]\\&=\left(a_1\,a_2-b_1\,b_2-c_1\,c_2-d_1\,d_2 \right )+i\,(a_2\,b_1+a_1\,b_2-d_1\,c_2+c_1\,d_2) \end{aligned}}

and

\displaystyle{\begin{aligned}z_1\,w_2+w_1\,\overline{z_2}&=\left(a_1+i\,b_1 \right )\,\left(c_2+i\,d_2 \right )+\left(c_1+i\,d_1 \right )\,\left(a_2-i\,b_2 \right )\\&=\left[(a_1\,c_2-b_1\,d_2)+i\,(d_2\,a_1+b_1\,c_2) \right ]+\left[(c_1\,a_2+d_1\,b_2)+i\,(d_1\,a_2-c_1\,b_2) \right ]\\&=\left(a_1\,c_2-b_1\,d_2+c_1\,a_2+d_1\,b_2 \right )+i\,(d_2\,a_1+b_1\,c_2+d_1\,a_2-c_1\,b_2) \end{aligned}}

Therefore,
\begin{align*}
A\,B\stackrel{f}{\mapsto} &(a_1\,a_2-b_1\,b_2-c_1\,c_2-d_1\,d_2,a_2\,b_1+a_1\,b_2-d_1\,c_2+c_1\,d_2,a_1\,\\
\end{align*}
If we define multiplication on $\displaystyle{\mathbb{R}^4}$ such that :

$\displaystyle{a=\left(a_1,b_1,c_1,d_1\right)\,,b=\left(a_2,b_2,c_2,d_2\right)\in\mathbb{R}^4}$, then :
\begin{align*}
a\cdot b&=\big(a_1\,a_2-b_1\,b_2-c_1\,c_2-d_1\,d_2,a_2\,b_1+a_1\,b_2-d_1\,c_2+c_1\,d_2,a_1\,\\
\end{align*}
we have that (easy but painful) the triplet $\displaystyle{\left(\mathbb{R}^4,+,\cdot\right)}$ is a ring with $\displaystyle{1=\left(1,0,0,0\right)}$ as its unity,

($\displaystyle{\left(1,0,0,0\right)=f\,(I_2)=f\,(1_{\mathbb{M_{2}}\,(\mathbb{C}))}}$ ) ,

isomorphic to $\displaystyle{\left(\mathbb{H},+,\cdot\right)}$

because the function $\displaystyle{f}$ maintains the operations of addition and multiplication ( painful) and additionally, is one to one and onto.

Therefore, $\displaystyle{\left(\mathbb{R}^4,+,\cdot\right)}$ is a division ring but not a field.

Let $\displaystyle{\left(a,b,c,d\right)\in\mathbb{R}^4-\left\{0\right\}}$ .

Then, $\displaystyle{\left(a,b,c,d\right)=f\,\left(\begin{pmatrix} a+i\,b && c+i\,d\\ -c+i\,d && a-i\,b \end{pmatrix}\right)}$ .

$\displaystyle{\left(\begin{pmatrix} a+i\,b && c+i\,d\\ -c+i\,d && a-i\,b \end{pmatrix}\right)^{-1}=\begin{pmatrix} \dfrac{a-i\,b}{a^2+b^2+c^2+d^2} && -\dfrac{c+i\,d}{a^2+b^2+c^2+d^2}\\ \dfrac{c-i\,d}{a^2+b^2+c^2+d^2} && \dfrac{a-i\,b}{a^2+b^2+c^2+d^2} \end{pmatrix}}$

so, $\displaystyle{\left(a,b,c,d\right)^{-1}=\dfrac{1}{a^2+b^2+c^2+d^2}\,\left(a,-b,-c,-d\right)}$ .

Top

 Post subject: Re: Division RingPosted: Sat Jun 25, 2016 11:51 am
 Team Member

Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426

Find, the subset $\displaystyle{Z\,(\mathbb{H})=\left\{r\in\mathbb{H}: r\cdot x=x\cdot r\,,\forall\,x\in\mathbb{H}\right\}}$,

that is the center of $\displaystyle{\left(\mathbb{H},+,\cdot\right)}$ and prove that it is a field.

Also, prove that $\displaystyle{\mathbb{H}}$ is a $\displaystyle{Z\,(\mathbb{H})}$ - vector space and find its dimension.

Top

 Post subject: Re: Division RingPosted: Sat Jun 25, 2016 11:52 am
 Team Member

Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426

Let $\displaystyle{x=\left(x_0,x_1,x_2,x_3,x_4\right)\in Z\,(\mathbb{R}^4)}$. Then,

$\displaystyle{x\cdot \left(0,1,0,0\right)=\left(0,1,0,1\right)\cdot x}$ and we have that :

$\displaystyle{\left(-x_1,x_0,x_3,-x_2\right)=\left(-x_1,x_0,-x_3,x_2\right)\implies x_3=x_2=0}$

so : $\displaystyle{x=\left(x_0,x_1,0,0\right)}$ . Also,

$\displaystyle{x\cdot \left(0,0,1,0\right)=\left(0,0,1,0\right)\cdot x}$ ad we get:

$\displaystyle{\left(0,0,x_0,x_1\right)=\left(0,0,x_0,-x_1\right)\implies x_0\in\mathbb{R}\,\land x_1=0}$ .

So, $\displaystyle{x=\left(x_0,0,0,0\right)}$. On the other hand, if $\displaystyle{\left(x,0,0,0\right)\in\mathbb{R}^4}$, then :

for each $\displaystyle{\left(a,b,c,d\right)\in\mathbb{R}^4}$ holds :

$\displaystyle{\left(x,0,0,0\right)\cdot \left(a,b,c,d\right)=\left(a,b,c,d\right)\cdot \left(x,0,0,0\right)=\left(x\,a,0,0,0\right)}$ .

Therefore,

$\displaystyle{Z\,(\mathbb{R}^4)=Z\,(\mathbb{H})=\left\{\left(x,0,0,0\right)\in\mathbb{R}^4: x\in\mathbb{R}\right\}\simeq \left\{\begin{pmatrix} z&0 \\ 0&z \end{pmatrix} : z\in\mathbb{R}\right\}}$

We define $\displaystyle{g:\mathbb{R}\longrightarrow Z\,(\mathbb{R}^4)}$ by $\displaystyle{g(x)=\left(x,0,0,0\right)}$

and we see that $\displaystyle{g}$ is an isomorphism

and thus $\displaystyle{\left(Z\,(\mathbb{R}^4),+,\cdot\right)\simeq \left(\mathbb{R},+,\cdot\right)}$ and it is a field.

Besides, since $\displaystyle{\left(\mathbb{R}^4,+,\cdot\right)}$ is a division ring, we have that $\displaystyle{Z\,(\mathbb{R}^4)}$

is a field. Check here: Vector space over the center of a division ring

The commutative additive group $\displaystyle{\left(\mathbb{R}^4,+\right)}$

equuiped with the scalar multiplication $\displaystyle{\star:Z\,(\mathbb{R}^4)\times \mathbb{R}^4\longrightarrow \mathbb{R}^4}$

$\displaystyle{\left(\left(x,0,0,0\right),\left(a,b,c,d\right)\right)\mapsto \left(x,0,0,0\right)\star \left(a,b,c,d\right)\mapsto \left(x\,a,x\,b,x,c,x,d\right)}$

is a left(right) $\displaystyle{S=Z\,(\mathbb{R}^4)}$ -module, or else a $\displaystyle{S=Z\,(\mathbb{R}^4)}$ - vector space, because :

if

$\displaystyle{\left(x,0,0,0\right)\,,\left(y,0,0,0\right)\in S\,\,,\left(a,b,c,d\right)\,,\left(e,f,g,h\right)\in\mathbb{R}^4}$, then :

\displaystyle{\begin{aligned} \left[(x,0,0,0)+(y,0,0,0)\right]\star (a,b,c,d)&=(x+y,0,0,0)\star(a,b,c,d)\\&=((x+y)\,a,(x+y)\,b,(x+y)\,c,(x+y)\,d)\\&=(x\,a,x\,b,x\,c,x\,d)+(y\,a,y\,b,y\,c,y\,d)\\&=(x,0,0,0)\star (a,b,c,d)+(y,0,0,0)\star (a,b,c,d)\end{aligned}}

\displaystyle{\begin{aligned} (x,0,0,0)\star \left[(a,b,c,d)+(e,f,g,h)\right]&=(x,0,0,0)\star (a+e,b+f,c+g,d+h)\\&=(x(a+e),x(b+f),x(c+g),x(d+h))\\&=(x\,a,x\,b,x\,c,x\,d)+(x\,e,x\,f,x\,g,x\,h)\\&=(x,0,0,0)\star (a,b,c,d)+(x,0,0,0)\star (e,f,g,h)\end{aligned}}

\displaystyle{\begin{aligned} \left[(x,0,0,0)\cdot (y,0,0,0)\right]\star (a,b,c,d)&=(xy,0,0,0)\star (a,b,c,d)\\&=((x\,y)\,a,(x\,y)\,b,(x\,y)\,c,(x\,y)\,d)\\&=(x,0,0,0)\star (y\,a,y\,b,y\,c,y\,d)\\&=(x,0,0,0)\star\left[(y,0,0,0)\star (a,b,c,d)\right]\end{aligned}}

$\displaystyle{1_{S}\star (a,b,c,d)=(1,0,0,0)\star (a,b,c,d)=(a,b,c,d)}$ .

Let now $\displaystyle{x=\left(x_0,x_1,x_2,x_3\right)\in\mathbb{R}^4}$. Then,

$\displaystyle{x=(x_0,0,0,0)+(0,x_1,0,0)+(0,0,x_2,0)+(0,0,0,x_3)=\sum_{i=1}^{4}(x_i,0,0,0)\star e_{i}}$, where :

$\displaystyle{e_1=(1,0,0,0)\,,e_2=(0,1,0,0)\,,e_3=(0,0,1,0)\,,e_4=(0,0,0,1)}$ .

Futhermore, if $\displaystyle{\left(a_{i},0,0,0\right)\in S\,,1\leq i\leq 4}$ such that $\displaystyle{\sum_{i=1}^{4}a_{i}\star e_{i}=\left(0,0,0,0\right)}$, then :

$\displaystyle{(a_1,0,0,0)\star (1,0,0,0)+(a_2,0,0,0)\star (0,1,0,0)+(a_3,0,0,0)\star (0,0,1,0)+(a_4,0,0,0)\star (0,0,0,1)=\left(0,0,0,0\right)}$

$\displaystyle{\implies (a_1,0,0,0)+(0,a_2,0,0)+(0,0,a_3,0)+(0,0,0,a_4)=\left(0,0,0,0\right)}$

$\displaystyle{\implies \left(a_1,a_2,a_3,a_4\right)=\left(0,0,0,0\right)}$

$\displaystyle{\implies \forall\,i\in\left\{1,2,3,4\right\}: a_{i}=0}$

$\displaystyle{\implies \forall\,i\in\left\{1,2,3,4\right\}: \left(a_{i},0,0,0\right)=\left(0,0,0,0\right)=0_{S}}$

In conclusion, the vectors $\displaystyle{e_{i}\,,1\leq i\leq 4}$ are linear-indepedent and thus the set

$\displaystyle{A=\left\{e_{i}\in\mathbb{R}^4: 1\leq i\leq 4\right\}}$ is a basis of $\displaystyle{\left(\mathbb{R}^4,+,\star\right)}$

and $\displaystyle{\dim_{S}\,\mathbb{R}^4=4\neq 2}$ .

The basis in $\displaystyle{\mathbb{H}=\left\{\begin{pmatrix} z&w \\ -\overline{w}&\overline{z} \end{pmatrix}: z\,,w\in\mathbb{C}\right\}}$

is (according to the isomorphism $\displaystyle{f}$ above) the set

$\displaystyle{A'=\left\{I_{2},\begin{pmatrix} i&0 \\ 0&-i \end{pmatrix}\,,\begin{pmatrix} 0&1 \\ -1 & 0 \end{pmatrix}\,,\begin{pmatrix} 0&i \\ i& 0 \end{pmatrix}\right\}}$

Top

 Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending
 Page 1 of 1 [ 6 posts ]

 All times are UTC [ DST ]

#### Mathimatikoi Online

Users browsing this forum: No registered users and 0 guests

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum

Search for:
 Jump to:  Select a forum ------------------ Algebra    Linear Algebra    Algebraic Structures    Homological Algebra Analysis    Real Analysis    Complex Analysis    Calculus    Multivariate Calculus    Functional Analysis    Measure and Integration Theory Geometry    Euclidean Geometry    Analytic Geometry    Projective Geometry, Solid Geometry    Differential Geometry Topology    General Topology    Algebraic Topology Category theory Algebraic Geometry Number theory Differential Equations    ODE    PDE Probability & Statistics Combinatorics General Mathematics Foundation Competitions Archives LaTeX    LaTeX & Mathjax    LaTeX code testings Meta