meta user
network profile
| bio | website | |
|---|---|---|
| location | ||
| age | 30 | |
| visits | member for | 2 years, 1 month |
| seen | 15 hours ago | |
| stats | profile views | 80 |
|
May 2 |
awarded | Nice Answer |
|
Apr 15 |
awarded | Enlightened |
|
Apr 15 |
awarded | Nice Answer |
|
Apr 2 |
awarded | Yearling |
|
Jun 2 |
comment |
Determining the derivative of $e^{5x}\tan(2x)$ shouldn't $g(x) = \tan(2x)$? |
|
May 30 |
comment |
product of two ideals @Andrew As an example, $2\mathbb{Z} + 3\mathbb{Z} = \mathbb{Z}$ because we can write $1 = 4 + (-3)$ and the first summand belongs to $2 \mathbb{Z}$ and the second one belongs to $3 \mathbb{Z}$. On the other hand, $4\mathbb{Z} + 6 \mathbb{Z} \neq \mathbb{Z}$. Any element of $4 \mathbb{Z} + 6\mathbb{Z}$ is of the form $4x+6y$ for some integers $x,y$; in particular, it is always even so it can never be equal to $1$. I hope this helps. |
|
May 30 |
comment |
product of two ideals @Andrew But $I+J = R$ is a condition on $I$ and $J$! In general, $I+J$ is an ideal of $R$ which may or may not be equal to $R$.An equivalent way to state this condition is that $1 = i + j$ for some $i \in I$ and some $j \in J$. That is, $I+J=R$ if and only if you can write $1$ as a sum of an element of $I$ and an element of $J$. |
|
May 29 |
comment |
product of two ideals @Dylan Moreland: Thanks for your comment! I just edited my answer. |
|
May 29 |
revised |
product of two ideals deleted 21 characters in body |
|
May 29 |
answered | product of two ideals |
|
May 29 |
answered | Simple Set Theory Question |
|
May 28 |
comment |
Duality of $L^p$ and $L^q$ @t.b. Thank you for your comment and for the link. I appreciate it. |
|
May 28 |
answered | Duality of $L^p$ and $L^q$ |
|
May 27 |
answered | Is this $\left|\left(\frac{a}{b}\right)^n-\left(\frac{a}{b}\right)^{n-1}\right|$ bounded? |
|
May 20 |
comment |
Is the determinant of a matrix lower when all its elements are lower? If $A$ and $A'$ are stochastic, then $A = A'$ as Marvis showed in the response above. |
|
May 20 |
comment |
Is the determinant of a matrix lower when all its elements are lower? It looks like we came up with the same example:-) |
|
May 20 |
answered | Is the determinant of a matrix lower when all its elements are lower? |
|
May 18 |
comment |
If $A$ an integral domain contains a field $K$ and $A$ over $K$ is a finite-dimensional vector space, then $A$ is a field. Yes, you are correct! |
|
May 18 |
answered | If $A$ an integral domain contains a field $K$ and $A$ over $K$ is a finite-dimensional vector space, then $A$ is a field. |
|
May 15 |
revised |
Square root of 1 is (not) -1 Added latex |
