3
votes
Accepted
Suppose $G$ is a group and $N$ is a subgroup of $G$ show the following operation is well defined.
As I've said before, the meaning of "the function is well-defined" is, ahem, not very well defined.
Essentially it means that what you have really is a function and the outputs are sensible. ...
- 383k
3
votes
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What's wrong in Method $2$ here?
Correction (which I think is irrelevant to the main question):
$\lim_{n\to\infty}\big(\frac{1}{(n+1)}-\frac1{(n+2)}+\frac{1}{(n+2)}-\frac1{(n+4)}+...+\frac{n}{2n}-\frac n{3n}\big)$
should be:
$\lim_{n\...
- 17.8k
3
votes
Accepted
A PID is a semisimple ring iff it is a field
This argument is fine but it's possible to avoid Artin-Wedderburn, as follows. Let $r \in R$ and consider the ideal $(r)$. This is an $R$-submodule of $R$ so by semisimplicity it has a complement, ...
- 409k
3
votes
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Automorphism group cyclic implies abelian group, do we have more?
If $G$ is abelian, then $\phi_g=\mathrm{id}_G$ for all $g\in G$.
Your error is in concluding that if $\phi_g = (\phi_x)^{\circ k}$ (that is, $\phi_g$ is $\phi_x$ composed with itself $k$ times), then ...
- 383k
2
votes
Confusion on Basic Logic Question about Contrapositive (simple probably)
A statement involving $\frac1x$ is not untrue for $x=0$, it is ill-formed because part of it is not defined (namely $\frac10$). An ill-formed sentence doesn't have a defined truth value.
However, we ...
- 10.2k
2
votes
Can someone check my induction proof of $n^2 > 2n + 1$ for all $ n \geq 3$.
Your proof is not a proof by induction !
In order to do a proof by induction you should be proving that
For $n = 3$
$$ n^2 > 2 n + 1$$
and for all $k \geq 3 :$ $$ k^2 >2k + 1 \implies (k+1)^2 &...
- 3,022
2
votes
Accepted
Can someone check my induction proof of $n^2 > 2n + 1$ for all $ n \geq 3$.
There is nothing seriously wrong with your proof, in your inductive step however you end up just proving the statement directly, but there is nothing wrong with this as it does still prove the ...
- 1,226
2
votes
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Prove with induction if $A= \begin{pmatrix}2&1\\-1&0 \end{pmatrix}$ $\forall n \in \mathbb{N}$ $A^n= \begin{pmatrix} n+1&n\\-n&1-n \end{pmatrix}$
To prove by induction that for the matrix
$$A = \begin{pmatrix}
2 & 1 \\
-1 & 0
\end{pmatrix}$$
for all natural numbers $n$,
$$A^n = \begin{pmatrix}
n + 1 & n \\
-n & 1 - n
\end{...
- 7,117
2
votes
Prove with induction if $A= \begin{pmatrix}2&1\\-1&0 \end{pmatrix}$ $\forall n \in \mathbb{N}$ $A^n= \begin{pmatrix} n+1&n\\-n&1-n \end{pmatrix}$
this question illustrates how finding the Jordan form, including the change of basis matrix, cleans things up.
$$\left(
\begin{array}{rr}
1 & 0 \\
-1 & 1 \\
\end{array}
\right)
\left(
\...
- 137k
1
vote
Accepted
Does this prove $X \times Y$ is countable if $X$ and $Y$ are countably infinite sets?
I suspect your confusion arises from conflating the act of defining an injection $f:X×Y->N$, with the process of actually evaluating $f((x,y))$ at every point of the infinite set.
Your argument is ...
- 671
1
vote
Accepted
Proof of $\operatorname{Aut}_G(G/H)\cong N_G(H)/H$
$\newcommand{\orb}{\mathsf{Orb}}$Your proof looks good to me.
Just for fun, consider the category $\orb'_G$ whose objects are subgroups of $G$ and arrows $H_1\to H_2$ are coset classes $[g]$ in $G/H_2$...
- 32.6k
1
vote
Accepted
Does this prove the union of countably many countable sets is countable?
It seems fine to me. The usual argument is pretty much the same: we assign to each set in the union some $i\in \mathbb N,$ and to each element in each set we assign some $j\in \mathbb N,$ so each ...
- 55.5k
1
vote
Accepted
Right way to examine if a set is a linear independent of $\mathcal L(V,W)$
Your solution is good. Better, in fact, than the provided solution, in my opinion.
The provided solution doesn't articulate it very well, but you have to remember that the $a_i$s can be essentially ...
- 48.8k
1
vote
What's wrong in Method $2$ here?
Note that $\frac{n}{(n+n)(n+2n)}=\frac{1}{2n}-\frac{1}{3n}\neq \frac{n}{2n}-\frac{n}{3n}$. But here is not the key.
You only need to expand more term in the formula
\begin{align}
&\sum_{r=1}^{n} ...
- 21
1
vote
I am trying to find $\int_{0}^{\infty} \cos(x) , dx$
This improper integral diverges.
$$
\int_0^T\cos(x)\;dx = \sin T,
\\
\text{the limit} \quad \lim_{T \to +\infty} \sin T \quad \text{does not exist}.
$$
However, the "mean value" does exist ...
- 107k
1
vote
Accepted
General solution of 2nd order linear DE with constant coefficients approaches zero as $x$ tends to infinity.
Case 1: $p^{2}> 2q$.
$-p\pm \sqrt {p^{2}-4q}<0$ since $\sqrt {p^{2}-4q} <p$. Thus, $m_1$ and $m_2$ are negative.
Case 2: $p^{2}< 2q$. Here, $m_1$ amd $m_2$ are complex numbers and the ...
- 28.6k
1
vote
Regarding the monotonicity through derivative
There are two ways to cope with your situation. Let us first recall your corollary (of the mean value theorem), and state another one.
Corollary 1. Assume $f:[a,b]\to\mathbb{R}$ is continuous on $[a,...
- 27.1k
1
vote
Can someone check my induction proof of $n^2 > 2n + 1$ for all $ n \geq 3$.
You are assuming the consequent:
If your induction assumption is:
$k^2>2k+1$
You can not start from:
$(k+1)^2>2(k+1)+1$
Since this is what you have to prove.
Start from your assumption:
$k^2>...
- 968
1
vote
Accepted
If $M$ is a finitely generated free $R$-module, then $Hom_R(M,R)$ is a free right $R$-module of the same rank
Yes, your reasoning is correct.
If $M$ is a finitely generated free $R$-module, then, for some $n \in \mathbb{N}_{>0}$, we have $$M \cong \bigoplus_{i = 1}^{n} R$$ so that $$\operatorname{Hom}_{R}(...
- 524
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vote
Decomposition of compact subset of a manifold
Following Moishe Kohan's suggestion.
In my book, a manifold $M$ is a second-countable, locally euclidean, Hausdorff space. In particular $M$ (and therefore $K\subseteq M$) is metrizable. As $K$ is ...
- 1,223
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vote
Prove that any positive integer greater than or equal to 9 can be written as a sum of the form a + b where a is a multiple of 5 and b is even.
Let N be an even number. a + b = N. We can split this up into 2 cases + check when the condition can't be passed because the numbers are too small.
5 doesn't pass since it's not even. 6 Doesn't pass ...
- 217
1
vote
Accepted
Negation of a statement validation
Your negated statement has a couple of issues. For one, the "if... then..." structure should dissappear and be replaced with an "... and..." structure. Furthermore, DeMorgan's Law ...
- 4,697
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