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. ...
Arturo Magidin's user avatar
3 votes
Accepted

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\...
Adam Rubinson's user avatar
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, ...
Qiaochu Yuan's user avatar
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 ...
Arturo Magidin's user avatar
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 ...
Karl's user avatar
  • 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 &...
Digitallis's user avatar
  • 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 ...
Carlyle's user avatar
  • 1,226
2 votes
Accepted

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{...
zeraoulia rafik's user avatar
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( \...
Will Jagy's user avatar
  • 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 ...
Julia Hayward's user avatar
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 $G/H_2$...
FShrike's user avatar
  • 32.6k
1 vote
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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 ...
spaceisdarkgreen's user avatar
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 ...
Theo Bendit's user avatar
  • 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} ...
frogpond The's user avatar
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 ...
GEdgar's user avatar
  • 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 ...
geetha290krm's user avatar
  • 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,...
Anne Bauval's user avatar
  • 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>...
ryaron's user avatar
  • 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}(...
Ben123's user avatar
  • 524
1 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 ...
Margaret's user avatar
  • 1,223
1 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 ...
Math and ML's user avatar
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 ...
RyRy the Fly Guy's user avatar

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