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May
25
comment Rephrasing the definition of a limit
Yes, that's it exactly.
May
25
comment Rephrasing the definition of a limit
No, that doesn't follow. Say $x<y<a$. It's possible that $f(x)$ is within $\epsilon$ of $L$ but $f(y)$ isn't, even in the case $f(x) \to L$.
May
25
comment Rephrasing the definition of a limit
I don't think "as close as you like by setting $x$ close enough to $a$" really captures $f(x) \to L$, because that to me sounds like "for each $\epsilon>0$ there's some $x$ close to $a$ such that $f(x)$ is within $\epsilon$ of $L$". But you need $f(x)$ close to $L$ on a neighbourhood of $a$ for the definition of $f(x)\to L$ to hold.
May
22
comment Does there exist a module structure over $\mathbb{C}^2$ as a $\mathbb{C}[x]$-module?
Every nonzero complex vector space carries infinitely many different $\mathbb{C}[x]$-module structures. You could try to construct a module structure on $M$ in which $x$ acts as 0.
May
20
reviewed Approve Inverse of two matrices multiplied
May
20
reviewed Approve Elliptic curve with prescribed lattice
May
19
comment $L\cap\left(M+\left(L\cap N\right)\right)=\left(L\cap M\right)+\left(L\cap N\right)$
Something in the lhs is in L and has the form $m+l$ for $m \in M$ and $l \in L\cap N$. If you could only show $m \in L\cap M$ you'd be done...
May
14
revised continuous functional calculus for nonunital $c^*$-algebras
unfortunate spelling error
May
14
comment How can I use limits in everyday life?
to use limits in your everyday life, try walking half of the way to school, then half of the distance remaining after that, then half of the way you still have to go, then....
May
11
reviewed Approve In a graph, is it always possible to construct a set of cycle basis, with each and every edge Is shared by at most 2 cycle bases?
May
4
answered Character of action of permutations on a subset
May
1
revised Prove that if G has a faithful complex irreducible representation
added 14 characters in body
May
1
answered Prove that if G has a faithful complex irreducible representation
Apr
29
comment Any 1-dimensional character $\otimes$ irreducible character is irreducible
The character of the tensor product of the two corresponding reps is the product $\chi_1\chi_2$. Try computing $\langle \chi_1\chi_2,\chi_1\chi_2\rangle = |G|^{-1}\sum_{g\in G} \cdots$ -- you want to show this equals 1, because then $\chi_1\chi_2$ is irreducible. You'll need to use the fact that $\chi_1(g)$ is a root of 1, so $|\chi_1(g)|^2 = 1$ for any $g$.
Apr
27
comment What are some good resources to study Cryptography?
If you enjoy abstract algebra, Koblitz's books Algebraic Aspects of Cryptography and A Course in Number Theory and Cryptography might be interesting to you.
Apr
26
awarded  Revival
Apr
25
comment What the's contradiction in showing the regular representation is indecomposable in characteristic $p$?
You can't prove that the two Ms must have trivial submodules otherwise
Apr
25
answered What the's contradiction in showing the regular representation is indecomposable in characteristic $p$?
Apr
25
answered A prime order group must be cyclic
Apr
25
revised Where is Cauchy's wrong proof?
deleted 93 characters in body