6,889 reputation
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bio website dedekindsparadise.wordpress.c…
location Sheffield, United Kingdom
age 26
visits member for 2 years, 11 months
seen yesterday

Hey, I am currently a PhD student at the University of Sheffield. My interests are in algebraic number theory and related fields.

Check out my academia page for some of my papers/talks. I welcome any suggestions on how to improve them.

(http://sheffield.academia.edu/DanielFretwell)


1d
comment differential of inner product of functions from $R^n \to R^n$
Does this work for any inner product?
1d
comment differential of inner product of functions from $R^n \to R^n$
It depends which inner product you are using...
Jan
22
answered $f(x) \in R[X]$ irreducible $\Rightarrow (f(x))$ ideal?
Jan
12
answered Is there an intuitive explanation for $ x^2+y^2=7 z^2 $ doesn't have any integer solution?
Jan
8
revised Whats wrong with this proof? (trying to prove a function is surjective)
added 800 characters in body
Jan
8
comment Whats wrong with this proof? (trying to prove a function is surjective)
Assuming you have learned your lesson I will add a proof in with my answer (although your question didn't ask for this).
Jan
8
revised Whats wrong with this proof? (trying to prove a function is surjective)
added 1 character in body
Jan
8
comment Whats wrong with this proof? (trying to prove a function is surjective)
But the question is asking why you can't prove by example, not how to prove the statement. In my opinion a symbolic proof by itself is just going to confuse the OP further.
Jan
8
answered Whats wrong with this proof? (trying to prove a function is surjective)
Jan
8
comment Whats wrong with this proof? (trying to prove a function is surjective)
This isn't answering the original question...
Jan
8
comment Whats wrong with this proof? (trying to prove a function is surjective)
Claim: All humans are male. Proof: Meet my friend Dr Bob. He is male, so all humans are male.
Jan
8
comment Whats wrong with this proof? (trying to prove a function is surjective)
Common undergrad error...prove something by example. In this case Dr Bob should be stripped of his title :p.
Jan
8
comment Diophantine equation $l^2+m^2+n^2=p^3+q^3$
It is known that a number $n$ is a sum of three squares if and only if $n \neq 4^a(8k+7)$ for some $a,k\geq 0$. So it is just a case of finding all such numbers below $10000$ (or as said above just do a computer search since each of $l,m,n$ must be less than $100$).
Jan
8
comment get 100 with the help of five 2
Well it is just a case of trying each of the finitely many operations in each "gap". Simple to just check each one.
Jan
8
comment get 100 with the help of five 2
I can't find a solution even allowing this! I can find an obvious one if we have $6$ of them though.
Jan
7
comment Showing irrationality of $\zeta(k)$ for some $k$ without calculating the value.
Is this answer satisfactory?
Jan
4
awarded  Explainer
Jan
4
answered Why $A=\{1,2\}$ is different of $B=\{\{1,2\}\}$?
Jan
4
comment Showing irrationality of $\zeta(k)$ for some $k$ without calculating the value.
This can also be tackled. Consider the weight $2$ Eisenstein series: $G_2(z) = 2\zeta(2) + 8\pi^2 \sum_{n=1}^{\infty}\sigma(n)q^n$. This is NOT a modular form but according to Diamond/Shurman p.$18$ the function $G_{2,N}(z) := G_{2}(z) - NG_{2}(Nz)$ is a weight $2$ modular form for $\Gamma_0(N)$ for any $N$. Choosing $N=p$ prime we find that the constant term is $(2-2p)\zeta(2) \neq 0$. Then we note that the Eisenstein subspace of $M_2(\Gamma_0(p))$ is known to be $1$-dimensional for any prime. The same argument as above applies. Obviously we are still assuming we know $\pi$ is transcendental
Jan
4
comment Showing irrationality of $\zeta(k)$ for some $k$ without calculating the value.
Sorry $k\geq 2$. Then it is just a case of showing that $\zeta(2)$ is irrational.