Comments for Mathematics Weblog

Comment on Eigenvalues without determinants by Angelos Sphyris

Angelos Sphyris — Tue, 13 Mar 2012 18:21:16 +0000

But how do you know that the vectors v, tv, t2v, t3v, …., tnv are all distinct to begin with?

Comment on Mathematics Summer School in Sirince in Turkey by Mathematics Village (Matematik Köyü) | Densytics

Mathematics Village (Matematik Köyü) | Densytics — Fri, 06 Jan 2012 01:30:26 +0000

[...] Those of us involved in mathematics education often whinge about … [...]

Comment on General Comments by Oliver

Oliver — Wed, 13 Apr 2011 12:07:16 +0000

As I see it, the set of residues of integers mod p^k, k a positive integer, meet the wikipedia criteria to form a commutative ring under residue addition and multiplication mod p^k.
I want to know if this set is a field.
For a ring to be a field, Wikipedia defines the requirement that every non zero element be invertible. In this set, every element that is not a factor of zero is invertible, so it is slightly different from the wikipedia requirement.

Comment on General Comments by Oliver

Oliver — Thu, 27 Jan 2011 17:15:26 +0000

Forget that question Steve. It looks like the binomial theorem for (a+b)^p will handle any type of value ( that I can think of) for a and b, including (cosx+sinx). I am thefore taking it that I can create my factorisation when a+b add to zero.

Comment on General Comments by Oliver

Oliver — Thu, 27 Jan 2011 13:50:15 +0000

After a marriage and a major house refurbishment I’m back to finish off my problem.
Having claimed victory over the devil, I find myself amusingly playing devil’s advocate to try and find a flaw in my paper. The binomial theorem works for negative integers. Does it work when one of the integers is 0?
My problem is to analyse the sum of two pth powers of integers, p an odd prime. We will do it with residues of these integers modulo p^k. The integers :a:b: are prime to p, but their sum has a factor p, and possibly a facot with higher powers of p.
One stratagem generally employed is to express
the sum as ((a+b)-b)^p +b^p and expand out the first term with the binomial theorem. The b^p terms cancel out leaving a factored expression (a+b)() and you prove the () factor has a single factor p.
Now I can prove what a and b must be for all the various combinations, but my factorisation is of the form (a+b)() where () does not have the factor p. The devil points this out, but I reply I can take my solution :a:b: mod p^k and exponentiate it using the stratagem and the result will be a residue expression mod p^k of (a+b)p(). Is this also valid if a+b=0 mod p^k, i.e. the stratagem forms (0)p() mod p^k? If not, I need to add another section.

Comment on Polynomial Division by John P

John P — Fri, 23 Apr 2010 00:14:42 +0000

What about if you want to display polynomial long division in modular.., for example, I am trying to long divide a polynomial in mod7

Comment on General Comments by Oliver

Oliver — Thu, 22 Apr 2010 15:51:34 +0000

Ignore the last paragraph above. What remains is my argument. (a+b) can only be zero. I don’t want to get into Dr. Faustus and the devil, but in dealing with the devil in mathematics he has two very powerful weapons, infinity and zero. I managed to control infinity, but zero almost beat me. Thanks for being able to bounce ideas off you. Over and out.

Comment on General Comments by Oliver

Oliver — Fri, 16 Apr 2010 07:18:17 +0000

Needless to say I cannot say a+b is zero mod p^3, so that is not the answer.

My last effort is as follows.
I form a+b residues mod p^2 on a 1:1 basis from other residues mod p^2. I end up with 10 say mod p^2. 5 of them don’t add to zero and can be rejected. The other 5 add to zero and are formed by adding two residues mod p^2. I can factor these as p.(a+b) where a+b is one of the rejected a+b pairs.
Since I am not factorising an integer, but a residues sum, I contend that the (a+b) factor in p.(a+b) is a residue mod p. I cannot make it an (a+b) residue mod p^2, and therefore I maintain my 1:1 relationship with a+b mod p^2.

Comment on General Comments by Oliver

Oliver — Thu, 08 Apr 2010 11:28:08 +0000

I think I’ve cracked it Steve. I have a certain type (class) of factor a+b,a:b: prime to p, which may be zero at some modular level, but eventually has to be non zero. Dealing at mod p^2 level above I cannot make my conclusion. The answer is to deal at mod p^3 level. From an a+b zero factor mod p^3 I have to derive other a+b type factors, either 0 or of form p(a+b). Analysing these factors mod p^2 it can only be that a+b = 0 mod p^2, which is enough for my problem.

Comment on General Comments by Oliver

Oliver — Wed, 07 Apr 2010 11:57:32 +0000

That’s a bit of a blow Steve. You are saying that a+b can factorise as p.p even though both a and b are prime to p. And don’t you have division by a factor of zero when you say hence c+d=p? I’ll leave it there. Thanks for your response.