Description:
Mathematical discussions and pursuits.
|
|
|
Categories for the Working Mathematician, page 61
|
| |
Am I boring with my questions? How to prove the corollary at the page 61 of "Categories for the Working Mathematician"? It is obvious that the mentioned natural transformation is a member of D(s,r). How to prove from this that it has the form D(h,-)?
|
|
|
Make Online Money
|
| |
This is very good online dating site for earn extra income to all house wife and un employeed [link]
|
|
|
ATTN: Group Moderator RE: JESSHC pissy behaviour
|
| |
Ever since the sloppy speller admitted "oooooh you got me" and I got him in my sig he's behavior has sunk lower and lower. Could this be an opportune time to test the Usenet Bypass Filter on other groups than aus.tv? Herc
|
|
|
JSH: Routinely beating mathematicians
 
|
| |
One thing worth mentioning--again--is that now I'm routinely beating even supposedly top mathematicians in their own fields when it comes to web search results. (Great fun of course for a supposed "crank" or "crackpot". I can beat them head-to-head, toe-to-toe. My research crushing theirs. THAT is sweet victory.)... more »
|
|
|
Comprehensive Test Banks and solution manuals
|
| |
Comprehensive Test Banks for student Email me at alltestbanks11[at]gmail.com if you need to buy any test banks listed below. All emails will be answered quickly. Use (CTRL + F) to search your subject/author. Or visit [link] for a latest list of resources. Anything not in the list can also be made available.... more »
|
|
|
The quartic or biquadratic equation
|
| |
I am trying to understand the discriminant of a quartic as it appears on page 12 of a book that i once held in my hand in Dun-Laoghaire public libray in Dublin Ireland. [link] I have gone back to that and other libraries in Dublin, Ireland to... more »
|
|
|
Categories for the Working Mathematician, page 52
|
| |
In "Categories for the Working Mathematician" Second Edition, page 52 is said: "... and one relation h = g o f". How h = g o f is a binary relation? It is not a relation, it is a formula. I don't understand, please explain.
|
|
|
FLT : (2x+1)^p + 2^p y^p = (2y+1)^p
|
| |
hi I would like to ask if there is a simple method to prove that for odd prime p , positive integers x,y,z The equation (2x+1)^p + 2^p y^p = (2y+1)^p has no solution. Or a reference for the proof of the exact same equation. tommy1729
|
|
|
|
