Difference between revisions of "Talk:Abc conjecture"

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(→‎Working Definition: That's an excellent explanation, and perhaps the simplest possible. Would you like to add it to the entry, replacing the short definition that is there?)
 
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:That's an excellent explanation, and perhaps the simplest possible.  Would you like to add it to the entry, replacing the short definition that is there?--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 16:54, 23 October 2015 (EDT)
 
:That's an excellent explanation, and perhaps the simplest possible.  Would you like to add it to the entry, replacing the short definition that is there?--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 16:54, 23 October 2015 (EDT)
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::Thank you! Please feel free to incorporate it! --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 17:04, 23 October 2015 (EDT)
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:::Just copying-and-pasting isn't exactly "incorporating": I tried to smooth the transitions, but the article needs still a lot of work! --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 08:59, 25 October 2015 (EDT)

Latest revision as of 12:59, October 25, 2015

Categories

Categories are capitalized. To avoid red links, one should write category:Number Theory instead of category:number theory. --AugustO (talk) 16:08, 23 October 2015 (EDT)

Thanks. The formulation, although commonly reported as stated, seems slightly off - if all the numbers are positive, then of-course the ratio will be positive also.--Andy Schlafly (talk) 16:11, 23 October 2015 (EDT)

Sorry, but you are misstating the formula: it is not square-free-part(abc) x r /c... Please revisit your sources.--AugustO (talk) 16:17, 23 October 2015 (EDT)

square-free-part(abc) times (r/c)

"square-free-part(abc) times (r/c) "

That is wrong. --AugustO (talk) 16:14, 23 October 2015 (EDT)

Working Definition

The abc-conjection doesn't deal with single cases, as it is a formula which holds for "almost all" triples of numbers with certain characteristics:

Let ε > 0 be fixed. Then for almost all triplets of co-prime integers a, b and c with a+b=c holds:

Here rad(a b c) is the "radical" of the product , i.e., the product of all unique primes dividing - or, in other words, the square-free-part.

"Almost all triplets" means that there is only a finite number of triplets which violates the condition. Some observations:

  • ε can be chosen as small as you like - as long it is greater than zero.
  • the finite number of exceptions obviously becomes smaller, the bigger ε is chosen.
  • There are infinitely many triplets violating - that's why isn't allowed.
That's an excellent explanation, and perhaps the simplest possible. Would you like to add it to the entry, replacing the short definition that is there?--Andy Schlafly (talk) 16:54, 23 October 2015 (EDT)
Thank you! Please feel free to incorporate it! --AugustO (talk) 17:04, 23 October 2015 (EDT)
Just copying-and-pasting isn't exactly "incorporating": I tried to smooth the transitions, but the article needs still a lot of work! --AugustO (talk) 08:59, 25 October 2015 (EDT)