Talk:Abc conjecture
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:
<math>rad(a b c)^{1+\epsilon} > c</math>
Here rad(a b c) is the "radical" of the product <math>a \times b \times c</math>, i.e., the product of all unique primes dividing <math>a \times b \times c</math> - 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 <math>rad(a b c) > c</math> - that's why <math>\epsilon = 0</math> 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)