Talk:Hilbert's tenth problem
|WikiProject Mathematics||(Rated B-class, Mid-priority)|
The article claims:
- The equation
- where is a polynomial of degree is solvable in rational numbers if and only if
- is solvable in natural numbers.
This cannot be true. x+1=0 is solvable in rational numbers, but x+z+1=0 is not solvable in natural numbers. 184.108.40.206 03:41, 21 January 2007 (UTC)
- I suspect the intent was to solve the original equation over the positive rationals. But I've changed "naturals" to "integers" in the article. Ben Standeven 05:14, 7 April 2007 (UTC)
- It didn't work that way either; in your version, the case z=0 caused problems. I think I've fixed it now. 220.127.116.11 12:28, 10 April 2007 (UTC)
Flaw in Article
There is no meaning for A student that knows about polinomials may understand the article until (s)he finds such a notation with no reference to its meaning and much less its discussion of "parameters"
"...with integer coefficients such that the set of values of a for which the equation
has solutions in natural numbers is not computable. So, not only is there no general algorithm for testing Diophantine equations for solvability, even for this one parameter family of equations, there is no algorithm ..."
For lack of refences (s)he simply gets lost.
- I've added a sentence to the intro, explaining this notation and the concept of Diophantine equations. That seems to be a prerequisite to understanding this article, so I don't think we need much detail.Ben Standeven (talk) 15:26, 31 March 2009 (UTC)
From the article:
- There exists a polynomial such that, given any Diophantine set there is a number such that