Timeline for Rational exponential expressions
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| S Aug 20, 2023 at 11:04 | history | suggested | C7X | CC BY-SA 4.0 |
Possible "f and f" typo, MathJaxify
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| Aug 20, 2023 at 5:13 | review | Suggested edits | |||
| S Aug 20, 2023 at 11:04 | |||||
| Jan 29, 2010 at 21:55 | comment | added | Charles Stewart | Got it! Thanks. Given the space of possible pairing/projection functions, the issue isn't easy to dismiss. | |
| Jan 29, 2010 at 20:05 | comment | added | Joel David Hamkins | What I am claiming is that for every multivariable integer domination problem f(a,b,c) <= g(a,b,c), there is an equivalent single variable problem F(n) <= G(n), using rational expressions involving the inverse pairing function, such that the original inequality holds except finitely often iff the new inequality holds except finitely often. And the proof of this is to think of n as coding the triple (a,b,c) via the polynomial coding function, and then replace all occurences of a with p(p(n)), all occurences of b with q(p(n)) and c with q(n). Your non-primeness set is not a domination problem. | |
| Jan 29, 2010 at 19:08 | comment | added | Charles Stewart | This doesn't deal with my worry. Consider the non-primeness Diophantine set characterised by (a+2)*(b+2)-n=0. The variable n plays a different role in determining the set of non-prime numbers: the a and b are existentially quantified, while the n is the argument to the predicate. I don't see how variables of different roles in this sense can be usefully simplified with pairing/projection functions. | |
| Jan 28, 2010 at 13:08 | comment | added | Joel David Hamkins | No, one variable will suffice, once we have the inverse pairing function p. First, it is easy to solve the equation to express also the other projection function q. Now, for example, one can think of triples (a,b,c) as coded by iterated pairs n=((a,b),c), and then extract the coordinates from the single variable n using a=p(p(n)), b=q(p(n)) and c=q(n). Thus, any expression in integer variables (a,b,c) can be thought of as an expression just in one variable n. | |
| Jan 28, 2010 at 10:40 | comment | added | Charles Stewart | I think that, if we add inverse pairing functions to Diophantines, we can only code up undecidable problems if we have two variables: we can't existentially quantify over the variable that ranges over the problem set. | |
| Jan 27, 2010 at 14:52 | history | answered | Joel David Hamkins | CC BY-SA 2.5 |