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This is math/philosophical question. Are all problems solvable? By solution, I also mean that if a problem has no solution, then that is still a solution. What I mean is that for every problem, is there always a solution? I was inspired to think about this after reading about the P vs NP problem. An example would be $x>x-1$ The fact that there is no solution for $x$ is, in fact, a solution for the problem as a whole.

Harnoor Lal
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Brothersquid
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  • You should take a look at Godel's Incompleteness Theorem (I don't understand it, but I think it cuts to the essence of this). – Archr Feb 02 '17 at 21:42
  • I think it is bordering on simple to find a suitable $x$ to satisfy $x > x - 1$ – Edward Evans Feb 02 '17 at 21:44
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    What you seem to be asking is not really if every problem is solvable but if every problem is in some sense decidable, i.e. categorization as solvable (with a solution) or unsolvable (but this still means the problem has been decided). Solvability in mathematics is always to be understood within a domain. $x^2 = -1$ has no solutions in $\mathbb{R}$ but it does in $\mathbb{C}$. – MM8 Feb 02 '17 at 21:48

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No. The Halting Problem is the quintessential example.

I don't understand Godel's Incompleteness Theorem, but it's also worth a look if you're interested in these sorts of questions.

Archr
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  • The thing is, OP said that he considers knowledge of "there is no solution" to be a solution in his sense (i.e. he just wants decidability in some sense). I think Gödel is more relevant to this than the Halting Problem. – MM8 Feb 02 '17 at 21:50
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    @TimonG.: We know there is no algorithm for the entire halting problem. But if we're looking at a particular Turing machine, it may either be that it can be proved to halt, or proved not to halt, or neither -- and in the latter case it can never be known for sure that it is indeed so. – hmakholm left over Monica Feb 02 '17 at 21:56
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It is known that there are polynomial equations in several variables with integer coefficients where it is impossible to determine in any systematic way whether or not there is an integer solution where one of the variables has a particular value. (This comes out of Hilbert's 10th problem).

But knowing that there is no solution is a solution too, you say? Well, how about the fact that it is also impossible to know for sure whether or not the equation you're looking at is one of the "impossible" ones?

We can't declare everything to be solvable just by moving goalposts. Depending on the precise meaning you choose for "problem" and "solution" (and even what you take it to mean that "there is" a solution), you'll be able to stave off the admission of unsolvability for some small number of steps by creative goalpost placement -- but sooner or later the buck has to stop.

hmakholm left over Monica
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  • You are stating as fact that there are absolutely unsolvable problems. That is a matter of opinion, not a proven fact. It is my opinion too, but I don't believe it's something anyone has proved, or is ever likely to prove, although Kalmár once gave some sort of philosophical argument for the existence of such problems. – bof Feb 02 '17 at 22:47
  • @bof: It's not a matter of opinion, and has been proven conclusively. Think very carefully about the generalized incompleteness theorem, which in layman terms says that no implementable formal system that can prove basic facts of arithmetic can prove or disprove every arithmetical statement. This also implies that if you play whack-a-mole and keep adding axioms for unprovable arithmetical statements, you will never finish if your system remains implementable. The incompleteness theorems are proven in very weak predicative systems too. – user21820 Feb 03 '17 at 07:04
  • @bof: To escape incompleteness, your formal system must be unable to prove basic facts of arithmetic (PA$^-$). So of course if you reject PA$^-$, you're free to assert that perhaps every problem is solvable. But before you can do that you'd have to describe clearly what are valid problems and what is valid reasoning, because it's no longer clear; the very notion of formal systems depends crucially on string manipulation, whose basic properties are essentially equivalent to PA$^-$... – user21820 Feb 03 '17 at 07:09
  • @bof: I accidentally omitted a condition: "no implementable formal system that can prove basic facts of arithmetic can prove or disprove every arithmetical statement unless it also proves that 0 = 1." =) – user21820 Feb 03 '17 at 07:17
  • @user21820 Who said anything about implementable formal systems? – bof Feb 03 '17 at 09:36
  • @bof: My point is that if you're not talking about an implementable formal system, then you would have to define precisely what you mean by "problem" and "solve", if not "absolutely unsolvable problem" is not even meaningful. – user21820 Feb 03 '17 at 09:41