# How is the non-existence of a solution proven?

I've been wondering how an argument that a solution to a particular problem doesn't exist is put together.

For instance "Pour-El and Richards found an ordinary differential equation $\phi'(t)=F(t,\phi(t))$ with $F$ computable and no computable solution."

Another puzzling example is the non-computability of Chaitin's constant.

I'm not talking about trivial examples, e.g. the DE $f(x)=f(x)+1$.

Could you give examples of outlines of such proofs? The only strategy I can think of is showing that the assumption that a solution exists leads to a contradiction, but surely there are more sophisticated ways than this?

EDIT: In case it wasn't clear (it probably wasn't), I also would examples of proof strategies involved in proving the non-existence of solutions to, for instance, DEs. The reason my examples focuses on computability is because I couldn't find any examples of the non-existence.

The easiest way to show that a function $f$ is incomputable is to exhibit - for each computable function $c$ - a value $n$ such that $f(n)\not=c(n)$. For example, let $\varphi_i$ be some canonical listing of all partial computable functions, and define $f(n)$ to be $\varphi_n(n)+1$, if $\varphi_n(n)$ is defined, and $0$ otherwise. Then immediately by definition $f$ is not computable!
Sometimes the argument along these lines is more involved - for instance, to show that the halting function $g$ (given by $g(n)=0$ if $\varphi_n(n)$ is defined, and $1$ otherwise) is not computable, we need to talk about constructing machines: if such a $g$ were computable, it would be $\varphi_k$ for some $k$. Now show that there is a computable $j$ such that $\varphi_j(i)$ is $1$ if $\varphi_k(i)=0$ and is undefined otherwise, and think about $\varphi_j(j)$ . . .
There is also coding. If I've already shown that some $f$ is not computable, then I can show that a function $\hat{f}$ is not computable by "reducing" $f$ to $\hat{f}$: showing how I can compute $f$ "relative to" $\hat{f}$. Since $f$ is not computable, neither is $\hat{f}$. This can be made precise - see "oracle Turing machines."
• @Thanks for the reply! I don't have any upvotes left for today, but I'll make sure to send you one your way! I have a question though: You say that $f(n)=\phi (n)+1$ if $\phi(n)$ is defined and $0$ otherwise, with the point (I guess) that $f(n)$ is always different from any $c(n)_n$ and because all the $c(n)_n$'s form a complete list of all computable functions, $f(n)$ is not computable? Is this correct? If so, that seems arbitrary, as you'd never be able to write down such a list? And I am also very interested in the non-existence of solution to DE's, so feel free to write about them as well! – Bobson Dugnutt Feb 22 '16 at 18:22
• @Lovsovs That's right, but there are lists of all (partial) computable functions - this is a crucial point in computability theory! We can't tell which ones are total, but we can effectively list all the partial ones. While this $f$ is indeed somewhat artificial, it's still an important example - and an important tool for showing that other functions aren't computable (via reduction as mentioned in my last paragraph). I don't know much about differential equations in computability, but if I think of anything I'll add it. – Noah Schweber Feb 22 '16 at 18:31