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prove that there are exists two partial non-computable functions $g(x)$ and $f(x)$ whose product is a computable function $h(x)$ = $f(x) * g(x)$ for $\forall$ $x \in N$

I thought that we can just take two undecidable sets in $f$ and $g$ will be their characteristic functions, but characteristic functions are total, when we need to find partial functions/

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  • $\begingroup$ @spaceisdarkgreen added my thoughts. Please remove -1 Upd: thnx $\endgroup$ Sep 24 '20 at 17:00
  • $\begingroup$ A total function is a (special case of a) partial function. And if you really need it to not be total for some reason, can always modify it ad hoc. $\endgroup$ Sep 24 '20 at 17:03
  • $\begingroup$ @spaceisdarkgreen do you mean i can take characteristic function and modify it's dom ? But how, i don't understand a little bit. $\endgroup$ Sep 24 '20 at 17:04
  • $\begingroup$ If $f$ is a total function, let $f^*(x)=f(x)$ if $x$ is not zero and undefined if it is zero. Then $f^*$ is partial, and probably close enough to $f$ that it's still usable. Or if you want to keep all the information from $f$, let $f^*(x+1)=f(x)$ and $f^*(0)$ undefined. Anyhow, I don't think this is important for this problem. I think the most common answer to this problem will have $f$ and $g$ total functions, and I don't think that's wrong since a total function is a partial function. $\endgroup$ Sep 24 '20 at 17:07
  • $\begingroup$ If $f,g$ are properly partial, then $h$ is also only partial, isn't it? Or is "zero times undefined" considered zero in the context? $\endgroup$ Sep 24 '20 at 17:12
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So we take undecidable set $A$. $\neg A$ is undecifable too.

$f$ returns $0$ if $x \in A$ else returns $1$

$g$ returns $0$ if $x \in \not A$ else returns $1$

So $h = g * f$ is a total computable function which always returns $0$

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