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For school, we have a proof that some functions are not Turing computable. The example is: $$ G(k) = \begin{cases} f_k(k) + 1 & \text{ if $f_k(k)$ is defined}, \\ 1 & \text{ otherwise}.\end{cases} $$

Claim: $G$ is non computable.

Proof: In view of obtaining a contradiction, let's say $G$ is computable, say by the $k$th Turing machine. Give the encoding of this $k$th Turing machine as an argument for $G$. This leads to a contradiction: if $f_k(k)$ is defined, then $f_k(k)$ is not equal to $g(k) = f_k(k) + 1$. Else $f_k(k)$ is undefined and not equal to $g(k) = 1$.

I don't understand the contradiction, help please...

Juho
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voluminat0
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1 Answers1

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The contradiction reached is that $0 = 1$ which violates one of Peano's axioms. Assume $G$ is computed by the $j$-th Turing machine. Observe that $G$ is everywhere defined. Then $$f_j(j) = G(j) = f_j(j) + 1$$ and by canceling $f_j(j)$ on both sides we get $$0 = 1.$$

Andrej Bauer
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  • Thank you! I think I get it now. I guess my teacher put a little too much effort in explaining the matter, which made it more difficult than it is. – voluminat0 Jan 18 '14 at 08:17