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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...

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    $\begingroup$ What part of the argument don't you understand? $\endgroup$ – Yuval Filmus Jan 17 '14 at 18:55
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    $\begingroup$ There is a good explanation at our reference question. $\endgroup$ – Raphael Jan 17 '14 at 21:39
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    $\begingroup$ If all you want to prove is that there exists an uncomputable function (rather than actually finding one), a simple counting argument suffices: there are countably many Turing machines but uncountably many functions. More details at the question linked by Raphael. $\endgroup$ – David Richerby Jan 17 '14 at 22:37
  • $\begingroup$ Do you realize that the $k$ in the proof is not the same $k$ as in the definition of $G$? $\endgroup$ – Andrej Bauer Jan 17 '14 at 23:14
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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 $$0 = 1.$$

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  • $\begingroup$ 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. $\endgroup$ – Voluminat0 Jan 18 '14 at 8:17

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