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  • $\begingroup$ Thanks, one more question. If $\sqrt{x}$ was not primitive recursive would this cause that $P(\sqrt{x})$ to be non primitive recursive ? $\endgroup$ Commented Feb 12, 2015 at 13:39
  • $\begingroup$ If $g(x)$ and $f(x)$ are primitive recursive, then $g(f(x))$ is too. If either of them isn't, then $g(f(x))$ is likely not primitive recursive either, although it could still happen to be, but it would have to be proven in some other way. $\sqrt x$ is not primitive recursive, so in general neither is $P(\sqrt x)$. However in your questions above, we're looking at $P([\sqrt x])$. $[\sqrt x]$ happens to be primitive recursive even though $\sqrt x$ isn't, which is an example of $g(f(x))$ being primitive recursive where $f(x)$ isn't (with $g(x) = [x]$ and $f(x) = \sqrt x$). $\endgroup$ Commented Feb 12, 2015 at 13:52
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    $\begingroup$ This link nayuki.io/page/primitive-recursive-functions says $\sqrt {x}$ is primitive recursive $\endgroup$ Commented Feb 12, 2015 at 14:10
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    $\begingroup$ Primitive recursive functions map to $\mathbb{N}$. The mathematical $\sqrt x$ is defined from $\mathbb{N}$ to $\mathbb{R}$ and as such can't be primitive recursive. Your references identifies $\sqrt x$ with it's integral part, in order to define $\sqrt x$ as a function that maps from $\mathbb{N}$ to $\mathbb{N}$ and is indeed primitive recursive. This function is what I would call $\lfloor\sqrt x\rfloor$, where $\lfloor x\rfloor$ represents taking the integral part (floor). So we're just using different notations. $\endgroup$ Commented Feb 12, 2015 at 14:20