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Is the set of non finitely describable real numbers closed under addition and squaring? If so, can someone give a proof? Thanks.

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non finitely describable real numbers = irrational? $\sqrt{2}^2 = 2$ ? – V-X Mar 23 '13 at 22:39
This has a good picture that explains it – User234 Mar 23 '13 at 22:39
What precisely do you mean by "finitely describable"? – Chris Eagle Mar 23 '13 at 22:42
I took it to mean "can be described for a fixed language by a finite sentence". – Ittay Weiss Mar 23 '13 at 22:49
@V-X, ChrisEagle: – anon Mar 23 '13 at 22:49

if $x$ is not finitely describable, then $-x$ is not finitely describable either. But $0=x+(-x)$ is finitely describable so no closure under addition.

Now if $x^2$ is finitely describable, then $x$ is finitely describable (it's described as "the square root (either positive or negative) of the number described by the description of $x$"). So, you do have closure under squaring.

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Isn't your last paragraph an argument for closure under squaring? – Ross Millikan Mar 23 '13 at 22:44
@RossMillikan oooops auto-pilot crashing. Thanks. I'm correcting it. – Ittay Weiss Mar 23 '13 at 22:45

If $x$ is non-describable, so is $-x$. So certainly we do not have closure under addition.

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I believe that the answer is "no," though I might have the wrong definition of "finitely describable."

Take any non-finitely-describable number $N$ and consider the number $K = 1 - N$. This number cannot be finitely describable, since if it were, we could describe $N$ as $1 - K = 1 - (1 - N) = N$. However, $N + K = N + (1 - N) = 1$, which is finitely describable. Thus the non-finitely-describable numbers are not closed under addition.

Hope this helps!

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Hum... if by non-finitely-describable you mean "can't construct a finite description (as e.g. a Turing machine)", they aren't closed with respect to addition: $a + b = 2$ if $a$ is one of yours, $b$ is too (if it wasn't, $a$ could be described). But 2 clearly isn't.

Squares I have no clue.

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