Mathematics Asked by user403504 on November 26, 2021
Suppose we take for example the smallest number strictly greater than 0 (which in the conventional real number system realm doesn’t exist but in some other number systems does (e.g the hypereals).
Well it’s clearly defined (we just have), but we can’t come up with an algorithm that will enumerate it’s decimals (0.00000…. is just 0 and if we come up with a 1 we can always find a smaller number)
I can’t seem to come up with any infinitesimal both definable and computable. Maybe it’s just the case they’re not computable.
The definition of a computable number is restricted to real numbers. As Ross Millikan has pointed out, only real numbers have decimal expansions.
However, it is possible to talk about computable complex numbers, for instance. If both the real part and the coefficient of $i$ in the imaginary part are computable real numbers, then it makes sense to say the complex number is also computable. A similar remark applies to other extensions of the real numbers.
Being less familiar with hyperreals, let me talk about the Surreal numbers instead. The ordinal numbers are part of the surreals, so $omega = {1, 2, 3, 4, ldots midphantom 0}$ is a surreal number, and it has an inverse $varepsilon := left{0mid frac 12, frac 13, frac 14, ldots right}$ which is infinitesimal.
These are well-specified non-real numbers. They are like $i$. They don't need to be calculated themselves, any more than one has to calculate $0$ or $1$. They both can be completely specifed by a countable number of symbols (as I just did). And just like infinite decimals, they can be well-approximated by the results of terminating after $n$ steps, which can be reached in a finite amount of time.
So any sensible extension of "computable" to surreals would surely consider them both computable. And because they are, so are $x +yomega + zvarepsilon$ for computable real numbers $x, y, z$.
Answered by Paul Sinclair on November 26, 2021
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