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throwawaymath | 6 years ago
I also don't really think you're using the right definition of computable here. You make it sound as though we're estimating, or truncating uncomputable numbers to make them computable when you say:
> From that definition, it's quite obvious that everything you compute has to be continuous, because you are never sure of what other decimals may be coming up, so whatever you compute has to be close enough.
It's not about being close enough or estimating, they're categorically different things. You can't obtain an uncomputable number, even by estimating, to any meaningful precision with a finite amount of time. So what are you saying here?
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