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DanArmak comments on Respond to what they probably meant - Less Wrong
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Comments (42)
Another reason to do this is to explicitly point out that what the other person means and what they are literally saying are incompatible. Whether this helps or impedes the conversation depends on how serious it is and how clear it was to everyone what was actually meant.
Example:
Speaker 1: "Pi has infinitely many digits".
Speaker 2: "So does every real number (e.g. 0 = 0.000....). What you mean is there's no way to write Pi with finitely many digits, in any basis."
pi=1 in base pi
... but that's not what you meant :)
As long as we're nitpicking, it should be 10.
A basis is normally defined as an integer. But you're right; there doesn't seem to be any reason not to extend this to reals. (Although what good is a non-computable basis to anyone I don't know...)
Wiki's discussion of base pi.
Wiki raises an interesting point, a little further down, talking about base sqrt(2). Any number that can be expressed in a finite number of digits in base 2 can also be expressed in a finite number of digits (albeit twice as many) in base sqrt(2); also, some irrational numbers (like sqrt(2)) can also be expressed in a finite number of digits in base sqrt(2).
This leads me to wonder whether there is any number base in which any number, rational or irrational, can be fully described as either a finite (potentially extremely large) number of digits or a repeating pattern (like 1/7 in decimal) or not...
There isn't. You can see this by considering the cardinality. The set of all (finite) sequences of digits in any finite alphabet (e.g. 0 to 9) is only countably large. So it can't map the irrational (real) numbers.
Actually, the set of all infinite sequences of digits in any finite alphabet with two or more symbols is uncountable - this can be shown via a diagonalization argument. I suspect you meant to say that the set of all finite sequences of digits is countably large.
Yes, that's right. Thanks!
I thought I might be able to get around that limit by permitting infinite repeating sequences of digits - but it turns out that that's equivalent to introducing a single new symbol to the notation (with the further restriction that it can only be used not more than once per number), and therefore the set of infinite repeating sequences is also countable; thus, in order to represent all real numbers, it remains insufficient.
Yup. Just for fun, some other ways to see that there are only countably many repeating sequences:
That's an interesting point. A computer program is itself a finite number of symbols chosen from a finite list (with certain restrictions that further reduce the number of programs that make sense).
The same can be said for the English language. In fact, the same can be said for any language that can be translated into English, or into any other language that has a finite alphabet.
And I don't know how a language with an infinite alphabet would work, but if it can be explained in English, then that implies that it can be translated to English.
This therefore implies that there must exist numbers which cannot be precisely specified at all.
Technical note: strictly this requires the Axiom of Choice, or at least some weaker version of it. For each of your countable sets, there is at least one way of counting it; but to count the whole lot you need to pick one way of counting each set. This is exactly the kind of thing that can fail to happen without Choice. You don't need "very much" Choice; e.g., the axiom of choice for countable collections is enough; but it turns out that the axiom of choice for countable collections of countable sets is not enough.
That was quick. Thank you.