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The idea that having variables which can take on an infinite range of real-valued values fascinates some people. But you can't, not really. Well that w

by xyby 12y ago

The idea that having variables which can take on an infinite range of real-valued values fascinates some people. But you can't, not really. Well that would mean it has an infinite amount of bits in memory, right? Because to set one variable to exactly pi would mean you set it to something that usually would need an infinite amount of bits.

zxcdw 12y ago
No. The Kolmogorov complexity (the information required to reproduce the information, e.g. the information required for the algorithm (divide a circumference of a circle by its diameter)) of Pi isn't infinite.
- lomnakkus 12y ago
  Unfortunately that just shifts the problem from one of space to one of time. (There's an algorithm for calculating the hexadecimal/binary digits of pi without calculating previous digits, but AFAIK none for the decimal digits, so YMMV depending on which base you're doing it in. That's just a particular feature of pi, though so it doesn't matter for the general point.)
  - hunnypot 12y ago
    Spigot!
algorias 12y ago
Your intuition is good, but pi is a bad example. There are ways to represent specific irrational numbers such as pi, e.g. a turing machine generating all its digits. For most numbers in R that isn't possible, however. The exceptions are called computable numbers and they form a tiny (countably infinite) subset of R.
- StandardFuture 12y ago
  Can you link to a proof for these "computable numbers" being a countably infinite subset of R? :)
  - alexbecker 12y ago
    Not the comment-er, but here's a rough sketch. Definition: A computable number is a real number x such that there exists a total computable function f:Q->Q for which, given any r>0, f(r) is within r of x. Proof: There are only countably many total computable functions, since each can be described as a finite string of symbols from a finite alphabet. Hence only countably many computable numbers.
  - avmich 12y ago
    Charles Petzold in "Annotated Turing" gives a good explanation of this and related subjects.

by xyby 12y ago

zxcdw 12y ago

No. The Kolmogorov complexity (the information required to reproduce the information, e.g. the information required for the algorithm (divide a circumference of a circle by its diameter)) of Pi isn't infinite.

lomnakkus 12y ago
Unfortunately that just shifts the problem from one of space to one of time. (There's an algorithm for calculating the hexadecimal/binary digits of pi without calculating previous digits, but AFAIK none for the decimal digits, so YMMV depending on which base you're doing it in. That's just a particular feature of pi, though so it doesn't matter for the general point.)
- hunnypot 12y ago
  Spigot!

algorias 12y ago

Your intuition is good, but pi is a bad example. There are ways to represent specific irrational numbers such as pi, e.g. a turing machine generating all its digits. For most numbers in R that isn't possible, however. The exceptions are called computable numbers and they form a tiny (countably infinite) subset of R.

StandardFuture 12y ago
Can you link to a proof for these "computable numbers" being a countably infinite subset of R? :)
- alexbecker 12y ago
  Not the comment-er, but here's a rough sketch. Definition: A computable number is a real number x such that there exists a total computable function f:Q->Q for which, given any r>0, f(r) is within r of x. Proof: There are only countably many total computable functions, since each can be described as a finite string of symbols from a finite alphabet. Hence only countably many computable numbers.
- avmich 12y ago
  Charles Petzold in "Annotated Turing" gives a good explanation of this and related subjects.