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> linguistic coincidence It's not a coincidence. The root of both ratio and rational come from latin ratio, gerund of reri, 'to calculate, to reckon, to think'

> linguistic coincidence It's not a coincidence. The root of both ratio and rational come from latin ratio, gerund of reri, 'to calculate, to reckon, to think'. To the ancients, just like us, computation, thinking, and reasoning are linked. Irrational numbers were literally numbers that 'could not be reckoned' in the normal sense. You can even go further with this analogy. The halting function is not computable because there is no finite procedure that tells whether any program halts. But this objection to infinity is also what the ancient Greeks had to continuity (Zeno's paradoxes) and irrational numbers, as there is no finite procedure that results in an exact answer. In fact, there is no bounded algorithm that can distinguish sqrt(2) on all possible numeric inputs.

xamuel 7y ago
Nonsense, the digits of sqrt(2) most definitely are computable and there are bounded algorithms that can check, given inputs n and k, whether or not n is the kth digit of sqrt(2).
- smallnamespace 7y ago
  I said 'all inputs' for a reason. If I give you an infinite sequence that is sqrt(2) you cannot in finite time say it actually is. Now in practice this isn't a problem, but it reveals the problem with finitism.
  - moefh 7y ago
    That doesn't seem to be a fair argument, you can also do that with rational numbers: if I give you an infinite sequence that is exactly 1/3, can you in finite time say it actually is? If you choose to represent numbers as infinite streams of digits, then obviously you can't compare them for equality in a finite amount of time. The issue is: is it possible to use real numbers in a way that sidesteps this problem? In general, it's impossible -- most real numbers are uncomputable! -- but for some useful subsets (beyond the rational numbers) it's possible.
    - smallnamespace 7y ago
      There's a fair point there, which is that some computations may look hard simply because we picked a difficult representation, but I'm not sure that applies here. There's finite, exact representation for 1/3 in whole numbers, namely itself, but as far as I know (?) there isn't one for sqrt(2) unless, say, your choice of representation is the root of some polynomial. Is there a bounded procedure that shows whether any two polynomials represent the same set of roots? Getting back to the original etymological question, my point is that irrational in the sense of 'can't be reckoned, computed' is very close to how we would see it. The Greeks simply had a different conception of computation, one grounded in finitism and constructing things geometrically.
      - messe 7y ago
        > Is there a bounded procedure that shows whether any two polynomials represent the same set of roots? Yes. Two polynomials share a root if their resultant vanishes.
    - btilly 7y ago
      It may not be a fair argument, but it hits one of the key questions in the philosophy of math. Classical math takes the attitude that absolute truth exists, and we can reason about reasoning fairly freely. In particular I can ask a question like, "Does this program halt?" and it will have a well-defined answer. Even if I don't know what it is. The set of programs that halt is a well-defined set, even if there is no procedure for that can always determine if a given program is in the set. Constructivists do not accept this point of view. To a constructivist, a question has 3 possible answers. True, false, and unknown. Talking about whether a program "really" halts when nobody has verified it one way or another is nonsensical. A construction that requires knowing something we can't find out, even in principle, is not a valid construction. Now in this point of view, we can carry out the construction of the real numbers as follows. A Cauchy sequence is a program that produces a sequence of numbers along with a proof that it converges to 0. Two programs define the same number if their sequences converge. Easy, peasy. But consider the following. A program that conducts a search for a proof or disproof of the Riemann conjecture, at each step of the search giving (-0.5)^n. If it finds a proof or disproof, it will continue giving (-0.5)^N where N is the step where it found that answer. If it doesn't, it continues. Now this is a Cauchy sequence. It converges to something. But to what? Is it positive or negative or 0? If there is a proof or disproof it will not be 0. It might be positive or negative. If neither proof nor disproof exist, it will be 0. To a classical mathematician, there must be an answer, we just don't know what it is. To a Constructivist, this is a question whose answer is unknown and therefore undefined. This number therefore cannot be categorized as positive, negative, or 0. Exactly because of the problem that you state, we have no way in guaranteed finite time to figure out whether this sequence becomes constant or forever approaches 0. I advocate learning constructivism. Not because it is useful - it is not. But because it shows that many things that mathematicians confidently claim do not actually follow by pure reasoning and cannot be proven. For example the existence of numbers that cannot be written down. To a constructivist, all numbers can be written down. We just cannot always tell them apart!
  - messe 7y ago
    The issue is your choice of expression. You're confusing the map for the territory. You don't have to express sqrt(2) as an infinite string of digits. All the algebraic numbers can be expressed by a finite number of symbols.

xamuel 7y ago

Nonsense, the digits of sqrt(2) most definitely are computable and there are bounded algorithms that can check, given inputs n and k, whether or not n is the kth digit of sqrt(2).

smallnamespace 7y ago
I said 'all inputs' for a reason. If I give you an infinite sequence that is sqrt(2) you cannot in finite time say it actually is. Now in practice this isn't a problem, but it reveals the problem with finitism.
- moefh 7y ago
  That doesn't seem to be a fair argument, you can also do that with rational numbers: if I give you an infinite sequence that is exactly 1/3, can you in finite time say it actually is? If you choose to represent numbers as infinite streams of digits, then obviously you can't compare them for equality in a finite amount of time. The issue is: is it possible to use real numbers in a way that sidesteps this problem? In general, it's impossible -- most real numbers are uncomputable! -- but for some useful subsets (beyond the rational numbers) it's possible.
  - smallnamespace 7y ago
    There's a fair point there, which is that some computations may look hard simply because we picked a difficult representation, but I'm not sure that applies here. There's finite, exact representation for 1/3 in whole numbers, namely itself, but as far as I know (?) there isn't one for sqrt(2) unless, say, your choice of representation is the root of some polynomial. Is there a bounded procedure that shows whether any two polynomials represent the same set of roots? Getting back to the original etymological question, my point is that irrational in the sense of 'can't be reckoned, computed' is very close to how we would see it. The Greeks simply had a different conception of computation, one grounded in finitism and constructing things geometrically.
    - messe 7y ago
      > Is there a bounded procedure that shows whether any two polynomials represent the same set of roots? Yes. Two polynomials share a root if their resultant vanishes.
  - btilly 7y ago
    It may not be a fair argument, but it hits one of the key questions in the philosophy of math. Classical math takes the attitude that absolute truth exists, and we can reason about reasoning fairly freely. In particular I can ask a question like, "Does this program halt?" and it will have a well-defined answer. Even if I don't know what it is. The set of programs that halt is a well-defined set, even if there is no procedure for that can always determine if a given program is in the set. Constructivists do not accept this point of view. To a constructivist, a question has 3 possible answers. True, false, and unknown. Talking about whether a program "really" halts when nobody has verified it one way or another is nonsensical. A construction that requires knowing something we can't find out, even in principle, is not a valid construction. Now in this point of view, we can carry out the construction of the real numbers as follows. A Cauchy sequence is a program that produces a sequence of numbers along with a proof that it converges to 0. Two programs define the same number if their sequences converge. Easy, peasy. But consider the following. A program that conducts a search for a proof or disproof of the Riemann conjecture, at each step of the search giving (-0.5)^n. If it finds a proof or disproof, it will continue giving (-0.5)^N where N is the step where it found that answer. If it doesn't, it continues. Now this is a Cauchy sequence. It converges to something. But to what? Is it positive or negative or 0? If there is a proof or disproof it will not be 0. It might be positive or negative. If neither proof nor disproof exist, it will be 0. To a classical mathematician, there must be an answer, we just don't know what it is. To a Constructivist, this is a question whose answer is unknown and therefore undefined. This number therefore cannot be categorized as positive, negative, or 0. Exactly because of the problem that you state, we have no way in guaranteed finite time to figure out whether this sequence becomes constant or forever approaches 0. I advocate learning constructivism. Not because it is useful - it is not. But because it shows that many things that mathematicians confidently claim do not actually follow by pure reasoning and cannot be proven. For example the existence of numbers that cannot be written down. To a constructivist, all numbers can be written down. We just cannot always tell them apart!
- messe 7y ago
  The issue is your choice of expression. You're confusing the map for the territory. You don't have to express sqrt(2) as an infinite string of digits. All the algebraic numbers can be expressed by a finite number of symbols.