4 ms·

> don't think of it as of a number! By the same token, you shouldn't think of a real "number" as a number, but as an infinite sequence of rational numbers. Yes

by derleth 14y ago

> don't think of it as of a number! By the same token, you shouldn't think of a real "number" as a number, but as an infinite sequence of rational numbers. Yes, all of the reals are infinite sequences of rationals. Also, a rational isn't a number but an ordered pair of integers. (Technically, the set of the rationals is the set of all ordered pairs of integers (excluding the ones with zero in the second position) with an equivalence class defined on them.) The integers are also not numbers, but the naturals with a sign adjoined to them. Finally, the naturals aren't numbers but sets of sets, with zero the empty set and all further integers built inductively from that. So... what does it really gain you to look 'inside' the numbers you use?

btilly 14y ago
One minor correction. The integers are more naturally defined as an equivalence class of pairs of natural numbers, with (a, b) equivalent to (c, d) if and only if a+d = b+c. (Your approach does not eliminate the equivalence class issue, because you have to specify that +0 and -0 are the same.) In this formulation, you define (a, b) + (c, d) to be (a+c, b+d). And you define (a, b) * (c, d) to be (a * c + b * d, a * d + b * c). After a little work you can prove that both operations send equivalence classes to equivalence classes. Then you can embed as follows. The natural n gets mapped to the integer (n, 0). The integer m gets mapped to the rational (m, 1). The rational r gets mapped to the real (r, r, r, r, ....). And the real x gets mapped onto the complex number x + 0i.
- psykotic 14y ago
  For those who find the above so-called Grothendieck construction somewhat puzzling, here's a little motivation and background. It is the group completion of a monoid. The simplest case is the group completion of the free monoid on one generator. The monoid has as elements all the finite strings (including the empty string) built from the symbol x. Its operation is concatenation. Any such string is uniquely defined by the number of x symbols that occur, so this monoid is isomorphic to the natural numbers. Its group completion must add an inverse for all elements. We will denote the formal inverse of x by the symbol x' with the relations xx' = x'x = 1 where 1 denotes the empty string. This cancellative concatenation is commutative because xx' and x'x both equal 1. Thus a string can be identified (though not uniquely) by a pair of natural numbers (m,n) that count the number of x and x' occurrences. To see the connection with the sign representation of integers, add an orientation to the relations xx' = x'x = 1 to get the length-shortening reduction rules xx' -> 1, x'x -> 1. It's then easy to see that every string has a unique normal form that is either empty or consists entirely of x symbols or entirely of x' symbols, corresponding to the cases 0, +n and -n, respectively. For if a string is not of this form it must have at least one x and one x' element. But then there must be at least one adjacent pair of x and x' elements. Hence the string admits a reduction and cannot be a normal form.
  - btilly 14y ago
    It should be noted for non-mathematical readers that a group has a single operation called multiplication, and the multiplicative identity is called 1. The natural numbers form a group with "multiplication" being addition, and "1" being 0. In which case the Grothendieck construction on the natural numbers gives the integers with addition. If calling the basic operation multiplication instead of addition is confusing, remember that one of the inspirations for group theory are permutations of a set of things, which can be represented by matrices using matrix multiplication to perform the permutations.
    - crntaylor 14y ago
      The naturals form a monoid, not a group. Or am I completely misinterpreting you?
      - btilly 14y ago
        Oops, I meant integers form a group. Minor mistake.
  - batista 14y ago
    >For those who find the above so-called Grothendieck construction somewhat puzzling, here's a little motivation and background. It is the group completion of a monoid. The simplest case is the group completion of the free monoid on one generator Yes, that clears it up totally...
    - psykotic 14y ago
      Hopefully you kept reading since that part was just the nomenclature for those interested in looking up this stuff.

by derleth 14y ago

btilly 14y ago

One minor correction. The integers are more naturally defined as an equivalence class of pairs of natural numbers, with (a, b) equivalent to (c, d) if and only if a+d = b+c. (Your approach does not eliminate the equivalence class issue, because you have to specify that +0 and -0 are the same.) In this formulation, you define (a, b) + (c, d) to be (a+c, b+d). And you define (a, b) * (c, d) to be (a * c + b * d, a * d + b * c). After a little work you can prove that both operations send equivalence classes to equivalence classes. Then you can embed as follows. The natural n gets mapped to the integer (n, 0). The integer m gets mapped to the rational (m, 1). The rational r gets mapped to the real (r, r, r, r, ....). And the real x gets mapped onto the complex number x + 0i.

psykotic 14y ago
For those who find the above so-called Grothendieck construction somewhat puzzling, here's a little motivation and background. It is the group completion of a monoid. The simplest case is the group completion of the free monoid on one generator. The monoid has as elements all the finite strings (including the empty string) built from the symbol x. Its operation is concatenation. Any such string is uniquely defined by the number of x symbols that occur, so this monoid is isomorphic to the natural numbers. Its group completion must add an inverse for all elements. We will denote the formal inverse of x by the symbol x' with the relations xx' = x'x = 1 where 1 denotes the empty string. This cancellative concatenation is commutative because xx' and x'x both equal 1. Thus a string can be identified (though not uniquely) by a pair of natural numbers (m,n) that count the number of x and x' occurrences. To see the connection with the sign representation of integers, add an orientation to the relations xx' = x'x = 1 to get the length-shortening reduction rules xx' -> 1, x'x -> 1. It's then easy to see that every string has a unique normal form that is either empty or consists entirely of x symbols or entirely of x' symbols, corresponding to the cases 0, +n and -n, respectively. For if a string is not of this form it must have at least one x and one x' element. But then there must be at least one adjacent pair of x and x' elements. Hence the string admits a reduction and cannot be a normal form.
- btilly 14y ago
  It should be noted for non-mathematical readers that a group has a single operation called multiplication, and the multiplicative identity is called 1. The natural numbers form a group with "multiplication" being addition, and "1" being 0. In which case the Grothendieck construction on the natural numbers gives the integers with addition. If calling the basic operation multiplication instead of addition is confusing, remember that one of the inspirations for group theory are permutations of a set of things, which can be represented by matrices using matrix multiplication to perform the permutations.
  - crntaylor 14y ago
    The naturals form a monoid, not a group. Or am I completely misinterpreting you?
    - btilly 14y ago
      Oops, I meant integers form a group. Minor mistake.
- batista 14y ago
  >For those who find the above so-called Grothendieck construction somewhat puzzling, here's a little motivation and background. It is the group completion of a monoid. The simplest case is the group completion of the free monoid on one generator Yes, that clears it up totally...
  - psykotic 14y ago
    Hopefully you kept reading since that part was just the nomenclature for those interested in looking up this stuff.