6 ms·

I always thought calculus should work equally well with rationals, because they are continuous (in that you can always get closer...)

Hermitian909 7y ago
The problem is that under the hood calculus is all about limits (or methods with a certain equivalence, we won't get into that). Take a derivative? that's a limit. Take an integral? That's a limit. And limits of seemingly rational expressions are sometimes irrational, case and point the limit of (1 + 1/n)^n as n approaches infinity is euler's constant, e. So if you're taking limits with rationals you're going to find yourself with undefined limits (because irrational numbers are no longer defined) breaking things constantly. Formally, the irrationals are dense but not complete.
- Hermitian909 7y ago
  er, too late to edit but the rationals are dense but not complete (well, the irrationals too, but that wasn't what I was getting at)
sseth 7y ago
I thought they were dense in the reals, but disconnected as between any two rationals there will always be an irrational number.
- mjcohen 7y ago
  But between any two irrationals (actually, any two reals) there are an infinite number of rationals.
nerdponx 7y ago
rationals, because they are continuous (in that you can always get closer...) You sort of can't get arbitrarily close. And that's the whole reason calculus doesn't work for rationals. Rationals fundamentally have "gaps", and correspond 1:1 with integers.
- jpmattia 7y ago
  > You sort of can't get arbitrarily close. Is there an obvious concrete example? I've never spent proper time studying Real Analysis, and I confess I have the same intuition as OP: That you could use rationals to approximate irrationals to arbitrary precision.
  - throw_away_1979 7y ago
    No, because that's not true, see e.g. https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_th... TL;DR: Every real number can be approximated by rationals. GP was being careless with language.
  - SamReidHughes 7y ago
    The operative words there are "sort of". You can in fact get arbitrarily close. Just as you can with other sets of numbers, like those the form m/10^n (decimal approximations), or the numbers tan(n) where n is an integer.
  - gianduja 7y ago
    You're right, you can. What's special about real numbers is that, if you have a sequence of reals for which the distance between two consecutive elements approaches zero, then there exists a real number that's the limit of the starting sequence. This isn't the case with rational numbers, eg. the sequence 1.4, 1.41, 1.414 ... (EDIT: these are increasing approximations of sqrt(2)) satisfies the hypothesis but there's no single rational number this sequence approaches. This property is called completeness. The real numbers are a complete topological space, whereas the rationals aren't.
    - Koshkin 7y ago
      To be fair, your sequence seems to approach 1.414141414... which is a rational number.
      - gianduja 7y ago
deleted 7y ago
[deleted]
throw_away_1979 7y ago
Continuity is a property of functions on topological spaces, not a property of sets of numbers. The property you are describing seems more like the Archimedean property or density of a set.
mcphage 7y ago
> rationals, because they are continuous (in that you can always get closer...) That means they're dense, but continuous is something else.
Jaxan 7y ago
Yeah, you would think so (and I thought the same for a while). All the epsilon delta stuff seems to work fine with rationals. But really most theorems of calculus (intermediate value thm, etc...) just are false, with very easy counter examples, as we see in the post.
deleted 7y ago
[deleted]
justinmeiners 7y ago
The missing property is that every bounded set has a least upper bounds or that every cauchy sequence converges. A consequence is that you can divide the rationals cleanly into disconnected parts. For example A = the set of rationals < sqrt(2) and B = the set of rationals > sqrt(2) https://en.wikipedia.org/wiki/Completeness_of_the_real_numbers https://en.wikipedia.org/wiki/Completeness_of_the_real_numbe...
joppy 7y ago
The problem with doing this is that you give up a lot of properties of functions that you might want, for example that continuous functions which cross the x-axis somewhere actually intercept the x-axis. This is not true if your x-axis is made of rational numbers: consider the function x^2 - 2.
- hyperpallium 7y ago
  Very nice example.

Hermitian909 7y ago

The problem is that under the hood calculus is all about limits (or methods with a certain equivalence, we won't get into that). Take a derivative? that's a limit. Take an integral? That's a limit. And limits of seemingly rational expressions are sometimes irrational, case and point the limit of (1 + 1/n)^n as n approaches infinity is euler's constant, e. So if you're taking limits with rationals you're going to find yourself with undefined limits (because irrational numbers are no longer defined) breaking things constantly. Formally, the irrationals are dense but not complete.

Hermitian909 7y ago
er, too late to edit but the rationals are dense but not complete (well, the irrationals too, but that wasn't what I was getting at)

sseth 7y ago

I thought they were dense in the reals, but disconnected as between any two rationals there will always be an irrational number.

mjcohen 7y ago
But between any two irrationals (actually, any two reals) there are an infinite number of rationals.

nerdponx 7y ago

rationals, because they are continuous (in that you can always get closer...) You sort of can't get arbitrarily close. And that's the whole reason calculus doesn't work for rationals. Rationals fundamentally have "gaps", and correspond 1:1 with integers.

jpmattia 7y ago
> You sort of can't get arbitrarily close. Is there an obvious concrete example? I've never spent proper time studying Real Analysis, and I confess I have the same intuition as OP: That you could use rationals to approximate irrationals to arbitrary precision.
- throw_away_1979 7y ago
  No, because that's not true, see e.g. https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_th... TL;DR: Every real number can be approximated by rationals. GP was being careless with language.
- SamReidHughes 7y ago
  The operative words there are "sort of". You can in fact get arbitrarily close. Just as you can with other sets of numbers, like those the form m/10^n (decimal approximations), or the numbers tan(n) where n is an integer.
- gianduja 7y ago
  You're right, you can. What's special about real numbers is that, if you have a sequence of reals for which the distance between two consecutive elements approaches zero, then there exists a real number that's the limit of the starting sequence. This isn't the case with rational numbers, eg. the sequence 1.4, 1.41, 1.414 ... (EDIT: these are increasing approximations of sqrt(2)) satisfies the hypothesis but there's no single rational number this sequence approaches. This property is called completeness. The real numbers are a complete topological space, whereas the rationals aren't.
  - Koshkin 7y ago
    To be fair, your sequence seems to approach 1.414141414... which is a rational number.
    - gianduja 7y ago

deleted 7y ago

[deleted]

throw_away_1979 7y ago

Continuity is a property of functions on topological spaces, not a property of sets of numbers. The property you are describing seems more like the Archimedean property or density of a set.

mcphage 7y ago

> rationals, because they are continuous (in that you can always get closer...) That means they're dense, but continuous is something else.

Jaxan 7y ago

Yeah, you would think so (and I thought the same for a while). All the epsilon delta stuff seems to work fine with rationals. But really most theorems of calculus (intermediate value thm, etc...) just are false, with very easy counter examples, as we see in the post.

deleted 7y ago

[deleted]

justinmeiners 7y ago

The missing property is that every bounded set has a least upper bounds or that every cauchy sequence converges. A consequence is that you can divide the rationals cleanly into disconnected parts. For example A = the set of rationals < sqrt(2) and B = the set of rationals > sqrt(2) https://en.wikipedia.org/wiki/Completeness_of_the_real_numbers https://en.wikipedia.org/wiki/Completeness_of_the_real_numbe...

joppy 7y ago

The problem with doing this is that you give up a lot of properties of functions that you might want, for example that continuous functions which cross the x-axis somewhere actually intercept the x-axis. This is not true if your x-axis is made of rational numbers: consider the function x^2 - 2.

hyperpallium 7y ago
Very nice example.