Proportion of an infinite set of numbers that satisfy a given property
$begingroup$
Let $P$ be a property of the real numbers. I am aware that one can formalise the notion that 'the proportion of the natural numbers that satisfy $P$ is $p$' by showing that
$${displaystylelim_{ntoinfty}dfrac{#{m in mathbb{N}_ {leq n} | m mathrm{satisifies} P}}{n}=p,}$$
but I am wondering whether this could be extended to, say, the rational numbers, or even an uncountable set like the irrational numbers. Since the rational numbers are countable you could try the same definition, but you would need to fix an enumeration first and then I would imagine that the result depends on the choice of enumeration. For the irrationals I have no idea how one would do this, or even if it makes sense. I know of Lebesgue measure, but I can't see how one could apply it in such a definition.
The main inspiration for this question was the following. It is easy to see that the sum of two irrational numbers is not necessarily irrational, though it seems that it is 'almost always' irrational - is there anyway to characterise this (and with irrational replaced by transcendental)?
real-analysis number-theory probability-theory
asked Jan 6 at 13:28
AlephNullAlephNull
3369
$endgroup$
add a comment |
$begingroup$
Let $P$ be a property of the real numbers. I am aware that one can formalise the notion that 'the proportion of the natural numbers that satisfy $P$ is $p$' by showing that
$${displaystylelim_{ntoinfty}dfrac{#{m in mathbb{N}_ {leq n} | m mathrm{satisifies} P}}{n}=p,}$$
but I am wondering whether this could be extended to, say, the rational numbers, or even an uncountable set like the irrational numbers. Since the rational numbers are countable you could try the same definition, but you would need to fix an enumeration first and then I would imagine that the result depends on the choice of enumeration. For the irrationals I have no idea how one would do this, or even if it makes sense. I know of Lebesgue measure, but I can't see how one could apply it in such a definition.
The main inspiration for this question was the following. It is easy to see that the sum of two irrational numbers is not necessarily irrational, though it seems that it is 'almost always' irrational - is there anyway to characterise this (and with irrational replaced by transcendental)?
real-analysis number-theory probability-theory
asked Jan 6 at 13:28
AlephNullAlephNull
3369
$endgroup$
add a comment |
$begingroup$
Let $P$ be a property of the real numbers. I am aware that one can formalise the notion that 'the proportion of the natural numbers that satisfy $P$ is $p$' by showing that
$${displaystylelim_{ntoinfty}dfrac{#{m in mathbb{N}_ {leq n} | m mathrm{satisifies} P}}{n}=p,}$$
but I am wondering whether this could be extended to, say, the rational numbers, or even an uncountable set like the irrational numbers. Since the rational numbers are countable you could try the same definition, but you would need to fix an enumeration first and then I would imagine that the result depends on the choice of enumeration. For the irrationals I have no idea how one would do this, or even if it makes sense. I know of Lebesgue measure, but I can't see how one could apply it in such a definition.
The main inspiration for this question was the following. It is easy to see that the sum of two irrational numbers is not necessarily irrational, though it seems that it is 'almost always' irrational - is there anyway to characterise this (and with irrational replaced by transcendental)?
real-analysis number-theory probability-theory
asked Jan 6 at 13:28
AlephNullAlephNull
3369
$endgroup$
Let $P$ be a property of the real numbers. I am aware that one can formalise the notion that 'the proportion of the natural numbers that satisfy $P$ is $p$' by showing that
$${displaystylelim_{ntoinfty}dfrac{#{m in mathbb{N}_ {leq n} | m mathrm{satisifies} P}}{n}=p,}$$
but I am wondering whether this could be extended to, say, the rational numbers, or even an uncountable set like the irrational numbers. Since the rational numbers are countable you could try the same definition, but you would need to fix an enumeration first and then I would imagine that the result depends on the choice of enumeration. For the irrationals I have no idea how one would do this, or even if it makes sense. I know of Lebesgue measure, but I can't see how one could apply it in such a definition.
The main inspiration for this question was the following. It is easy to see that the sum of two irrational numbers is not necessarily irrational, though it seems that it is 'almost always' irrational - is there anyway to characterise this (and with irrational replaced by transcendental)?
real-analysis number-theory probability-theory
real-analysis number-theory probability-theory
asked Jan 6 at 13:28
AlephNullAlephNull
3369
asked Jan 6 at 13:28
AlephNullAlephNull
3369
asked Jan 6 at 13:28
AlephNullAlephNull
3369
asked Jan 6 at 13:28
AlephNullAlephNull
3369
asked Jan 6 at 13:28
AlephNullAlephNull
3369
3369
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes, I think a variant of the (Lebesgue) density will do, intuitively: if $mu$ is the Lebesgue measure check, and $P$ is the set of reals with property $P$ (and assume it to be measurable) whether
$$lim_{n to infty} frac{m(P cap [-n,n])}{2n}$$ exists, and that's your proportion. For the irrationals we'd get $1$ of course, the rationals $0$.
Locally we always have density $0$ or $1$ a.e. as stated in the Lebesgue density theorem.
answered Jan 6 at 13:39
Henno BrandsmaHenno Brandsma
109k347115
$endgroup$
$begingroup$
Oh yes, that's rather obvious. I suppose I haven't highlighted the main question enough, so I'll make a new question asking that.
$endgroup$
– AlephNull
Jan 6 at 14:05
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes, I think a variant of the (Lebesgue) density will do, intuitively: if $mu$ is the Lebesgue measure check, and $P$ is the set of reals with property $P$ (and assume it to be measurable) whether
$$lim_{n to infty} frac{m(P cap [-n,n])}{2n}$$ exists, and that's your proportion. For the irrationals we'd get $1$ of course, the rationals $0$.
Locally we always have density $0$ or $1$ a.e. as stated in the Lebesgue density theorem.
answered Jan 6 at 13:39
Henno BrandsmaHenno Brandsma
109k347115
$endgroup$
$begingroup$
Oh yes, that's rather obvious. I suppose I haven't highlighted the main question enough, so I'll make a new question asking that.
$endgroup$
– AlephNull
Jan 6 at 14:05
add a comment |
$begingroup$
Yes, I think a variant of the (Lebesgue) density will do, intuitively: if $mu$ is the Lebesgue measure check, and $P$ is the set of reals with property $P$ (and assume it to be measurable) whether
$$lim_{n to infty} frac{m(P cap [-n,n])}{2n}$$ exists, and that's your proportion. For the irrationals we'd get $1$ of course, the rationals $0$.
Locally we always have density $0$ or $1$ a.e. as stated in the Lebesgue density theorem.
answered Jan 6 at 13:39
Henno BrandsmaHenno Brandsma
109k347115
$endgroup$
$begingroup$
Oh yes, that's rather obvious. I suppose I haven't highlighted the main question enough, so I'll make a new question asking that.
$endgroup$
– AlephNull
Jan 6 at 14:05
add a comment |
$begingroup$
Yes, I think a variant of the (Lebesgue) density will do, intuitively: if $mu$ is the Lebesgue measure check, and $P$ is the set of reals with property $P$ (and assume it to be measurable) whether
$$lim_{n to infty} frac{m(P cap [-n,n])}{2n}$$ exists, and that's your proportion. For the irrationals we'd get $1$ of course, the rationals $0$.
Locally we always have density $0$ or $1$ a.e. as stated in the Lebesgue density theorem.
answered Jan 6 at 13:39
Henno BrandsmaHenno Brandsma
109k347115
$endgroup$
Yes, I think a variant of the (Lebesgue) density will do, intuitively: if $mu$ is the Lebesgue measure check, and $P$ is the set of reals with property $P$ (and assume it to be measurable) whether
$$lim_{n to infty} frac{m(P cap [-n,n])}{2n}$$ exists, and that's your proportion. For the irrationals we'd get $1$ of course, the rationals $0$.
Locally we always have density $0$ or $1$ a.e. as stated in the Lebesgue density theorem.
answered Jan 6 at 13:39
Henno BrandsmaHenno Brandsma
109k347115
answered Jan 6 at 13:39
Henno BrandsmaHenno Brandsma
109k347115
answered Jan 6 at 13:39
Henno BrandsmaHenno Brandsma
109k347115
answered Jan 6 at 13:39
Henno BrandsmaHenno Brandsma
109k347115
109k347115
$begingroup$
Oh yes, that's rather obvious. I suppose I haven't highlighted the main question enough, so I'll make a new question asking that.
$endgroup$
– AlephNull
Jan 6 at 14:05
add a comment |
$begingroup$
Oh yes, that's rather obvious. I suppose I haven't highlighted the main question enough, so I'll make a new question asking that.
$endgroup$
– AlephNull
Jan 6 at 14:05
$begingroup$
Oh yes, that's rather obvious. I suppose I haven't highlighted the main question enough, so I'll make a new question asking that.
$endgroup$
– AlephNull
Jan 6 at 14:05
$begingroup$
Oh yes, that's rather obvious. I suppose I haven't highlighted the main question enough, so I'll make a new question asking that.
$endgroup$
– AlephNull
Jan 6 at 14:05
add a comment |
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