Is the sum of two irrational numbers almost always irrational?
Clearly the sum of two irrational numbers is not necessarily irrational. But is it true that it is 'almost always' irrational, in the sense that $$displaystylelim_{xtoinfty}dfrac{lambda(Pcap B(x))}{lambda(Rcap B(x))}=1$$ where $lambda$ is Lebesgue measure, $Rsubset mathbb{R}^2$ is the set of points with irrational coordinates and $Psubset R$ is the set of those points the sum of whose coordinates is irrational (and $B(x)$ the disc of radius $x$ centred at the origin)? And I guess the same question applies to transcendental numbers. Intuitively it seems true, but I don't know how one would prove this. If not, then there is the question of whether the limit exists and if it does what is it.
real-analysis number-theory probability-theory
asked yesterday
AlephNullAlephNull
2209
add a comment |
Clearly the sum of two irrational numbers is not necessarily irrational. But is it true that it is 'almost always' irrational, in the sense that $$displaystylelim_{xtoinfty}dfrac{lambda(Pcap B(x))}{lambda(Rcap B(x))}=1$$ where $lambda$ is Lebesgue measure, $Rsubset mathbb{R}^2$ is the set of points with irrational coordinates and $Psubset R$ is the set of those points the sum of whose coordinates is irrational (and $B(x)$ the disc of radius $x$ centred at the origin)? And I guess the same question applies to transcendental numbers. Intuitively it seems true, but I don't know how one would prove this. If not, then there is the question of whether the limit exists and if it does what is it.
real-analysis number-theory probability-theory
asked yesterday
AlephNullAlephNull
2209
1
Related: Is the sum and difference of two irrationals always irrational?
– mrtaurho
yesterday
I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
– Milo Brandt
yesterday
add a comment |
Clearly the sum of two irrational numbers is not necessarily irrational. But is it true that it is 'almost always' irrational, in the sense that $$displaystylelim_{xtoinfty}dfrac{lambda(Pcap B(x))}{lambda(Rcap B(x))}=1$$ where $lambda$ is Lebesgue measure, $Rsubset mathbb{R}^2$ is the set of points with irrational coordinates and $Psubset R$ is the set of those points the sum of whose coordinates is irrational (and $B(x)$ the disc of radius $x$ centred at the origin)? And I guess the same question applies to transcendental numbers. Intuitively it seems true, but I don't know how one would prove this. If not, then there is the question of whether the limit exists and if it does what is it.
real-analysis number-theory probability-theory
asked yesterday
AlephNullAlephNull
2209
Clearly the sum of two irrational numbers is not necessarily irrational. But is it true that it is 'almost always' irrational, in the sense that $$displaystylelim_{xtoinfty}dfrac{lambda(Pcap B(x))}{lambda(Rcap B(x))}=1$$ where $lambda$ is Lebesgue measure, $Rsubset mathbb{R}^2$ is the set of points with irrational coordinates and $Psubset R$ is the set of those points the sum of whose coordinates is irrational (and $B(x)$ the disc of radius $x$ centred at the origin)? And I guess the same question applies to transcendental numbers. Intuitively it seems true, but I don't know how one would prove this. If not, then there is the question of whether the limit exists and if it does what is it.
real-analysis number-theory probability-theory
real-analysis number-theory probability-theory
asked yesterday
AlephNullAlephNull
2209
asked yesterday
AlephNullAlephNull
2209
edited yesterday
AlephNull
asked yesterday
AlephNullAlephNull
2209
asked yesterday
AlephNullAlephNull
2209
asked yesterday
AlephNullAlephNull
2209
2209
1
Related: Is the sum and difference of two irrationals always irrational?
– mrtaurho
yesterday
I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
– Milo Brandt
yesterday
add a comment |
1
Related: Is the sum and difference of two irrationals always irrational?
– mrtaurho
yesterday
I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
– Milo Brandt
yesterday
1
1
Related: Is the sum and difference of two irrationals always irrational?
– mrtaurho
yesterday
Related: Is the sum and difference of two irrationals always irrational?
– mrtaurho
yesterday
I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
– Milo Brandt
yesterday
I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
– Milo Brandt
yesterday
add a comment |
1 Answer
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Let $NP$ be the set of pairs whose sum is rational. I think its easier to prove $lambda(NPcap B(x))=0$. In fact since $NPcap B(x)subset NP$, we just prove $lambda(NP)=0$ and we are done. Let $$NP_x={(x,y)|x+yin mathbb{Q}}$$
Notice that the restriction addition to this set is translation by $x$ which is measure preserving, hence the inverse image of the rational numbers has measure zero. However, there is a weak form of Fubini's theorem that says that if a subset of a product measure space ( which $mathbb{R}^2$ is) has the property that its intersection with each slice has measure zero then the set has measure zero. Hence $lambda(NP)=0$.
To bring this back to the specific question you are asking, $NPcup P=mathbb{R}^2$, so for any open ball $B(x)$ $lambda(Pcap B(x))=1$.
On the other hand the set of points whose coordinates are irrational is the complement of a set of measure zero, so $lambda(Rcap B(x))=1$. Hence you are taking the limit of $1/1$.
answered yesterday
Charlie FrohmanCharlie Frohman
1,468913
Nice answer (I suppose you mean area $pi x^2$ of $B(x$) as opposed to 1?). And I suppose the same argument would go through for transcendentals because $overline{mathbb{Q}}$ is countable too.
– AlephNull
yesterday
1
Fubini's theorem is a bit overkill - if you can prove that a single line has measure zero (not so hard by explicit construction of covers by open balls), then this is a countable union of lines, so has measure zero.
– Milo Brandt
yesterday
I guess the answer is YES? (notice that the question was actually answerable by a simple Yes/No... I read this answer and I think it says Yes but I am sincerely not sure)
– Stijn de Witt
21 hours ago
add a comment |
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Let $NP$ be the set of pairs whose sum is rational. I think its easier to prove $lambda(NPcap B(x))=0$. In fact since $NPcap B(x)subset NP$, we just prove $lambda(NP)=0$ and we are done. Let $$NP_x={(x,y)|x+yin mathbb{Q}}$$
Notice that the restriction addition to this set is translation by $x$ which is measure preserving, hence the inverse image of the rational numbers has measure zero. However, there is a weak form of Fubini's theorem that says that if a subset of a product measure space ( which $mathbb{R}^2$ is) has the property that its intersection with each slice has measure zero then the set has measure zero. Hence $lambda(NP)=0$.
To bring this back to the specific question you are asking, $NPcup P=mathbb{R}^2$, so for any open ball $B(x)$ $lambda(Pcap B(x))=1$.
On the other hand the set of points whose coordinates are irrational is the complement of a set of measure zero, so $lambda(Rcap B(x))=1$. Hence you are taking the limit of $1/1$.
answered yesterday
Charlie FrohmanCharlie Frohman
1,468913
Nice answer (I suppose you mean area $pi x^2$ of $B(x$) as opposed to 1?). And I suppose the same argument would go through for transcendentals because $overline{mathbb{Q}}$ is countable too.
– AlephNull
yesterday
1
Fubini's theorem is a bit overkill - if you can prove that a single line has measure zero (not so hard by explicit construction of covers by open balls), then this is a countable union of lines, so has measure zero.
– Milo Brandt
yesterday
I guess the answer is YES? (notice that the question was actually answerable by a simple Yes/No... I read this answer and I think it says Yes but I am sincerely not sure)
– Stijn de Witt
21 hours ago
add a comment |
Let $NP$ be the set of pairs whose sum is rational. I think its easier to prove $lambda(NPcap B(x))=0$. In fact since $NPcap B(x)subset NP$, we just prove $lambda(NP)=0$ and we are done. Let $$NP_x={(x,y)|x+yin mathbb{Q}}$$
Notice that the restriction addition to this set is translation by $x$ which is measure preserving, hence the inverse image of the rational numbers has measure zero. However, there is a weak form of Fubini's theorem that says that if a subset of a product measure space ( which $mathbb{R}^2$ is) has the property that its intersection with each slice has measure zero then the set has measure zero. Hence $lambda(NP)=0$.
To bring this back to the specific question you are asking, $NPcup P=mathbb{R}^2$, so for any open ball $B(x)$ $lambda(Pcap B(x))=1$.
On the other hand the set of points whose coordinates are irrational is the complement of a set of measure zero, so $lambda(Rcap B(x))=1$. Hence you are taking the limit of $1/1$.
answered yesterday
Charlie FrohmanCharlie Frohman
1,468913
Nice answer (I suppose you mean area $pi x^2$ of $B(x$) as opposed to 1?). And I suppose the same argument would go through for transcendentals because $overline{mathbb{Q}}$ is countable too.
– AlephNull
yesterday
1
Fubini's theorem is a bit overkill - if you can prove that a single line has measure zero (not so hard by explicit construction of covers by open balls), then this is a countable union of lines, so has measure zero.
– Milo Brandt
yesterday
I guess the answer is YES? (notice that the question was actually answerable by a simple Yes/No... I read this answer and I think it says Yes but I am sincerely not sure)
– Stijn de Witt
21 hours ago
add a comment |
Let $NP$ be the set of pairs whose sum is rational. I think its easier to prove $lambda(NPcap B(x))=0$. In fact since $NPcap B(x)subset NP$, we just prove $lambda(NP)=0$ and we are done. Let $$NP_x={(x,y)|x+yin mathbb{Q}}$$
Notice that the restriction addition to this set is translation by $x$ which is measure preserving, hence the inverse image of the rational numbers has measure zero. However, there is a weak form of Fubini's theorem that says that if a subset of a product measure space ( which $mathbb{R}^2$ is) has the property that its intersection with each slice has measure zero then the set has measure zero. Hence $lambda(NP)=0$.
To bring this back to the specific question you are asking, $NPcup P=mathbb{R}^2$, so for any open ball $B(x)$ $lambda(Pcap B(x))=1$.
On the other hand the set of points whose coordinates are irrational is the complement of a set of measure zero, so $lambda(Rcap B(x))=1$. Hence you are taking the limit of $1/1$.
answered yesterday
Charlie FrohmanCharlie Frohman
1,468913
Let $NP$ be the set of pairs whose sum is rational. I think its easier to prove $lambda(NPcap B(x))=0$. In fact since $NPcap B(x)subset NP$, we just prove $lambda(NP)=0$ and we are done. Let $$NP_x={(x,y)|x+yin mathbb{Q}}$$
Notice that the restriction addition to this set is translation by $x$ which is measure preserving, hence the inverse image of the rational numbers has measure zero. However, there is a weak form of Fubini's theorem that says that if a subset of a product measure space ( which $mathbb{R}^2$ is) has the property that its intersection with each slice has measure zero then the set has measure zero. Hence $lambda(NP)=0$.
To bring this back to the specific question you are asking, $NPcup P=mathbb{R}^2$, so for any open ball $B(x)$ $lambda(Pcap B(x))=1$.
On the other hand the set of points whose coordinates are irrational is the complement of a set of measure zero, so $lambda(Rcap B(x))=1$. Hence you are taking the limit of $1/1$.
answered yesterday
Charlie FrohmanCharlie Frohman
1,468913
edited yesterday
answered yesterday
Charlie FrohmanCharlie Frohman
1,468913
answered yesterday
Charlie FrohmanCharlie Frohman
1,468913
answered yesterday
Charlie FrohmanCharlie Frohman
1,468913
1,468913
Nice answer (I suppose you mean area $pi x^2$ of $B(x$) as opposed to 1?). And I suppose the same argument would go through for transcendentals because $overline{mathbb{Q}}$ is countable too.
– AlephNull
yesterday
1
Fubini's theorem is a bit overkill - if you can prove that a single line has measure zero (not so hard by explicit construction of covers by open balls), then this is a countable union of lines, so has measure zero.
– Milo Brandt
yesterday
I guess the answer is YES? (notice that the question was actually answerable by a simple Yes/No... I read this answer and I think it says Yes but I am sincerely not sure)
– Stijn de Witt
21 hours ago
add a comment |
Nice answer (I suppose you mean area $pi x^2$ of $B(x$) as opposed to 1?). And I suppose the same argument would go through for transcendentals because $overline{mathbb{Q}}$ is countable too.
– AlephNull
yesterday
1
Fubini's theorem is a bit overkill - if you can prove that a single line has measure zero (not so hard by explicit construction of covers by open balls), then this is a countable union of lines, so has measure zero.
– Milo Brandt
yesterday
I guess the answer is YES? (notice that the question was actually answerable by a simple Yes/No... I read this answer and I think it says Yes but I am sincerely not sure)
– Stijn de Witt
21 hours ago
Nice answer (I suppose you mean area $pi x^2$ of $B(x$) as opposed to 1?). And I suppose the same argument would go through for transcendentals because $overline{mathbb{Q}}$ is countable too.
– AlephNull
yesterday
Nice answer (I suppose you mean area $pi x^2$ of $B(x$) as opposed to 1?). And I suppose the same argument would go through for transcendentals because $overline{mathbb{Q}}$ is countable too.
– AlephNull
yesterday
1
1
Fubini's theorem is a bit overkill - if you can prove that a single line has measure zero (not so hard by explicit construction of covers by open balls), then this is a countable union of lines, so has measure zero.
– Milo Brandt
yesterday
Fubini's theorem is a bit overkill - if you can prove that a single line has measure zero (not so hard by explicit construction of covers by open balls), then this is a countable union of lines, so has measure zero.
– Milo Brandt
yesterday
I guess the answer is YES? (notice that the question was actually answerable by a simple Yes/No... I read this answer and I think it says Yes but I am sincerely not sure)
– Stijn de Witt
21 hours ago
I guess the answer is YES? (notice that the question was actually answerable by a simple Yes/No... I read this answer and I think it says Yes but I am sincerely not sure)
– Stijn de Witt
21 hours ago
add a comment |
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Related: Is the sum and difference of two irrationals always irrational?
– mrtaurho
yesterday
I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
– Milo Brandt
yesterday