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Is the sum of two irrational numbers almost always irrational?
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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 Jan 6 at 14:25
AlephNullAlephNull
2529
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add a comment |
$begingroup$
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 Jan 6 at 14:25
AlephNullAlephNull
2529
$endgroup$
1
$begingroup$
Related: Is the sum and difference of two irrationals always irrational?
$endgroup$
– mrtaurho
Jan 6 at 14:45
$begingroup$
I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
$endgroup$
– Milo Brandt
Jan 6 at 16:46
add a comment |
$begingroup$
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 Jan 6 at 14:25
AlephNullAlephNull
2529
$endgroup$
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 Jan 6 at 14:25
AlephNullAlephNull
2529
asked Jan 6 at 14:25
AlephNullAlephNull
2529
edited Jan 6 at 14:37
AlephNull
asked Jan 6 at 14:25
AlephNullAlephNull
2529
asked Jan 6 at 14:25
AlephNullAlephNull
2529
asked Jan 6 at 14:25
AlephNullAlephNull
2529
2529
1
$begingroup$
Related: Is the sum and difference of two irrationals always irrational?
$endgroup$
– mrtaurho
Jan 6 at 14:45
$begingroup$
I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
$endgroup$
– Milo Brandt
Jan 6 at 16:46
add a comment |
1
$begingroup$
Related: Is the sum and difference of two irrationals always irrational?
$endgroup$
– mrtaurho
Jan 6 at 14:45
$begingroup$
I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
$endgroup$
– Milo Brandt
Jan 6 at 16:46
1
1
$begingroup$
Related: Is the sum and difference of two irrationals always irrational?
$endgroup$
– mrtaurho
Jan 6 at 14:45
$begingroup$
Related: Is the sum and difference of two irrationals always irrational?
$endgroup$
– mrtaurho
Jan 6 at 14:45
$begingroup$
I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
$endgroup$
– Milo Brandt
Jan 6 at 16:46
$begingroup$
I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
$endgroup$
– Milo Brandt
Jan 6 at 16:46
add a comment |
1 Answer
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active
oldest
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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 Jan 6 at 14:37
Charlie FrohmanCharlie Frohman
1,478913
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$begingroup$
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.
$endgroup$
– AlephNull
Jan 6 at 15:03
1
$begingroup$
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.
$endgroup$
– Milo Brandt
Jan 6 at 16:49
$begingroup$
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)
$endgroup$
– Stijn de Witt
Jan 6 at 23:00
add a comment |
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1 Answer
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$begingroup$
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 Jan 6 at 14:37
Charlie FrohmanCharlie Frohman
1,478913
$endgroup$
$begingroup$
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.
$endgroup$
– AlephNull
Jan 6 at 15:03
1
$begingroup$
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.
$endgroup$
– Milo Brandt
Jan 6 at 16:49
$begingroup$
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)
$endgroup$
– Stijn de Witt
Jan 6 at 23:00
add a comment |
$begingroup$
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 Jan 6 at 14:37
Charlie FrohmanCharlie Frohman
1,478913
$endgroup$
$begingroup$
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.
$endgroup$
– AlephNull
Jan 6 at 15:03
1
$begingroup$
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.
$endgroup$
– Milo Brandt
Jan 6 at 16:49
$begingroup$
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)
$endgroup$
– Stijn de Witt
Jan 6 at 23:00
add a comment |
$begingroup$
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 Jan 6 at 14:37
Charlie FrohmanCharlie Frohman
1,478913
$endgroup$
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 Jan 6 at 14:37
Charlie FrohmanCharlie Frohman
1,478913
edited Jan 6 at 14:46
answered Jan 6 at 14:37
Charlie FrohmanCharlie Frohman
1,478913
answered Jan 6 at 14:37
Charlie FrohmanCharlie Frohman
1,478913
answered Jan 6 at 14:37
Charlie FrohmanCharlie Frohman
1,478913
1,478913
$begingroup$
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.
$endgroup$
– AlephNull
Jan 6 at 15:03
1
$begingroup$
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.
$endgroup$
– Milo Brandt
Jan 6 at 16:49
$begingroup$
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)
$endgroup$
– Stijn de Witt
Jan 6 at 23:00
add a comment |
$begingroup$
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.
$endgroup$
– AlephNull
Jan 6 at 15:03
1
$begingroup$
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.
$endgroup$
– Milo Brandt
Jan 6 at 16:49
$begingroup$
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)
$endgroup$
– Stijn de Witt
Jan 6 at 23:00
$begingroup$
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.
$endgroup$
– AlephNull
Jan 6 at 15:03
$begingroup$
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.
$endgroup$
– AlephNull
Jan 6 at 15:03
1
1
$begingroup$
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.
$endgroup$
– Milo Brandt
Jan 6 at 16:49
$begingroup$
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.
$endgroup$
– Milo Brandt
Jan 6 at 16:49
$begingroup$
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)
$endgroup$
– Stijn de Witt
Jan 6 at 23:00
$begingroup$
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)
$endgroup$
– Stijn de Witt
Jan 6 at 23:00
add a comment |
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$begingroup$
Related: Is the sum and difference of two irrationals always irrational?
$endgroup$
– mrtaurho
Jan 6 at 14:45
$begingroup$
I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
$endgroup$
– Milo Brandt
Jan 6 at 16:46