Two Subsets of Squares of Reciprocals of Primes with Equal Sums












2












$begingroup$


Let $$A={frac{1}{2^2},frac{1}{3^2},frac{1}{5^2},...}$$ be the set of squares of the reciprocals of prime numbers.



We have
$$sum_{xin A}x < infty$$
Do there exist $B subset A$, $C subset A$, $B cap C = emptyset$, such that
$$sum_{xin B}x = sum_{xin C}x ?$$
It is important that we deal with primes and not with all natural numbers, otherwise we would have infinitely many solutions, as described in Wikipedia.






number-theory prime-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The answer is pathologically yes; take $B=C=varnothing$. These are the only finite sets with this property. There are infinitely many solutions with $B$ and $C$ infinite.
    $endgroup$
    – Servaes
    12 hours ago


















2












$begingroup$


Let $$A={frac{1}{2^2},frac{1}{3^2},frac{1}{5^2},...}$$ be the set of squares of the reciprocals of prime numbers.



We have
$$sum_{xin A}x < infty$$
Do there exist $B subset A$, $C subset A$, $B cap C = emptyset$, such that
$$sum_{xin B}x = sum_{xin C}x ?$$
It is important that we deal with primes and not with all natural numbers, otherwise we would have infinitely many solutions, as described in Wikipedia.






number-theory prime-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The answer is pathologically yes; take $B=C=varnothing$. These are the only finite sets with this property. There are infinitely many solutions with $B$ and $C$ infinite.
    $endgroup$
    – Servaes
    12 hours ago
















2












2








2





$begingroup$


Let $$A={frac{1}{2^2},frac{1}{3^2},frac{1}{5^2},...}$$ be the set of squares of the reciprocals of prime numbers.



We have
$$sum_{xin A}x < infty$$
Do there exist $B subset A$, $C subset A$, $B cap C = emptyset$, such that
$$sum_{xin B}x = sum_{xin C}x ?$$
It is important that we deal with primes and not with all natural numbers, otherwise we would have infinitely many solutions, as described in Wikipedia.






number-theory prime-numbers







share|cite|improve this question











$endgroup$




Let $$A={frac{1}{2^2},frac{1}{3^2},frac{1}{5^2},...}$$ be the set of squares of the reciprocals of prime numbers.



We have
$$sum_{xin A}x < infty$$
Do there exist $B subset A$, $C subset A$, $B cap C = emptyset$, such that
$$sum_{xin B}x = sum_{xin C}x ?$$
It is important that we deal with primes and not with all natural numbers, otherwise we would have infinitely many solutions, as described in Wikipedia.






number-theory prime-numbers




number-theory prime-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 12 hours ago







co.sine

















asked 13 hours ago









co.sineco.sine

372112




372112








  • 1




    $begingroup$
    The answer is pathologically yes; take $B=C=varnothing$. These are the only finite sets with this property. There are infinitely many solutions with $B$ and $C$ infinite.
    $endgroup$
    – Servaes
    12 hours ago
















  • 1




    $begingroup$
    The answer is pathologically yes; take $B=C=varnothing$. These are the only finite sets with this property. There are infinitely many solutions with $B$ and $C$ infinite.
    $endgroup$
    – Servaes
    12 hours ago










1




1




$begingroup$
The answer is pathologically yes; take $B=C=varnothing$. These are the only finite sets with this property. There are infinitely many solutions with $B$ and $C$ infinite.
$endgroup$
– Servaes
12 hours ago






$begingroup$
The answer is pathologically yes; take $B=C=varnothing$. These are the only finite sets with this property. There are infinitely many solutions with $B$ and $C$ infinite.
$endgroup$
– Servaes
12 hours ago












2 Answers
2






active

oldest

votes


















4












$begingroup$

The answer is negative. Let $C=sum_{p}frac{1}{p^2}$. This constant is approximately $frac{1}{2}logfrac{5}{2}$ due to Euler's product, leading to:
$$ sum_{p}frac{1}{p^2}approx sum_pfrac{1}{2}logleft(frac{1+frac{1}{p^2}}{1-frac{1}{p^2}}right)=frac{1}{2}logfrac{zeta(2)^2}{zeta(4)}. $$
Assume that the set of prime numbers can be partitioned as $Ucup V$ with $Ucap V=emptyset$ and $U,Vneqemptyset$.
We may assume without loss of generality that $2in U$, hence
$$ sum_{pin U}frac{1}{p^2}geq frac{1}{4} > frac{1}{2}sum_{pin Ucup V}frac{1}{p^2}=frac{C}{2} $$
which contradicts
$$ sum_{pin U}frac{1}{p^2}=sum_{pin V}frac{1}{p^2}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @co.sine: actually I was completely wrong before, I have fixed the answer.
    $endgroup$
    – Jack D'Aurizio
    11 hours ago










  • $begingroup$
    BRILLIANT, Jack D'Aurizio. For the constant $C$: mathoverflow.net/a/53444/102549
    $endgroup$
    – co.sine
    11 hours ago










  • $begingroup$
    What about the case $U cup V subsetneq {text{primes}}$?
    $endgroup$
    – co.sine
    11 hours ago












  • $begingroup$
    I'd rather lean to believe your original answer: we do not have to partition the set of all primes, but only want to find two disjoint subsets thereof.
    $endgroup$
    – W-t-P
    11 hours ago










  • $begingroup$
    For $U$, the sum must be at least $1/4$. $C<1/2$, so the maximum sum for $V$ must be less than $1/4$.
    $endgroup$
    – qwr
    8 hours ago



















3












$begingroup$

Assuming that $B$ and $C$ (are finite and) have the required property, denote by $P$ the product of all primes $p$ with $1/p^2in B$, and by $Q$ the product of all primes $p$ with $1/p^2in C$. Then the LHS of $sum_{xin B}x=sum_{xin C}x$ would be a rational number which, reduced to its lower terms, has its denominator equal to $prod_{pin P} p^2$, while the RHS has the denominator equal to $prod_{pin Q}p^2$. By the uniqueness of prime decomposition, we would then have $P=Q$, a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What if $B$ and $C$ are empty or infinite? The latter is the interesting part of the question really, as your argument shows.
    $endgroup$
    – Servaes
    12 hours ago












  • $begingroup$
    @Servaes: I believe that in this case equality is possible, but this would be a little technical to explain in details.
    $endgroup$
    – W-t-P
    12 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

The answer is negative. Let $C=sum_{p}frac{1}{p^2}$. This constant is approximately $frac{1}{2}logfrac{5}{2}$ due to Euler's product, leading to:
$$ sum_{p}frac{1}{p^2}approx sum_pfrac{1}{2}logleft(frac{1+frac{1}{p^2}}{1-frac{1}{p^2}}right)=frac{1}{2}logfrac{zeta(2)^2}{zeta(4)}. $$
Assume that the set of prime numbers can be partitioned as $Ucup V$ with $Ucap V=emptyset$ and $U,Vneqemptyset$.
We may assume without loss of generality that $2in U$, hence
$$ sum_{pin U}frac{1}{p^2}geq frac{1}{4} > frac{1}{2}sum_{pin Ucup V}frac{1}{p^2}=frac{C}{2} $$
which contradicts
$$ sum_{pin U}frac{1}{p^2}=sum_{pin V}frac{1}{p^2}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @co.sine: actually I was completely wrong before, I have fixed the answer.
    $endgroup$
    – Jack D'Aurizio
    11 hours ago










  • $begingroup$
    BRILLIANT, Jack D'Aurizio. For the constant $C$: mathoverflow.net/a/53444/102549
    $endgroup$
    – co.sine
    11 hours ago










  • $begingroup$
    What about the case $U cup V subsetneq {text{primes}}$?
    $endgroup$
    – co.sine
    11 hours ago












  • $begingroup$
    I'd rather lean to believe your original answer: we do not have to partition the set of all primes, but only want to find two disjoint subsets thereof.
    $endgroup$
    – W-t-P
    11 hours ago










  • $begingroup$
    For $U$, the sum must be at least $1/4$. $C<1/2$, so the maximum sum for $V$ must be less than $1/4$.
    $endgroup$
    – qwr
    8 hours ago
















4












$begingroup$

The answer is negative. Let $C=sum_{p}frac{1}{p^2}$. This constant is approximately $frac{1}{2}logfrac{5}{2}$ due to Euler's product, leading to:
$$ sum_{p}frac{1}{p^2}approx sum_pfrac{1}{2}logleft(frac{1+frac{1}{p^2}}{1-frac{1}{p^2}}right)=frac{1}{2}logfrac{zeta(2)^2}{zeta(4)}. $$
Assume that the set of prime numbers can be partitioned as $Ucup V$ with $Ucap V=emptyset$ and $U,Vneqemptyset$.
We may assume without loss of generality that $2in U$, hence
$$ sum_{pin U}frac{1}{p^2}geq frac{1}{4} > frac{1}{2}sum_{pin Ucup V}frac{1}{p^2}=frac{C}{2} $$
which contradicts
$$ sum_{pin U}frac{1}{p^2}=sum_{pin V}frac{1}{p^2}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @co.sine: actually I was completely wrong before, I have fixed the answer.
    $endgroup$
    – Jack D'Aurizio
    11 hours ago










  • $begingroup$
    BRILLIANT, Jack D'Aurizio. For the constant $C$: mathoverflow.net/a/53444/102549
    $endgroup$
    – co.sine
    11 hours ago










  • $begingroup$
    What about the case $U cup V subsetneq {text{primes}}$?
    $endgroup$
    – co.sine
    11 hours ago












  • $begingroup$
    I'd rather lean to believe your original answer: we do not have to partition the set of all primes, but only want to find two disjoint subsets thereof.
    $endgroup$
    – W-t-P
    11 hours ago










  • $begingroup$
    For $U$, the sum must be at least $1/4$. $C<1/2$, so the maximum sum for $V$ must be less than $1/4$.
    $endgroup$
    – qwr
    8 hours ago














4












4








4





$begingroup$

The answer is negative. Let $C=sum_{p}frac{1}{p^2}$. This constant is approximately $frac{1}{2}logfrac{5}{2}$ due to Euler's product, leading to:
$$ sum_{p}frac{1}{p^2}approx sum_pfrac{1}{2}logleft(frac{1+frac{1}{p^2}}{1-frac{1}{p^2}}right)=frac{1}{2}logfrac{zeta(2)^2}{zeta(4)}. $$
Assume that the set of prime numbers can be partitioned as $Ucup V$ with $Ucap V=emptyset$ and $U,Vneqemptyset$.
We may assume without loss of generality that $2in U$, hence
$$ sum_{pin U}frac{1}{p^2}geq frac{1}{4} > frac{1}{2}sum_{pin Ucup V}frac{1}{p^2}=frac{C}{2} $$
which contradicts
$$ sum_{pin U}frac{1}{p^2}=sum_{pin V}frac{1}{p^2}.$$






share|cite|improve this answer











$endgroup$



The answer is negative. Let $C=sum_{p}frac{1}{p^2}$. This constant is approximately $frac{1}{2}logfrac{5}{2}$ due to Euler's product, leading to:
$$ sum_{p}frac{1}{p^2}approx sum_pfrac{1}{2}logleft(frac{1+frac{1}{p^2}}{1-frac{1}{p^2}}right)=frac{1}{2}logfrac{zeta(2)^2}{zeta(4)}. $$
Assume that the set of prime numbers can be partitioned as $Ucup V$ with $Ucap V=emptyset$ and $U,Vneqemptyset$.
We may assume without loss of generality that $2in U$, hence
$$ sum_{pin U}frac{1}{p^2}geq frac{1}{4} > frac{1}{2}sum_{pin Ucup V}frac{1}{p^2}=frac{C}{2} $$
which contradicts
$$ sum_{pin U}frac{1}{p^2}=sum_{pin V}frac{1}{p^2}.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 11 hours ago

























answered 12 hours ago









Jack D'AurizioJack D'Aurizio

290k33282662




290k33282662












  • $begingroup$
    @co.sine: actually I was completely wrong before, I have fixed the answer.
    $endgroup$
    – Jack D'Aurizio
    11 hours ago










  • $begingroup$
    BRILLIANT, Jack D'Aurizio. For the constant $C$: mathoverflow.net/a/53444/102549
    $endgroup$
    – co.sine
    11 hours ago










  • $begingroup$
    What about the case $U cup V subsetneq {text{primes}}$?
    $endgroup$
    – co.sine
    11 hours ago












  • $begingroup$
    I'd rather lean to believe your original answer: we do not have to partition the set of all primes, but only want to find two disjoint subsets thereof.
    $endgroup$
    – W-t-P
    11 hours ago










  • $begingroup$
    For $U$, the sum must be at least $1/4$. $C<1/2$, so the maximum sum for $V$ must be less than $1/4$.
    $endgroup$
    – qwr
    8 hours ago


















  • $begingroup$
    @co.sine: actually I was completely wrong before, I have fixed the answer.
    $endgroup$
    – Jack D'Aurizio
    11 hours ago










  • $begingroup$
    BRILLIANT, Jack D'Aurizio. For the constant $C$: mathoverflow.net/a/53444/102549
    $endgroup$
    – co.sine
    11 hours ago










  • $begingroup$
    What about the case $U cup V subsetneq {text{primes}}$?
    $endgroup$
    – co.sine
    11 hours ago












  • $begingroup$
    I'd rather lean to believe your original answer: we do not have to partition the set of all primes, but only want to find two disjoint subsets thereof.
    $endgroup$
    – W-t-P
    11 hours ago










  • $begingroup$
    For $U$, the sum must be at least $1/4$. $C<1/2$, so the maximum sum for $V$ must be less than $1/4$.
    $endgroup$
    – qwr
    8 hours ago
















$begingroup$
@co.sine: actually I was completely wrong before, I have fixed the answer.
$endgroup$
– Jack D'Aurizio
11 hours ago




$begingroup$
@co.sine: actually I was completely wrong before, I have fixed the answer.
$endgroup$
– Jack D'Aurizio
11 hours ago












$begingroup$
BRILLIANT, Jack D'Aurizio. For the constant $C$: mathoverflow.net/a/53444/102549
$endgroup$
– co.sine
11 hours ago




$begingroup$
BRILLIANT, Jack D'Aurizio. For the constant $C$: mathoverflow.net/a/53444/102549
$endgroup$
– co.sine
11 hours ago












$begingroup$
What about the case $U cup V subsetneq {text{primes}}$?
$endgroup$
– co.sine
11 hours ago






$begingroup$
What about the case $U cup V subsetneq {text{primes}}$?
$endgroup$
– co.sine
11 hours ago














$begingroup$
I'd rather lean to believe your original answer: we do not have to partition the set of all primes, but only want to find two disjoint subsets thereof.
$endgroup$
– W-t-P
11 hours ago




$begingroup$
I'd rather lean to believe your original answer: we do not have to partition the set of all primes, but only want to find two disjoint subsets thereof.
$endgroup$
– W-t-P
11 hours ago












$begingroup$
For $U$, the sum must be at least $1/4$. $C<1/2$, so the maximum sum for $V$ must be less than $1/4$.
$endgroup$
– qwr
8 hours ago




$begingroup$
For $U$, the sum must be at least $1/4$. $C<1/2$, so the maximum sum for $V$ must be less than $1/4$.
$endgroup$
– qwr
8 hours ago











3












$begingroup$

Assuming that $B$ and $C$ (are finite and) have the required property, denote by $P$ the product of all primes $p$ with $1/p^2in B$, and by $Q$ the product of all primes $p$ with $1/p^2in C$. Then the LHS of $sum_{xin B}x=sum_{xin C}x$ would be a rational number which, reduced to its lower terms, has its denominator equal to $prod_{pin P} p^2$, while the RHS has the denominator equal to $prod_{pin Q}p^2$. By the uniqueness of prime decomposition, we would then have $P=Q$, a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What if $B$ and $C$ are empty or infinite? The latter is the interesting part of the question really, as your argument shows.
    $endgroup$
    – Servaes
    12 hours ago












  • $begingroup$
    @Servaes: I believe that in this case equality is possible, but this would be a little technical to explain in details.
    $endgroup$
    – W-t-P
    12 hours ago
















3












$begingroup$

Assuming that $B$ and $C$ (are finite and) have the required property, denote by $P$ the product of all primes $p$ with $1/p^2in B$, and by $Q$ the product of all primes $p$ with $1/p^2in C$. Then the LHS of $sum_{xin B}x=sum_{xin C}x$ would be a rational number which, reduced to its lower terms, has its denominator equal to $prod_{pin P} p^2$, while the RHS has the denominator equal to $prod_{pin Q}p^2$. By the uniqueness of prime decomposition, we would then have $P=Q$, a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What if $B$ and $C$ are empty or infinite? The latter is the interesting part of the question really, as your argument shows.
    $endgroup$
    – Servaes
    12 hours ago












  • $begingroup$
    @Servaes: I believe that in this case equality is possible, but this would be a little technical to explain in details.
    $endgroup$
    – W-t-P
    12 hours ago














3












3








3





$begingroup$

Assuming that $B$ and $C$ (are finite and) have the required property, denote by $P$ the product of all primes $p$ with $1/p^2in B$, and by $Q$ the product of all primes $p$ with $1/p^2in C$. Then the LHS of $sum_{xin B}x=sum_{xin C}x$ would be a rational number which, reduced to its lower terms, has its denominator equal to $prod_{pin P} p^2$, while the RHS has the denominator equal to $prod_{pin Q}p^2$. By the uniqueness of prime decomposition, we would then have $P=Q$, a contradiction.






share|cite|improve this answer











$endgroup$



Assuming that $B$ and $C$ (are finite and) have the required property, denote by $P$ the product of all primes $p$ with $1/p^2in B$, and by $Q$ the product of all primes $p$ with $1/p^2in C$. Then the LHS of $sum_{xin B}x=sum_{xin C}x$ would be a rational number which, reduced to its lower terms, has its denominator equal to $prod_{pin P} p^2$, while the RHS has the denominator equal to $prod_{pin Q}p^2$. By the uniqueness of prime decomposition, we would then have $P=Q$, a contradiction.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 12 hours ago

























answered 12 hours ago









W-t-PW-t-P

1,102610




1,102610












  • $begingroup$
    What if $B$ and $C$ are empty or infinite? The latter is the interesting part of the question really, as your argument shows.
    $endgroup$
    – Servaes
    12 hours ago












  • $begingroup$
    @Servaes: I believe that in this case equality is possible, but this would be a little technical to explain in details.
    $endgroup$
    – W-t-P
    12 hours ago


















  • $begingroup$
    What if $B$ and $C$ are empty or infinite? The latter is the interesting part of the question really, as your argument shows.
    $endgroup$
    – Servaes
    12 hours ago












  • $begingroup$
    @Servaes: I believe that in this case equality is possible, but this would be a little technical to explain in details.
    $endgroup$
    – W-t-P
    12 hours ago
















$begingroup$
What if $B$ and $C$ are empty or infinite? The latter is the interesting part of the question really, as your argument shows.
$endgroup$
– Servaes
12 hours ago






$begingroup$
What if $B$ and $C$ are empty or infinite? The latter is the interesting part of the question really, as your argument shows.
$endgroup$
– Servaes
12 hours ago














$begingroup$
@Servaes: I believe that in this case equality is possible, but this would be a little technical to explain in details.
$endgroup$
– W-t-P
12 hours ago




$begingroup$
@Servaes: I believe that in this case equality is possible, but this would be a little technical to explain in details.
$endgroup$
– W-t-P
12 hours ago


















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Alcedinidae

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Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout ...
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Origin of the phrase “under your belt”?

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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 17 1 Today, I found myself discussing what students should have "under their belts" during a lecture, and I wondered to myself if there was some inappropriate undertone here that I might be unaware of. Question : What is the origin of the phrase "under your belt"? A google search revealed some webpages (e.g. [1]) that assert its origin is related to consumption of food -- once a meal has been eaten, it's under your belt (which is a relief). However, with such websites, I have no way to determine fact from "random guy on the internet making stuff up". etymology idioms ...
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