Partitioning the positive integers into finitely many arithmetic progressions
$begingroup$
From Bóna's A Walk through Combinatorics:
Prove or disprove that if we partition the positive integers into finitely many arithmetic progressions then there will be at least one arithmetic progression whose difference and initial term are equal.
A variant of this problems asks to prove that there will be at least one arithmetic progression whose difference divides the initial term, which has an elementary proof. I suspect the same arithmetic progression will have the initial term equal to the difference. However, I cannot prove it.
arithmetic-progression elementary-proofs
edited 5 hours ago
Rodrigo de Azevedo
1,8422719
asked 13 hours ago
JohnJohn
194
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add a comment |
$begingroup$
From Bóna's A Walk through Combinatorics:
Prove or disprove that if we partition the positive integers into finitely many arithmetic progressions then there will be at least one arithmetic progression whose difference and initial term are equal.
A variant of this problems asks to prove that there will be at least one arithmetic progression whose difference divides the initial term, which has an elementary proof. I suspect the same arithmetic progression will have the initial term equal to the difference. However, I cannot prove it.
arithmetic-progression elementary-proofs
edited 5 hours ago
Rodrigo de Azevedo
1,8422719
asked 13 hours ago
JohnJohn
194
New contributor
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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2
$begingroup$
What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
$endgroup$
– Fedor Petrov
12 hours ago
1
$begingroup$
Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
$endgroup$
– John
11 hours ago
add a comment |
$begingroup$
From Bóna's A Walk through Combinatorics:
Prove or disprove that if we partition the positive integers into finitely many arithmetic progressions then there will be at least one arithmetic progression whose difference and initial term are equal.
A variant of this problems asks to prove that there will be at least one arithmetic progression whose difference divides the initial term, which has an elementary proof. I suspect the same arithmetic progression will have the initial term equal to the difference. However, I cannot prove it.
arithmetic-progression elementary-proofs
edited 5 hours ago
Rodrigo de Azevedo
1,8422719
asked 13 hours ago
JohnJohn
194
New contributor
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
From Bóna's A Walk through Combinatorics:
Prove or disprove that if we partition the positive integers into finitely many arithmetic progressions then there will be at least one arithmetic progression whose difference and initial term are equal.
A variant of this problems asks to prove that there will be at least one arithmetic progression whose difference divides the initial term, which has an elementary proof. I suspect the same arithmetic progression will have the initial term equal to the difference. However, I cannot prove it.
arithmetic-progression elementary-proofs
arithmetic-progression elementary-proofs
edited 5 hours ago
Rodrigo de Azevedo
1,8422719
asked 13 hours ago
JohnJohn
194
New contributor
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 5 hours ago
Rodrigo de Azevedo
1,8422719
asked 13 hours ago
JohnJohn
194
New contributor
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 5 hours ago
Rodrigo de Azevedo
1,8422719
edited 5 hours ago
Rodrigo de Azevedo
1,8422719
edited 5 hours ago
Rodrigo de Azevedo
1,8422719
1,8422719
asked 13 hours ago
JohnJohn
194
New contributor
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 13 hours ago
JohnJohn
194
asked 13 hours ago
JohnJohn
194
194
New contributor
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
John is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
$endgroup$
– Fedor Petrov
12 hours ago
1
$begingroup$
Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
$endgroup$
– John
11 hours ago
add a comment |
2
$begingroup$
What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
$endgroup$
– Fedor Petrov
12 hours ago
1
$begingroup$
Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
$endgroup$
– John
11 hours ago
2
2
$begingroup$
What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
$endgroup$
– Fedor Petrov
12 hours ago
$begingroup$
What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
$endgroup$
– Fedor Petrov
12 hours ago
1
1
$begingroup$
Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
$endgroup$
– John
11 hours ago
$begingroup$
Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
$endgroup$
– John
11 hours ago
add a comment |
2 Answers
2
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oldest
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$begingroup$
Any progression (if they are all infinite, of course, but otherwise the statement is clearly wrong) should have its initial term $a$ not greater than the difference $d$. Indeed, if $a>d$, and $a-d$ is covered by another progression $P$ with difference $d_1$, then $a-d+dd_1$ is covered twice --- a contradiction. Therefore if $a$ is divisible by $d$, we must have $a=d$. And such $a$ exists as you noted in OP (a short proof: take a progression which contains the product of all differences).
answered 10 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
$endgroup$
add a comment |
$begingroup$
Counter example where there are an infinite number of infinite arithmetic progressions and NO progression has the difference equal to the initial term.
for k from 1 to inf {
S(k) = 2^k * n + 2 ^ (k-1) for n from 0 to inf }
S(1) is odd integers;
S(2) is 2, 6, 10 - twice an odd integer;
and continues for powers of 2.
In all cases the initial term is the difference / 2.
answered 4 hours ago
Judah GreenblattJudah Greenblatt
1
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Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$begingroup$
The question does specifically say that the partition is into finitely many arithmetic progressions, which seems essential.
$endgroup$
– user44191
4 hours ago
add a comment |
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2 Answers
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$begingroup$
Any progression (if they are all infinite, of course, but otherwise the statement is clearly wrong) should have its initial term $a$ not greater than the difference $d$. Indeed, if $a>d$, and $a-d$ is covered by another progression $P$ with difference $d_1$, then $a-d+dd_1$ is covered twice --- a contradiction. Therefore if $a$ is divisible by $d$, we must have $a=d$. And such $a$ exists as you noted in OP (a short proof: take a progression which contains the product of all differences).
answered 10 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
$endgroup$
add a comment |
$begingroup$
Any progression (if they are all infinite, of course, but otherwise the statement is clearly wrong) should have its initial term $a$ not greater than the difference $d$. Indeed, if $a>d$, and $a-d$ is covered by another progression $P$ with difference $d_1$, then $a-d+dd_1$ is covered twice --- a contradiction. Therefore if $a$ is divisible by $d$, we must have $a=d$. And such $a$ exists as you noted in OP (a short proof: take a progression which contains the product of all differences).
answered 10 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
$endgroup$
add a comment |
$begingroup$
Any progression (if they are all infinite, of course, but otherwise the statement is clearly wrong) should have its initial term $a$ not greater than the difference $d$. Indeed, if $a>d$, and $a-d$ is covered by another progression $P$ with difference $d_1$, then $a-d+dd_1$ is covered twice --- a contradiction. Therefore if $a$ is divisible by $d$, we must have $a=d$. And such $a$ exists as you noted in OP (a short proof: take a progression which contains the product of all differences).
answered 10 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
$endgroup$
Any progression (if they are all infinite, of course, but otherwise the statement is clearly wrong) should have its initial term $a$ not greater than the difference $d$. Indeed, if $a>d$, and $a-d$ is covered by another progression $P$ with difference $d_1$, then $a-d+dd_1$ is covered twice --- a contradiction. Therefore if $a$ is divisible by $d$, we must have $a=d$. And such $a$ exists as you noted in OP (a short proof: take a progression which contains the product of all differences).
answered 10 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
edited 7 hours ago
answered 10 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
answered 10 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
answered 10 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
49.6k5114230
add a comment |
add a comment |
$begingroup$
Counter example where there are an infinite number of infinite arithmetic progressions and NO progression has the difference equal to the initial term.
for k from 1 to inf {
S(k) = 2^k * n + 2 ^ (k-1) for n from 0 to inf }
S(1) is odd integers;
S(2) is 2, 6, 10 - twice an odd integer;
and continues for powers of 2.
In all cases the initial term is the difference / 2.
answered 4 hours ago
Judah GreenblattJudah Greenblatt
1
New contributor
Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
$begingroup$
The question does specifically say that the partition is into finitely many arithmetic progressions, which seems essential.
$endgroup$
– user44191
4 hours ago
add a comment |
$begingroup$
Counter example where there are an infinite number of infinite arithmetic progressions and NO progression has the difference equal to the initial term.
for k from 1 to inf {
S(k) = 2^k * n + 2 ^ (k-1) for n from 0 to inf }
S(1) is odd integers;
S(2) is 2, 6, 10 - twice an odd integer;
and continues for powers of 2.
In all cases the initial term is the difference / 2.
answered 4 hours ago
Judah GreenblattJudah Greenblatt
1
New contributor
Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
The question does specifically say that the partition is into finitely many arithmetic progressions, which seems essential.
$endgroup$
– user44191
4 hours ago
add a comment |
$begingroup$
Counter example where there are an infinite number of infinite arithmetic progressions and NO progression has the difference equal to the initial term.
for k from 1 to inf {
S(k) = 2^k * n + 2 ^ (k-1) for n from 0 to inf }
S(1) is odd integers;
S(2) is 2, 6, 10 - twice an odd integer;
and continues for powers of 2.
In all cases the initial term is the difference / 2.
answered 4 hours ago
Judah GreenblattJudah Greenblatt
1
New contributor
Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Counter example where there are an infinite number of infinite arithmetic progressions and NO progression has the difference equal to the initial term.
for k from 1 to inf {
S(k) = 2^k * n + 2 ^ (k-1) for n from 0 to inf }
S(1) is odd integers;
S(2) is 2, 6, 10 - twice an odd integer;
and continues for powers of 2.
In all cases the initial term is the difference / 2.
answered 4 hours ago
Judah GreenblattJudah Greenblatt
1
New contributor
Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
Judah GreenblattJudah Greenblatt
1
New contributor
Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
Judah GreenblattJudah Greenblatt
1
answered 4 hours ago
Judah GreenblattJudah Greenblatt
1
1
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Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Judah Greenblatt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$begingroup$
The question does specifically say that the partition is into finitely many arithmetic progressions, which seems essential.
$endgroup$
– user44191
4 hours ago
add a comment |
$begingroup$
The question does specifically say that the partition is into finitely many arithmetic progressions, which seems essential.
$endgroup$
– user44191
4 hours ago
$begingroup$
The question does specifically say that the partition is into finitely many arithmetic progressions, which seems essential.
$endgroup$
– user44191
4 hours ago
$begingroup$
The question does specifically say that the partition is into finitely many arithmetic progressions, which seems essential.
$endgroup$
– user44191
4 hours ago
add a comment |
John is a new contributor. Be nice, and check out our Code of Conduct.
John is a new contributor. Be nice, and check out our Code of Conduct.
John is a new contributor. Be nice, and check out our Code of Conduct.
John is a new contributor. Be nice, and check out our Code of Conduct.
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2
$begingroup$
What is the origin of this question? Usually if you ask to "prove" something you should have some strong evidence that the proof exists.
$endgroup$
– Fedor Petrov
12 hours ago
1
$begingroup$
Sorry I should’ve asked for counter example also. I have edited the question now. The source is the variant mentioned below the main question which is an exercise from “A Walk Through Combinatorics”.
$endgroup$
– John
11 hours ago