A question about partitioning positivie integers into finitely many arithmetic progresions
$begingroup$
Prove or disprove that if we partition 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 AP whose difference divides the initial term, which has an elementary proof. I suspect the same AP will have the initial term equal to the difference. But I cannot prove it.
arithmetic-progression elementary-proofs
asked 7 hours ago
JohnJohn
164
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add a comment |
$begingroup$
Prove or disprove that if we partition 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 AP whose difference divides the initial term, which has an elementary proof. I suspect the same AP will have the initial term equal to the difference. But I cannot prove it.
arithmetic-progression elementary-proofs
asked 7 hours ago
JohnJohn
164
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
6 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
5 hours ago
add a comment |
$begingroup$
Prove or disprove that if we partition 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 AP whose difference divides the initial term, which has an elementary proof. I suspect the same AP will have the initial term equal to the difference. But I cannot prove it.
arithmetic-progression elementary-proofs
asked 7 hours ago
JohnJohn
164
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.
$endgroup$
Prove or disprove that if we partition 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 AP whose difference divides the initial term, which has an elementary proof. I suspect the same AP will have the initial term equal to the difference. But I cannot prove it.
arithmetic-progression elementary-proofs
arithmetic-progression elementary-proofs
asked 7 hours ago
JohnJohn
164
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 7 hours ago
JohnJohn
164
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
John
asked 7 hours ago
JohnJohn
164
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 7 hours ago
JohnJohn
164
asked 7 hours ago
JohnJohn
164
164
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.
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
6 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
5 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
6 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
5 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
6 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
6 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
5 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
5 hours ago
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
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 4 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
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add a comment |
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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 4 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 4 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 4 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 4 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
edited 1 hour ago
answered 4 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
answered 4 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
answered 4 hours ago
Fedor PetrovFedor Petrov
49.6k5114230
49.6k5114230
add a comment |
add a comment |
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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
6 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
5 hours ago