Partitioning the positive integers into finitely many arithmetic progressions
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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 Feb 25 at 21:02
Rodrigo de Azevedo
1,8422719
asked Feb 25 at 13:28
JohnJohn
194
$endgroup$
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 Feb 25 at 21:02
Rodrigo de Azevedo
1,8422719
asked Feb 25 at 13:28
JohnJohn
194
$endgroup$
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
Feb 25 at 14:30
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
Feb 25 at 15:27
$begingroup$
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.
$endgroup$
– Judah Greenblatt
Feb 25 at 21:43
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 Feb 25 at 21:02
Rodrigo de Azevedo
1,8422719
asked Feb 25 at 13:28
JohnJohn
194
$endgroup$
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 Feb 25 at 21:02
Rodrigo de Azevedo
1,8422719
asked Feb 25 at 13:28
JohnJohn
194
edited Feb 25 at 21:02
Rodrigo de Azevedo
1,8422719
asked Feb 25 at 13:28
JohnJohn
194
edited Feb 25 at 21:02
Rodrigo de Azevedo
1,8422719
edited Feb 25 at 21:02
Rodrigo de Azevedo
1,8422719
edited Feb 25 at 21:02
Rodrigo de Azevedo
1,8422719
1,8422719
asked Feb 25 at 13:28
JohnJohn
194
asked Feb 25 at 13:28
JohnJohn
194
asked Feb 25 at 13:28
JohnJohn
194
194
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
Feb 25 at 14:30
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
Feb 25 at 15:27
$begingroup$
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.
$endgroup$
– Judah Greenblatt
Feb 25 at 21:43
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
Feb 25 at 14:30
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
Feb 25 at 15:27
$begingroup$
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.
$endgroup$
– Judah Greenblatt
Feb 25 at 21:43
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
Feb 25 at 14:30
$begingroup$
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
Feb 25 at 14:30
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
Feb 25 at 15:27
$begingroup$
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
Feb 25 at 15:27
$begingroup$
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.
$endgroup$
– Judah Greenblatt
Feb 25 at 21:43
$begingroup$
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.
$endgroup$
– Judah Greenblatt
Feb 25 at 21:43
add a comment |
1 Answer
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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 Feb 25 at 15:43
Fedor PetrovFedor Petrov
51.1k6118235
$endgroup$
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 Feb 25 at 15:43
Fedor PetrovFedor Petrov
51.1k6118235
$endgroup$
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 Feb 25 at 15:43
Fedor PetrovFedor Petrov
51.1k6118235
$endgroup$
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 Feb 25 at 15:43
Fedor PetrovFedor Petrov
51.1k6118235
$endgroup$
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 Feb 25 at 15:43
Fedor PetrovFedor Petrov
51.1k6118235
edited Feb 25 at 18:51
answered Feb 25 at 15:43
Fedor PetrovFedor Petrov
51.1k6118235
answered Feb 25 at 15:43
Fedor PetrovFedor Petrov
51.1k6118235
answered Feb 25 at 15:43
Fedor PetrovFedor Petrov
51.1k6118235
51.1k6118235
add a comment |
add a comment |
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$begingroup$
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
Feb 25 at 14:30
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
Feb 25 at 15:27
$begingroup$
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.
$endgroup$
– Judah Greenblatt
Feb 25 at 21:43