Counting equivalence classes
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Given any $n$ integers $m_1,m_2,ldots m_nin mathbb{N}$ if we define an equivalence relation $sim$ on $mathbb{N}^n$ as follows:
$$(a_1,a_2,ldots a_n)sim (b_1,b_2,ldots b_n)iff text{For every integer }1leq kleq ntext{ we have }a_kequiv b_kbmod m_k$$
Then what is the cardinality of the quotient set $mathbb{N}^n/sim$?
For the special case when $m_1,m_2,ldots m_n$ are pairwise coprime, I know $|mathbb{N}^n/sim|=m_1m_2cdots m_n$. However what about in general when $m_1,m_2,ldots m_n$ might have common factors?
combinatorics modular-arithmetic equivalence-relations
asked Dec 5 '18 at 0:43
user3865391user3865391
6141215
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add a comment |
$begingroup$
Given any $n$ integers $m_1,m_2,ldots m_nin mathbb{N}$ if we define an equivalence relation $sim$ on $mathbb{N}^n$ as follows:
$$(a_1,a_2,ldots a_n)sim (b_1,b_2,ldots b_n)iff text{For every integer }1leq kleq ntext{ we have }a_kequiv b_kbmod m_k$$
Then what is the cardinality of the quotient set $mathbb{N}^n/sim$?
For the special case when $m_1,m_2,ldots m_n$ are pairwise coprime, I know $|mathbb{N}^n/sim|=m_1m_2cdots m_n$. However what about in general when $m_1,m_2,ldots m_n$ might have common factors?
combinatorics modular-arithmetic equivalence-relations
asked Dec 5 '18 at 0:43
user3865391user3865391
6141215
$endgroup$
add a comment |
$begingroup$
Given any $n$ integers $m_1,m_2,ldots m_nin mathbb{N}$ if we define an equivalence relation $sim$ on $mathbb{N}^n$ as follows:
$$(a_1,a_2,ldots a_n)sim (b_1,b_2,ldots b_n)iff text{For every integer }1leq kleq ntext{ we have }a_kequiv b_kbmod m_k$$
Then what is the cardinality of the quotient set $mathbb{N}^n/sim$?
For the special case when $m_1,m_2,ldots m_n$ are pairwise coprime, I know $|mathbb{N}^n/sim|=m_1m_2cdots m_n$. However what about in general when $m_1,m_2,ldots m_n$ might have common factors?
combinatorics modular-arithmetic equivalence-relations
asked Dec 5 '18 at 0:43
user3865391user3865391
6141215
$endgroup$
Given any $n$ integers $m_1,m_2,ldots m_nin mathbb{N}$ if we define an equivalence relation $sim$ on $mathbb{N}^n$ as follows:
$$(a_1,a_2,ldots a_n)sim (b_1,b_2,ldots b_n)iff text{For every integer }1leq kleq ntext{ we have }a_kequiv b_kbmod m_k$$
Then what is the cardinality of the quotient set $mathbb{N}^n/sim$?
For the special case when $m_1,m_2,ldots m_n$ are pairwise coprime, I know $|mathbb{N}^n/sim|=m_1m_2cdots m_n$. However what about in general when $m_1,m_2,ldots m_n$ might have common factors?
combinatorics modular-arithmetic equivalence-relations
combinatorics modular-arithmetic equivalence-relations
asked Dec 5 '18 at 0:43
user3865391user3865391
6141215
asked Dec 5 '18 at 0:43
user3865391user3865391
6141215
asked Dec 5 '18 at 0:43
user3865391user3865391
6141215
asked Dec 5 '18 at 0:43
user3865391user3865391
6141215
asked Dec 5 '18 at 0:43
user3865391user3865391
6141215
6141215
add a comment |
add a comment |
1 Answer
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The answer is unconditionally $m_1...m_n$. You did not place any restrictions between the entries of the $n$-tuple.
answered Dec 5 '18 at 4:27
zoidbergzoidberg
1,065113
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I don't understand what you mean by restrictions.
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– user3865391
Dec 6 '18 at 20:11
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You say you have an argument when $m_i$ are pairwise coprime. Is there anything that goes wrong with the proof when for instance $n=2$ and $m_1=2,m_2=4$?
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– zoidberg
Dec 6 '18 at 20:28
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Please don't downvote posts simply because you don't understand.
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– zoidberg
Dec 6 '18 at 20:47
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
The answer is unconditionally $m_1...m_n$. You did not place any restrictions between the entries of the $n$-tuple.
answered Dec 5 '18 at 4:27
zoidbergzoidberg
1,065113
$endgroup$
$begingroup$
I don't understand what you mean by restrictions.
$endgroup$
– user3865391
Dec 6 '18 at 20:11
$begingroup$
You say you have an argument when $m_i$ are pairwise coprime. Is there anything that goes wrong with the proof when for instance $n=2$ and $m_1=2,m_2=4$?
$endgroup$
– zoidberg
Dec 6 '18 at 20:28
$begingroup$
Please don't downvote posts simply because you don't understand.
$endgroup$
– zoidberg
Dec 6 '18 at 20:47
add a comment |
$begingroup$
The answer is unconditionally $m_1...m_n$. You did not place any restrictions between the entries of the $n$-tuple.
answered Dec 5 '18 at 4:27
zoidbergzoidberg
1,065113
$endgroup$
$begingroup$
I don't understand what you mean by restrictions.
$endgroup$
– user3865391
Dec 6 '18 at 20:11
$begingroup$
You say you have an argument when $m_i$ are pairwise coprime. Is there anything that goes wrong with the proof when for instance $n=2$ and $m_1=2,m_2=4$?
$endgroup$
– zoidberg
Dec 6 '18 at 20:28
$begingroup$
Please don't downvote posts simply because you don't understand.
$endgroup$
– zoidberg
Dec 6 '18 at 20:47
add a comment |
$begingroup$
The answer is unconditionally $m_1...m_n$. You did not place any restrictions between the entries of the $n$-tuple.
answered Dec 5 '18 at 4:27
zoidbergzoidberg
1,065113
$endgroup$
The answer is unconditionally $m_1...m_n$. You did not place any restrictions between the entries of the $n$-tuple.
answered Dec 5 '18 at 4:27
zoidbergzoidberg
1,065113
answered Dec 5 '18 at 4:27
zoidbergzoidberg
1,065113
answered Dec 5 '18 at 4:27
zoidbergzoidberg
1,065113
answered Dec 5 '18 at 4:27
zoidbergzoidberg
1,065113
1,065113
$begingroup$
I don't understand what you mean by restrictions.
$endgroup$
– user3865391
Dec 6 '18 at 20:11
$begingroup$
You say you have an argument when $m_i$ are pairwise coprime. Is there anything that goes wrong with the proof when for instance $n=2$ and $m_1=2,m_2=4$?
$endgroup$
– zoidberg
Dec 6 '18 at 20:28
$begingroup$
Please don't downvote posts simply because you don't understand.
$endgroup$
– zoidberg
Dec 6 '18 at 20:47
add a comment |
$begingroup$
I don't understand what you mean by restrictions.
$endgroup$
– user3865391
Dec 6 '18 at 20:11
$begingroup$
You say you have an argument when $m_i$ are pairwise coprime. Is there anything that goes wrong with the proof when for instance $n=2$ and $m_1=2,m_2=4$?
$endgroup$
– zoidberg
Dec 6 '18 at 20:28
$begingroup$
Please don't downvote posts simply because you don't understand.
$endgroup$
– zoidberg
Dec 6 '18 at 20:47
$begingroup$
I don't understand what you mean by restrictions.
$endgroup$
– user3865391
Dec 6 '18 at 20:11
$begingroup$
I don't understand what you mean by restrictions.
$endgroup$
– user3865391
Dec 6 '18 at 20:11
$begingroup$
You say you have an argument when $m_i$ are pairwise coprime. Is there anything that goes wrong with the proof when for instance $n=2$ and $m_1=2,m_2=4$?
$endgroup$
– zoidberg
Dec 6 '18 at 20:28
$begingroup$
You say you have an argument when $m_i$ are pairwise coprime. Is there anything that goes wrong with the proof when for instance $n=2$ and $m_1=2,m_2=4$?
$endgroup$
– zoidberg
Dec 6 '18 at 20:28
$begingroup$
Please don't downvote posts simply because you don't understand.
$endgroup$
– zoidberg
Dec 6 '18 at 20:47
$begingroup$
Please don't downvote posts simply because you don't understand.
$endgroup$
– zoidberg
Dec 6 '18 at 20:47
add a comment |
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