Applying set operations on congruent mod relation
$begingroup$
I've been asked a question to solve about congruent modulo. But the question is very different than another congruent modulo questions I have seen so far. It wants me to apply set operations on it. Can you give me a hint how to solve this question, not the answer. Thanks in advance...
Question:
https://ibb.co/SKQYCys
discrete-mathematics modular-arithmetic congruence-relations
asked Dec 11 '18 at 10:18
Ozan YurtseverOzan Yurtsever
1031
$endgroup$
add a comment |
$begingroup$
I've been asked a question to solve about congruent modulo. But the question is very different than another congruent modulo questions I have seen so far. It wants me to apply set operations on it. Can you give me a hint how to solve this question, not the answer. Thanks in advance...
Question:
https://ibb.co/SKQYCys
discrete-mathematics modular-arithmetic congruence-relations
asked Dec 11 '18 at 10:18
Ozan YurtseverOzan Yurtsever
1031
$endgroup$
add a comment |
$begingroup$
I've been asked a question to solve about congruent modulo. But the question is very different than another congruent modulo questions I have seen so far. It wants me to apply set operations on it. Can you give me a hint how to solve this question, not the answer. Thanks in advance...
Question:
https://ibb.co/SKQYCys
discrete-mathematics modular-arithmetic congruence-relations
asked Dec 11 '18 at 10:18
Ozan YurtseverOzan Yurtsever
1031
$endgroup$
I've been asked a question to solve about congruent modulo. But the question is very different than another congruent modulo questions I have seen so far. It wants me to apply set operations on it. Can you give me a hint how to solve this question, not the answer. Thanks in advance...
Question:
https://ibb.co/SKQYCys
discrete-mathematics modular-arithmetic congruence-relations
discrete-mathematics modular-arithmetic congruence-relations
asked Dec 11 '18 at 10:18
Ozan YurtseverOzan Yurtsever
1031
asked Dec 11 '18 at 10:18
Ozan YurtseverOzan Yurtsever
1031
asked Dec 11 '18 at 10:18
Ozan YurtseverOzan Yurtsever
1031
asked Dec 11 '18 at 10:18
Ozan YurtseverOzan Yurtsever
1031
asked Dec 11 '18 at 10:18
Ozan YurtseverOzan Yurtsever
1031
1031
add a comment |
add a comment |
1 Answer
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$begingroup$
Well, there's not much you can say to describe the union and the set-theoretic difference, other than to refer to them as such. I must confess I don't understand what the symbol denotes in d. (perhaps a 'direct product' of relations on the direct product' of supporting sets).
The interesting things start happening at b. It is a fact that an arbitrary (non-empty) intersection of equivalences remains an equivalence. Further more, the exercise you mention deals not merely with abstract equivalences, but with special ones that exhibit additional compatibility with the algebraic operations usually defined on the integers (such equivalences are generally called congruences).
It is worth stating the following result. For arbitrary $n in mathbb{N}$ let us denote the congruence modulo $n$ by $R_n$; it then holds that:
$R_m cap R_n=R_{[m,n]}$
where $[m,n]$ denotes the least common multiple of $m$ and $n$.
answered Dec 11 '18 at 11:58
ΑΘΩΑΘΩ
2463
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add a comment |
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1 Answer
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$begingroup$
Well, there's not much you can say to describe the union and the set-theoretic difference, other than to refer to them as such. I must confess I don't understand what the symbol denotes in d. (perhaps a 'direct product' of relations on the direct product' of supporting sets).
The interesting things start happening at b. It is a fact that an arbitrary (non-empty) intersection of equivalences remains an equivalence. Further more, the exercise you mention deals not merely with abstract equivalences, but with special ones that exhibit additional compatibility with the algebraic operations usually defined on the integers (such equivalences are generally called congruences).
It is worth stating the following result. For arbitrary $n in mathbb{N}$ let us denote the congruence modulo $n$ by $R_n$; it then holds that:
$R_m cap R_n=R_{[m,n]}$
where $[m,n]$ denotes the least common multiple of $m$ and $n$.
answered Dec 11 '18 at 11:58
ΑΘΩΑΘΩ
2463
$endgroup$
add a comment |
$begingroup$
Well, there's not much you can say to describe the union and the set-theoretic difference, other than to refer to them as such. I must confess I don't understand what the symbol denotes in d. (perhaps a 'direct product' of relations on the direct product' of supporting sets).
The interesting things start happening at b. It is a fact that an arbitrary (non-empty) intersection of equivalences remains an equivalence. Further more, the exercise you mention deals not merely with abstract equivalences, but with special ones that exhibit additional compatibility with the algebraic operations usually defined on the integers (such equivalences are generally called congruences).
It is worth stating the following result. For arbitrary $n in mathbb{N}$ let us denote the congruence modulo $n$ by $R_n$; it then holds that:
$R_m cap R_n=R_{[m,n]}$
where $[m,n]$ denotes the least common multiple of $m$ and $n$.
answered Dec 11 '18 at 11:58
ΑΘΩΑΘΩ
2463
$endgroup$
add a comment |
$begingroup$
Well, there's not much you can say to describe the union and the set-theoretic difference, other than to refer to them as such. I must confess I don't understand what the symbol denotes in d. (perhaps a 'direct product' of relations on the direct product' of supporting sets).
The interesting things start happening at b. It is a fact that an arbitrary (non-empty) intersection of equivalences remains an equivalence. Further more, the exercise you mention deals not merely with abstract equivalences, but with special ones that exhibit additional compatibility with the algebraic operations usually defined on the integers (such equivalences are generally called congruences).
It is worth stating the following result. For arbitrary $n in mathbb{N}$ let us denote the congruence modulo $n$ by $R_n$; it then holds that:
$R_m cap R_n=R_{[m,n]}$
where $[m,n]$ denotes the least common multiple of $m$ and $n$.
answered Dec 11 '18 at 11:58
ΑΘΩΑΘΩ
2463
$endgroup$
Well, there's not much you can say to describe the union and the set-theoretic difference, other than to refer to them as such. I must confess I don't understand what the symbol denotes in d. (perhaps a 'direct product' of relations on the direct product' of supporting sets).
The interesting things start happening at b. It is a fact that an arbitrary (non-empty) intersection of equivalences remains an equivalence. Further more, the exercise you mention deals not merely with abstract equivalences, but with special ones that exhibit additional compatibility with the algebraic operations usually defined on the integers (such equivalences are generally called congruences).
It is worth stating the following result. For arbitrary $n in mathbb{N}$ let us denote the congruence modulo $n$ by $R_n$; it then holds that:
$R_m cap R_n=R_{[m,n]}$
where $[m,n]$ denotes the least common multiple of $m$ and $n$.
answered Dec 11 '18 at 11:58
ΑΘΩΑΘΩ
2463
answered Dec 11 '18 at 11:58
ΑΘΩΑΘΩ
2463
answered Dec 11 '18 at 11:58
ΑΘΩΑΘΩ
2463
answered Dec 11 '18 at 11:58
ΑΘΩΑΘΩ
2463
2463
add a comment |
add a comment |
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