Characterization of the inclusion $a + mathbb N cdot b subseteq bigcup_{i=1}^n c_i + mathbb N cdot d_i$
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Let $a,b,c,d in mathbb N$, where $mathbb N = {0,1,2,ldots}$. Then we have
$$
a + mathbb N cdot b subseteq c + mathbb N cdot d
$$
iff $a - c$ and $b$ are dividable by $d$.
Proof. Suppose we have $a + mathbb N cdot b subseteq c + mathbb N cdot d$, then $a - c in mathbb Ncdot d$, hence $d$ divides $a-c$. Then choose some $n$ not dividable by $d$ and the equality $a - c + ncdot b in mathbb Ncdot d$ implies that $b$ must be dividable by $d$. $square$
So now generalize this in the following manner, if we have $2n$ numbers $c_1, ldots, c_n$ and $d_1,ldots, d_n$. Exactly when do we have
$$
a + mathbb N cdot b subseteq bigcup_{i=1}^n c_i + mathbb N cdot d_i.
$$
and the selection of those numbers is minimal in the sense that if we take some $c_j, d_j$
away, then $a + mathbb Ncdot b$ is no longer in the union of the remaining ones, i.e. every number is mandatory. Is there any easy criterion like the one above in terms of these numbers?
abstract-algebra number-theory elementary-set-theory
asked Nov 20 at 15:57
StefanH
8,05552161
add a comment |
up vote
1
down vote
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Let $a,b,c,d in mathbb N$, where $mathbb N = {0,1,2,ldots}$. Then we have
$$
a + mathbb N cdot b subseteq c + mathbb N cdot d
$$
iff $a - c$ and $b$ are dividable by $d$.
Proof. Suppose we have $a + mathbb N cdot b subseteq c + mathbb N cdot d$, then $a - c in mathbb Ncdot d$, hence $d$ divides $a-c$. Then choose some $n$ not dividable by $d$ and the equality $a - c + ncdot b in mathbb Ncdot d$ implies that $b$ must be dividable by $d$. $square$
So now generalize this in the following manner, if we have $2n$ numbers $c_1, ldots, c_n$ and $d_1,ldots, d_n$. Exactly when do we have
$$
a + mathbb N cdot b subseteq bigcup_{i=1}^n c_i + mathbb N cdot d_i.
$$
and the selection of those numbers is minimal in the sense that if we take some $c_j, d_j$
away, then $a + mathbb Ncdot b$ is no longer in the union of the remaining ones, i.e. every number is mandatory. Is there any easy criterion like the one above in terms of these numbers?
abstract-algebra number-theory elementary-set-theory
asked Nov 20 at 15:57
StefanH
8,05552161
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $a,b,c,d in mathbb N$, where $mathbb N = {0,1,2,ldots}$. Then we have
$$
a + mathbb N cdot b subseteq c + mathbb N cdot d
$$
iff $a - c$ and $b$ are dividable by $d$.
Proof. Suppose we have $a + mathbb N cdot b subseteq c + mathbb N cdot d$, then $a - c in mathbb Ncdot d$, hence $d$ divides $a-c$. Then choose some $n$ not dividable by $d$ and the equality $a - c + ncdot b in mathbb Ncdot d$ implies that $b$ must be dividable by $d$. $square$
So now generalize this in the following manner, if we have $2n$ numbers $c_1, ldots, c_n$ and $d_1,ldots, d_n$. Exactly when do we have
$$
a + mathbb N cdot b subseteq bigcup_{i=1}^n c_i + mathbb N cdot d_i.
$$
and the selection of those numbers is minimal in the sense that if we take some $c_j, d_j$
away, then $a + mathbb Ncdot b$ is no longer in the union of the remaining ones, i.e. every number is mandatory. Is there any easy criterion like the one above in terms of these numbers?
abstract-algebra number-theory elementary-set-theory
asked Nov 20 at 15:57
StefanH
8,05552161
Let $a,b,c,d in mathbb N$, where $mathbb N = {0,1,2,ldots}$. Then we have
$$
a + mathbb N cdot b subseteq c + mathbb N cdot d
$$
iff $a - c$ and $b$ are dividable by $d$.
Proof. Suppose we have $a + mathbb N cdot b subseteq c + mathbb N cdot d$, then $a - c in mathbb Ncdot d$, hence $d$ divides $a-c$. Then choose some $n$ not dividable by $d$ and the equality $a - c + ncdot b in mathbb Ncdot d$ implies that $b$ must be dividable by $d$. $square$
So now generalize this in the following manner, if we have $2n$ numbers $c_1, ldots, c_n$ and $d_1,ldots, d_n$. Exactly when do we have
$$
a + mathbb N cdot b subseteq bigcup_{i=1}^n c_i + mathbb N cdot d_i.
$$
and the selection of those numbers is minimal in the sense that if we take some $c_j, d_j$
away, then $a + mathbb Ncdot b$ is no longer in the union of the remaining ones, i.e. every number is mandatory. Is there any easy criterion like the one above in terms of these numbers?
abstract-algebra number-theory elementary-set-theory
abstract-algebra number-theory elementary-set-theory
asked Nov 20 at 15:57
StefanH
8,05552161
asked Nov 20 at 15:57
StefanH
8,05552161
edited Nov 21 at 12:50
asked Nov 20 at 15:57
StefanH
8,05552161
asked Nov 20 at 15:57
StefanH
8,05552161
asked Nov 20 at 15:57
StefanH
8,05552161
8,05552161
add a comment |
add a comment |
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I don't understand what did you mean with "easy criterion" but when we have $2n$ numbers $c_1, dots, c_n$ and $d_1, dots, d_n$ and $$ a+ mathbb{N}cdot b subseteq cup_{i=1}^{n}c_i+ mathbb{N}cdot d_i$$ and then we add 1 more $(c_{n+1},d_{n+1})$ which doesn't "relate" to $a$ and $b$ but it's still $$ a+ mathbb{N}cdot b subseteq cup_{i=1}^{n+1}c_i+ mathbb{N}cdot d_i$$ It means if you add more $(c,d)$ in the problem then it makes the problem more "weak". I think it will be more accurate when $$ a+ mathbb{N}cdot b subseteq cap_{i=1}^{n}c_i+ mathbb{N}cdot d_i$$
answered Nov 21 at 10:02
Tung N. Dinh
111
No, I meant union. The elements of $a + mathbb N cdot b$ could be distributed among the sets from the union. Different wording, when is it the case that we can always find some $x = a + nb$ which is not in any of the $c_i + mathbb N cdot d_i$. Concerning your objection, yes they must be "minimal" in the sense that taking one away will destroy the inclusion relation. Thanks for that, I will update my question. Hope that makes more sense then.
– StefanH
Nov 21 at 12:48
add a comment |
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1 Answer
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up vote
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I don't understand what did you mean with "easy criterion" but when we have $2n$ numbers $c_1, dots, c_n$ and $d_1, dots, d_n$ and $$ a+ mathbb{N}cdot b subseteq cup_{i=1}^{n}c_i+ mathbb{N}cdot d_i$$ and then we add 1 more $(c_{n+1},d_{n+1})$ which doesn't "relate" to $a$ and $b$ but it's still $$ a+ mathbb{N}cdot b subseteq cup_{i=1}^{n+1}c_i+ mathbb{N}cdot d_i$$ It means if you add more $(c,d)$ in the problem then it makes the problem more "weak". I think it will be more accurate when $$ a+ mathbb{N}cdot b subseteq cap_{i=1}^{n}c_i+ mathbb{N}cdot d_i$$
answered Nov 21 at 10:02
Tung N. Dinh
111
No, I meant union. The elements of $a + mathbb N cdot b$ could be distributed among the sets from the union. Different wording, when is it the case that we can always find some $x = a + nb$ which is not in any of the $c_i + mathbb N cdot d_i$. Concerning your objection, yes they must be "minimal" in the sense that taking one away will destroy the inclusion relation. Thanks for that, I will update my question. Hope that makes more sense then.
– StefanH
Nov 21 at 12:48
add a comment |
up vote
1
down vote
I don't understand what did you mean with "easy criterion" but when we have $2n$ numbers $c_1, dots, c_n$ and $d_1, dots, d_n$ and $$ a+ mathbb{N}cdot b subseteq cup_{i=1}^{n}c_i+ mathbb{N}cdot d_i$$ and then we add 1 more $(c_{n+1},d_{n+1})$ which doesn't "relate" to $a$ and $b$ but it's still $$ a+ mathbb{N}cdot b subseteq cup_{i=1}^{n+1}c_i+ mathbb{N}cdot d_i$$ It means if you add more $(c,d)$ in the problem then it makes the problem more "weak". I think it will be more accurate when $$ a+ mathbb{N}cdot b subseteq cap_{i=1}^{n}c_i+ mathbb{N}cdot d_i$$
answered Nov 21 at 10:02
Tung N. Dinh
111
No, I meant union. The elements of $a + mathbb N cdot b$ could be distributed among the sets from the union. Different wording, when is it the case that we can always find some $x = a + nb$ which is not in any of the $c_i + mathbb N cdot d_i$. Concerning your objection, yes they must be "minimal" in the sense that taking one away will destroy the inclusion relation. Thanks for that, I will update my question. Hope that makes more sense then.
– StefanH
Nov 21 at 12:48
add a comment |
up vote
1
down vote
up vote
1
down vote
I don't understand what did you mean with "easy criterion" but when we have $2n$ numbers $c_1, dots, c_n$ and $d_1, dots, d_n$ and $$ a+ mathbb{N}cdot b subseteq cup_{i=1}^{n}c_i+ mathbb{N}cdot d_i$$ and then we add 1 more $(c_{n+1},d_{n+1})$ which doesn't "relate" to $a$ and $b$ but it's still $$ a+ mathbb{N}cdot b subseteq cup_{i=1}^{n+1}c_i+ mathbb{N}cdot d_i$$ It means if you add more $(c,d)$ in the problem then it makes the problem more "weak". I think it will be more accurate when $$ a+ mathbb{N}cdot b subseteq cap_{i=1}^{n}c_i+ mathbb{N}cdot d_i$$
answered Nov 21 at 10:02
Tung N. Dinh
111
I don't understand what did you mean with "easy criterion" but when we have $2n$ numbers $c_1, dots, c_n$ and $d_1, dots, d_n$ and $$ a+ mathbb{N}cdot b subseteq cup_{i=1}^{n}c_i+ mathbb{N}cdot d_i$$ and then we add 1 more $(c_{n+1},d_{n+1})$ which doesn't "relate" to $a$ and $b$ but it's still $$ a+ mathbb{N}cdot b subseteq cup_{i=1}^{n+1}c_i+ mathbb{N}cdot d_i$$ It means if you add more $(c,d)$ in the problem then it makes the problem more "weak". I think it will be more accurate when $$ a+ mathbb{N}cdot b subseteq cap_{i=1}^{n}c_i+ mathbb{N}cdot d_i$$
answered Nov 21 at 10:02
Tung N. Dinh
111
answered Nov 21 at 10:02
Tung N. Dinh
111
answered Nov 21 at 10:02
Tung N. Dinh
111
answered Nov 21 at 10:02
Tung N. Dinh
111
111
No, I meant union. The elements of $a + mathbb N cdot b$ could be distributed among the sets from the union. Different wording, when is it the case that we can always find some $x = a + nb$ which is not in any of the $c_i + mathbb N cdot d_i$. Concerning your objection, yes they must be "minimal" in the sense that taking one away will destroy the inclusion relation. Thanks for that, I will update my question. Hope that makes more sense then.
– StefanH
Nov 21 at 12:48
add a comment |
No, I meant union. The elements of $a + mathbb N cdot b$ could be distributed among the sets from the union. Different wording, when is it the case that we can always find some $x = a + nb$ which is not in any of the $c_i + mathbb N cdot d_i$. Concerning your objection, yes they must be "minimal" in the sense that taking one away will destroy the inclusion relation. Thanks for that, I will update my question. Hope that makes more sense then.
– StefanH
Nov 21 at 12:48
No, I meant union. The elements of $a + mathbb N cdot b$ could be distributed among the sets from the union. Different wording, when is it the case that we can always find some $x = a + nb$ which is not in any of the $c_i + mathbb N cdot d_i$. Concerning your objection, yes they must be "minimal" in the sense that taking one away will destroy the inclusion relation. Thanks for that, I will update my question. Hope that makes more sense then.
– StefanH
Nov 21 at 12:48
No, I meant union. The elements of $a + mathbb N cdot b$ could be distributed among the sets from the union. Different wording, when is it the case that we can always find some $x = a + nb$ which is not in any of the $c_i + mathbb N cdot d_i$. Concerning your objection, yes they must be "minimal" in the sense that taking one away will destroy the inclusion relation. Thanks for that, I will update my question. Hope that makes more sense then.
– StefanH
Nov 21 at 12:48
add a comment |
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