If $amid b$, then $forall x(bin mathbb z) exists y(a in mathbb Z/amathbb Z([x]bsubseteq [y]a.$
$begingroup$
I've got a question on how to prove the following statement:
If $amid b$, then $forall big([x]_bin mathbb Z, exists ([y]_a in mathbb Z,([x]_bsubseteq [y]_a)big).$
If additionally, $aneq b$, then "$subseteq$" in the statement above can be substituted with "$subsetneq$".
So for example let $a = 3,; b = 6, ;x = 1,; y = 1$.
- $3mid 6$
$[1]_6 = [7,13,19,25,31,37, ...]$ and $[1]_3 = [4,7,10,13,16,19, ...]$
- So $[1]_6 subseteq [1]_3$ seems to be true but how would one prove that?
It makes totally sense but unfortunately I don't know how to prove it. Any tips/ideas?
proof-writing divisibility
asked Jan 16 at 21:52
GreenGreen
62
$endgroup$
add a comment |
$begingroup$
I've got a question on how to prove the following statement:
If $amid b$, then $forall big([x]_bin mathbb Z, exists ([y]_a in mathbb Z,([x]_bsubseteq [y]_a)big).$
If additionally, $aneq b$, then "$subseteq$" in the statement above can be substituted with "$subsetneq$".
So for example let $a = 3,; b = 6, ;x = 1,; y = 1$.
- $3mid 6$
$[1]_6 = [7,13,19,25,31,37, ...]$ and $[1]_3 = [4,7,10,13,16,19, ...]$
- So $[1]_6 subseteq [1]_3$ seems to be true but how would one prove that?
It makes totally sense but unfortunately I don't know how to prove it. Any tips/ideas?
proof-writing divisibility
asked Jan 16 at 21:52
GreenGreen
62
$endgroup$
$begingroup$
Did you already search here for answers at this site? "To contain is to divide", i.e, $amid b$ means $(a)supseteq (b)$.
$endgroup$
– Dietrich Burde
Jan 16 at 21:55
$begingroup$
Yes I tried but couldn't find a similiar statement like this.
$endgroup$
– Green
Jan 16 at 22:14
$begingroup$
See this question and use Sergio's proof.
$endgroup$
– Dietrich Burde
Jan 16 at 22:19
$begingroup$
Let $,b = na.,$ Add $,x,$ to $, bBbb Z = naBbb Z, subseteq, aBbb Z $
$endgroup$
– Bill Dubuque
Jan 17 at 0:58
add a comment |
$begingroup$
I've got a question on how to prove the following statement:
If $amid b$, then $forall big([x]_bin mathbb Z, exists ([y]_a in mathbb Z,([x]_bsubseteq [y]_a)big).$
If additionally, $aneq b$, then "$subseteq$" in the statement above can be substituted with "$subsetneq$".
So for example let $a = 3,; b = 6, ;x = 1,; y = 1$.
- $3mid 6$
$[1]_6 = [7,13,19,25,31,37, ...]$ and $[1]_3 = [4,7,10,13,16,19, ...]$
- So $[1]_6 subseteq [1]_3$ seems to be true but how would one prove that?
It makes totally sense but unfortunately I don't know how to prove it. Any tips/ideas?
proof-writing divisibility
asked Jan 16 at 21:52
GreenGreen
62
$endgroup$
I've got a question on how to prove the following statement:
If $amid b$, then $forall big([x]_bin mathbb Z, exists ([y]_a in mathbb Z,([x]_bsubseteq [y]_a)big).$
If additionally, $aneq b$, then "$subseteq$" in the statement above can be substituted with "$subsetneq$".
So for example let $a = 3,; b = 6, ;x = 1,; y = 1$.
- $3mid 6$
$[1]_6 = [7,13,19,25,31,37, ...]$ and $[1]_3 = [4,7,10,13,16,19, ...]$
- So $[1]_6 subseteq [1]_3$ seems to be true but how would one prove that?
It makes totally sense but unfortunately I don't know how to prove it. Any tips/ideas?
proof-writing divisibility
proof-writing divisibility
asked Jan 16 at 21:52
GreenGreen
62
asked Jan 16 at 21:52
GreenGreen
62
edited Jan 17 at 11:04
Green
asked Jan 16 at 21:52
GreenGreen
62
asked Jan 16 at 21:52
GreenGreen
62
asked Jan 16 at 21:52
GreenGreen
62
62
$begingroup$
Did you already search here for answers at this site? "To contain is to divide", i.e, $amid b$ means $(a)supseteq (b)$.
$endgroup$
– Dietrich Burde
Jan 16 at 21:55
$begingroup$
Yes I tried but couldn't find a similiar statement like this.
$endgroup$
– Green
Jan 16 at 22:14
$begingroup$
See this question and use Sergio's proof.
$endgroup$
– Dietrich Burde
Jan 16 at 22:19
$begingroup$
Let $,b = na.,$ Add $,x,$ to $, bBbb Z = naBbb Z, subseteq, aBbb Z $
$endgroup$
– Bill Dubuque
Jan 17 at 0:58
add a comment |
$begingroup$
Did you already search here for answers at this site? "To contain is to divide", i.e, $amid b$ means $(a)supseteq (b)$.
$endgroup$
– Dietrich Burde
Jan 16 at 21:55
$begingroup$
Yes I tried but couldn't find a similiar statement like this.
$endgroup$
– Green
Jan 16 at 22:14
$begingroup$
See this question and use Sergio's proof.
$endgroup$
– Dietrich Burde
Jan 16 at 22:19
$begingroup$
Let $,b = na.,$ Add $,x,$ to $, bBbb Z = naBbb Z, subseteq, aBbb Z $
$endgroup$
– Bill Dubuque
Jan 17 at 0:58
$begingroup$
Did you already search here for answers at this site? "To contain is to divide", i.e, $amid b$ means $(a)supseteq (b)$.
$endgroup$
– Dietrich Burde
Jan 16 at 21:55
$begingroup$
Did you already search here for answers at this site? "To contain is to divide", i.e, $amid b$ means $(a)supseteq (b)$.
$endgroup$
– Dietrich Burde
Jan 16 at 21:55
$begingroup$
Yes I tried but couldn't find a similiar statement like this.
$endgroup$
– Green
Jan 16 at 22:14
$begingroup$
Yes I tried but couldn't find a similiar statement like this.
$endgroup$
– Green
Jan 16 at 22:14
$begingroup$
See this question and use Sergio's proof.
$endgroup$
– Dietrich Burde
Jan 16 at 22:19
$begingroup$
See this question and use Sergio's proof.
$endgroup$
– Dietrich Burde
Jan 16 at 22:19
$begingroup$
Let $,b = na.,$ Add $,x,$ to $, bBbb Z = naBbb Z, subseteq, aBbb Z $
$endgroup$
– Bill Dubuque
Jan 17 at 0:58
$begingroup$
Let $,b = na.,$ Add $,x,$ to $, bBbb Z = naBbb Z, subseteq, aBbb Z $
$endgroup$
– Bill Dubuque
Jan 17 at 0:58
add a comment |
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$begingroup$
Did you already search here for answers at this site? "To contain is to divide", i.e, $amid b$ means $(a)supseteq (b)$.
$endgroup$
– Dietrich Burde
Jan 16 at 21:55
$begingroup$
Yes I tried but couldn't find a similiar statement like this.
$endgroup$
– Green
Jan 16 at 22:14
$begingroup$
See this question and use Sergio's proof.
$endgroup$
– Dietrich Burde
Jan 16 at 22:19
$begingroup$
Let $,b = na.,$ Add $,x,$ to $, bBbb Z = naBbb Z, subseteq, aBbb Z $
$endgroup$
– Bill Dubuque
Jan 17 at 0:58