Mixing
Find all equivalence classes?
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a) Let S be : S={3,4,5,6,7,8} and the relation ~ defined as m~n if m^2 ≡ n^2 (mod 5).
b) Let S be : S={1,2,3,4,5,6,7,8} and the relation ~ defined as m~n if m ≡ n (mod 4).
Can someone help me with this ?
discrete-mathematics equivalence-relations
asked Feb 2 at 15:15
LaylaLayla
244
$endgroup$
add a comment |
$begingroup$
a) Let S be : S={3,4,5,6,7,8} and the relation ~ defined as m~n if m^2 ≡ n^2 (mod 5).
b) Let S be : S={1,2,3,4,5,6,7,8} and the relation ~ defined as m~n if m ≡ n (mod 4).
Can someone help me with this ?
discrete-mathematics equivalence-relations
asked Feb 2 at 15:15
LaylaLayla
244
$endgroup$
$begingroup$
It shouldn't be to complicate to find them... what is unclear for you ?
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– idm
Feb 2 at 15:17
$begingroup$
@idm How to solve them, by rules.
$endgroup$
– Layla
Feb 2 at 15:19
$begingroup$
The set is quite small, perhaps the question just wants you to find them manually?
$endgroup$
– Displayname
Feb 2 at 15:21
add a comment |
$begingroup$
a) Let S be : S={3,4,5,6,7,8} and the relation ~ defined as m~n if m^2 ≡ n^2 (mod 5).
b) Let S be : S={1,2,3,4,5,6,7,8} and the relation ~ defined as m~n if m ≡ n (mod 4).
Can someone help me with this ?
discrete-mathematics equivalence-relations
asked Feb 2 at 15:15
LaylaLayla
244
$endgroup$
a) Let S be : S={3,4,5,6,7,8} and the relation ~ defined as m~n if m^2 ≡ n^2 (mod 5).
b) Let S be : S={1,2,3,4,5,6,7,8} and the relation ~ defined as m~n if m ≡ n (mod 4).
Can someone help me with this ?
discrete-mathematics equivalence-relations
discrete-mathematics equivalence-relations
asked Feb 2 at 15:15
LaylaLayla
244
asked Feb 2 at 15:15
LaylaLayla
244
edited Feb 2 at 17:30
Layla
asked Feb 2 at 15:15
LaylaLayla
244
asked Feb 2 at 15:15
LaylaLayla
244
asked Feb 2 at 15:15
LaylaLayla
244
244
$begingroup$
It shouldn't be to complicate to find them... what is unclear for you ?
$endgroup$
– idm
Feb 2 at 15:17
$begingroup$
@idm How to solve them, by rules.
$endgroup$
– Layla
Feb 2 at 15:19
$begingroup$
The set is quite small, perhaps the question just wants you to find them manually?
$endgroup$
– Displayname
Feb 2 at 15:21
add a comment |
$begingroup$
It shouldn't be to complicate to find them... what is unclear for you ?
$endgroup$
– idm
Feb 2 at 15:17
$begingroup$
@idm How to solve them, by rules.
$endgroup$
– Layla
Feb 2 at 15:19
$begingroup$
The set is quite small, perhaps the question just wants you to find them manually?
$endgroup$
– Displayname
Feb 2 at 15:21
$begingroup$
It shouldn't be to complicate to find them... what is unclear for you ?
$endgroup$
– idm
Feb 2 at 15:17
$begingroup$
It shouldn't be to complicate to find them... what is unclear for you ?
$endgroup$
– idm
Feb 2 at 15:17
$begingroup$
@idm How to solve them, by rules.
$endgroup$
– Layla
Feb 2 at 15:19
$begingroup$
@idm How to solve them, by rules.
$endgroup$
– Layla
Feb 2 at 15:19
$begingroup$
The set is quite small, perhaps the question just wants you to find them manually?
$endgroup$
– Displayname
Feb 2 at 15:21
$begingroup$
The set is quite small, perhaps the question just wants you to find them manually?
$endgroup$
– Displayname
Feb 2 at 15:21
add a comment |
1 Answer
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That looks pretty straight forward- especially since the set is small enough to just calculate everything. The given base set is {3, 4, 5, 6, 7, 8}. Since the equivalence relation involves squares modulo 7, I would first calculate those: 3^2= 9= 4 (mod 5), 4^2= 16= 1 (mod 5), 5^2= 25= 0 (mod 5), 6^2= 36= 1 (mod 5), 7^2= 49= 4 (mod 5), 8^2= 64= 4 (mod 5). 3, 7, and 8 all have squares equal to 4 modulo 5 so are "equivalent". One equivalence class is {3, 7, 8}. 4 and 6 have squares equal to 1 modulo 5 so are "equivalent" so another equivalence class is {4, 6}. 5 is alone in its equivalence class- the third equivalence class is the singleton set {5}.
For the second problem you have "m mod n (mod 4)". I have no idea what that means! Do you intend "m= n (mod 4)"? If so, then 3= 3 (mod 4), 4= 0 (mod 4), 5= 1 (mod 4), 6= 2 (mod 4), 7= 3 (mod 4), and 8= 0 (mod 4). 3 and 7 are both equal to 3 (mod 4) so one equivalence class is {3, 7}. 4 and 8 are both equal to 0 (mod 4) so another equivalence class is {4, 8}. 5 and 6 form equivalence classes of their, {5}, and {6}.
answered Feb 2 at 15:51
user247327user247327
11.6k1516
$endgroup$
$begingroup$
Thank you so much.. :) m~n is m^2 ≡ n^2 (mod 5) for a) and m~n is m ≡ n (mod 4) for b)
$endgroup$
– Layla
Feb 2 at 17:08
add a comment |
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$begingroup$
That looks pretty straight forward- especially since the set is small enough to just calculate everything. The given base set is {3, 4, 5, 6, 7, 8}. Since the equivalence relation involves squares modulo 7, I would first calculate those: 3^2= 9= 4 (mod 5), 4^2= 16= 1 (mod 5), 5^2= 25= 0 (mod 5), 6^2= 36= 1 (mod 5), 7^2= 49= 4 (mod 5), 8^2= 64= 4 (mod 5). 3, 7, and 8 all have squares equal to 4 modulo 5 so are "equivalent". One equivalence class is {3, 7, 8}. 4 and 6 have squares equal to 1 modulo 5 so are "equivalent" so another equivalence class is {4, 6}. 5 is alone in its equivalence class- the third equivalence class is the singleton set {5}.
For the second problem you have "m mod n (mod 4)". I have no idea what that means! Do you intend "m= n (mod 4)"? If so, then 3= 3 (mod 4), 4= 0 (mod 4), 5= 1 (mod 4), 6= 2 (mod 4), 7= 3 (mod 4), and 8= 0 (mod 4). 3 and 7 are both equal to 3 (mod 4) so one equivalence class is {3, 7}. 4 and 8 are both equal to 0 (mod 4) so another equivalence class is {4, 8}. 5 and 6 form equivalence classes of their, {5}, and {6}.
answered Feb 2 at 15:51
user247327user247327
11.6k1516
$endgroup$
$begingroup$
Thank you so much.. :) m~n is m^2 ≡ n^2 (mod 5) for a) and m~n is m ≡ n (mod 4) for b)
$endgroup$
– Layla
Feb 2 at 17:08
add a comment |
$begingroup$
That looks pretty straight forward- especially since the set is small enough to just calculate everything. The given base set is {3, 4, 5, 6, 7, 8}. Since the equivalence relation involves squares modulo 7, I would first calculate those: 3^2= 9= 4 (mod 5), 4^2= 16= 1 (mod 5), 5^2= 25= 0 (mod 5), 6^2= 36= 1 (mod 5), 7^2= 49= 4 (mod 5), 8^2= 64= 4 (mod 5). 3, 7, and 8 all have squares equal to 4 modulo 5 so are "equivalent". One equivalence class is {3, 7, 8}. 4 and 6 have squares equal to 1 modulo 5 so are "equivalent" so another equivalence class is {4, 6}. 5 is alone in its equivalence class- the third equivalence class is the singleton set {5}.
For the second problem you have "m mod n (mod 4)". I have no idea what that means! Do you intend "m= n (mod 4)"? If so, then 3= 3 (mod 4), 4= 0 (mod 4), 5= 1 (mod 4), 6= 2 (mod 4), 7= 3 (mod 4), and 8= 0 (mod 4). 3 and 7 are both equal to 3 (mod 4) so one equivalence class is {3, 7}. 4 and 8 are both equal to 0 (mod 4) so another equivalence class is {4, 8}. 5 and 6 form equivalence classes of their, {5}, and {6}.
answered Feb 2 at 15:51
user247327user247327
11.6k1516
$endgroup$
$begingroup$
Thank you so much.. :) m~n is m^2 ≡ n^2 (mod 5) for a) and m~n is m ≡ n (mod 4) for b)
$endgroup$
– Layla
Feb 2 at 17:08
add a comment |
$begingroup$
That looks pretty straight forward- especially since the set is small enough to just calculate everything. The given base set is {3, 4, 5, 6, 7, 8}. Since the equivalence relation involves squares modulo 7, I would first calculate those: 3^2= 9= 4 (mod 5), 4^2= 16= 1 (mod 5), 5^2= 25= 0 (mod 5), 6^2= 36= 1 (mod 5), 7^2= 49= 4 (mod 5), 8^2= 64= 4 (mod 5). 3, 7, and 8 all have squares equal to 4 modulo 5 so are "equivalent". One equivalence class is {3, 7, 8}. 4 and 6 have squares equal to 1 modulo 5 so are "equivalent" so another equivalence class is {4, 6}. 5 is alone in its equivalence class- the third equivalence class is the singleton set {5}.
For the second problem you have "m mod n (mod 4)". I have no idea what that means! Do you intend "m= n (mod 4)"? If so, then 3= 3 (mod 4), 4= 0 (mod 4), 5= 1 (mod 4), 6= 2 (mod 4), 7= 3 (mod 4), and 8= 0 (mod 4). 3 and 7 are both equal to 3 (mod 4) so one equivalence class is {3, 7}. 4 and 8 are both equal to 0 (mod 4) so another equivalence class is {4, 8}. 5 and 6 form equivalence classes of their, {5}, and {6}.
answered Feb 2 at 15:51
user247327user247327
11.6k1516
$endgroup$
That looks pretty straight forward- especially since the set is small enough to just calculate everything. The given base set is {3, 4, 5, 6, 7, 8}. Since the equivalence relation involves squares modulo 7, I would first calculate those: 3^2= 9= 4 (mod 5), 4^2= 16= 1 (mod 5), 5^2= 25= 0 (mod 5), 6^2= 36= 1 (mod 5), 7^2= 49= 4 (mod 5), 8^2= 64= 4 (mod 5). 3, 7, and 8 all have squares equal to 4 modulo 5 so are "equivalent". One equivalence class is {3, 7, 8}. 4 and 6 have squares equal to 1 modulo 5 so are "equivalent" so another equivalence class is {4, 6}. 5 is alone in its equivalence class- the third equivalence class is the singleton set {5}.
For the second problem you have "m mod n (mod 4)". I have no idea what that means! Do you intend "m= n (mod 4)"? If so, then 3= 3 (mod 4), 4= 0 (mod 4), 5= 1 (mod 4), 6= 2 (mod 4), 7= 3 (mod 4), and 8= 0 (mod 4). 3 and 7 are both equal to 3 (mod 4) so one equivalence class is {3, 7}. 4 and 8 are both equal to 0 (mod 4) so another equivalence class is {4, 8}. 5 and 6 form equivalence classes of their, {5}, and {6}.
answered Feb 2 at 15:51
user247327user247327
11.6k1516
answered Feb 2 at 15:51
user247327user247327
11.6k1516
answered Feb 2 at 15:51
user247327user247327
11.6k1516
answered Feb 2 at 15:51
user247327user247327
11.6k1516
11.6k1516
$begingroup$
Thank you so much.. :) m~n is m^2 ≡ n^2 (mod 5) for a) and m~n is m ≡ n (mod 4) for b)
$endgroup$
– Layla
Feb 2 at 17:08
add a comment |
$begingroup$
Thank you so much.. :) m~n is m^2 ≡ n^2 (mod 5) for a) and m~n is m ≡ n (mod 4) for b)
$endgroup$
– Layla
Feb 2 at 17:08
$begingroup$
Thank you so much.. :) m~n is m^2 ≡ n^2 (mod 5) for a) and m~n is m ≡ n (mod 4) for b)
$endgroup$
– Layla
Feb 2 at 17:08
$begingroup$
Thank you so much.. :) m~n is m^2 ≡ n^2 (mod 5) for a) and m~n is m ≡ n (mod 4) for b)
$endgroup$
– Layla
Feb 2 at 17:08
add a comment |
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$begingroup$
It shouldn't be to complicate to find them... what is unclear for you ?
$endgroup$
– idm
Feb 2 at 15:17
$begingroup$
@idm How to solve them, by rules.
$endgroup$
– Layla
Feb 2 at 15:19
$begingroup$
The set is quite small, perhaps the question just wants you to find them manually?
$endgroup$
– Displayname
Feb 2 at 15:21