Prove, that $M={(a,b)mid a,b in mathbb{N_0} land (a-b) text{ mod } 4 = 0} $ is a equivalence relation
Prove, that $M={(a,b)mid a,b in mathbb{N_0} land (a-b) text{ mod } 4 = 0} $ is an equivalence-relation.
Refl.: $a-a=0 text{ mod } 4 =0$
Sym.: $forall x,y in M: (x,y) implies (y,x)$ (Not sure about this.)
Trans.: $left((a-b)+(b-c)right)text{ mod } 4 = 0 iff (a-c)text{ mod } 4 =0$
Can you help me with proving, that the relation is symmetric and is everything else correct?
discrete-mathematics relations equivalence-relations symmetry
asked Nov 27 '18 at 15:30
Doesbaddel
19910
add a comment |
Prove, that $M={(a,b)mid a,b in mathbb{N_0} land (a-b) text{ mod } 4 = 0} $ is an equivalence-relation.
Refl.: $a-a=0 text{ mod } 4 =0$
Sym.: $forall x,y in M: (x,y) implies (y,x)$ (Not sure about this.)
Trans.: $left((a-b)+(b-c)right)text{ mod } 4 = 0 iff (a-c)text{ mod } 4 =0$
Can you help me with proving, that the relation is symmetric and is everything else correct?
discrete-mathematics relations equivalence-relations symmetry
asked Nov 27 '18 at 15:30
Doesbaddel
19910
For the symmetry, you just need to realize that $-0 = 0$. Transivity is obvious since $(a-b)+(b-c) = a-c$ (so mod should be the same). It might help to realize that $a$ and $b$ are related iff $a$ and $b$ have the same remainder when divided by $4$ which makes three properties pretty obvious
– user160738
Nov 27 '18 at 15:36
Yeah, I've already finished r and t, just needed some help for s. Thanks, that helped!
– Doesbaddel
Nov 27 '18 at 18:47
Can I write: $(a-b) text{ mod } 4=0 iff {mid a-b mid} text{ mod } 4=0$. Therefore, $(b-a) text{ mod } 4=0$ must be ${mid b-a mid} text{ mod } 4=0$ and because ${mid a-b mid} = {mid b-a mid} text{ mod } 4 =0 $ the relation is symmetric?
– Doesbaddel
Nov 27 '18 at 19:03
add a comment |
Prove, that $M={(a,b)mid a,b in mathbb{N_0} land (a-b) text{ mod } 4 = 0} $ is an equivalence-relation.
Refl.: $a-a=0 text{ mod } 4 =0$
Sym.: $forall x,y in M: (x,y) implies (y,x)$ (Not sure about this.)
Trans.: $left((a-b)+(b-c)right)text{ mod } 4 = 0 iff (a-c)text{ mod } 4 =0$
Can you help me with proving, that the relation is symmetric and is everything else correct?
discrete-mathematics relations equivalence-relations symmetry
asked Nov 27 '18 at 15:30
Doesbaddel
19910
Prove, that $M={(a,b)mid a,b in mathbb{N_0} land (a-b) text{ mod } 4 = 0} $ is an equivalence-relation.
Refl.: $a-a=0 text{ mod } 4 =0$
Sym.: $forall x,y in M: (x,y) implies (y,x)$ (Not sure about this.)
Trans.: $left((a-b)+(b-c)right)text{ mod } 4 = 0 iff (a-c)text{ mod } 4 =0$
Can you help me with proving, that the relation is symmetric and is everything else correct?
discrete-mathematics relations equivalence-relations symmetry
discrete-mathematics relations equivalence-relations symmetry
asked Nov 27 '18 at 15:30
Doesbaddel
19910
asked Nov 27 '18 at 15:30
Doesbaddel
19910
asked Nov 27 '18 at 15:30
Doesbaddel
19910
asked Nov 27 '18 at 15:30
Doesbaddel
19910
asked Nov 27 '18 at 15:30
Doesbaddel
19910
19910
For the symmetry, you just need to realize that $-0 = 0$. Transivity is obvious since $(a-b)+(b-c) = a-c$ (so mod should be the same). It might help to realize that $a$ and $b$ are related iff $a$ and $b$ have the same remainder when divided by $4$ which makes three properties pretty obvious
– user160738
Nov 27 '18 at 15:36
Yeah, I've already finished r and t, just needed some help for s. Thanks, that helped!
– Doesbaddel
Nov 27 '18 at 18:47
Can I write: $(a-b) text{ mod } 4=0 iff {mid a-b mid} text{ mod } 4=0$. Therefore, $(b-a) text{ mod } 4=0$ must be ${mid b-a mid} text{ mod } 4=0$ and because ${mid a-b mid} = {mid b-a mid} text{ mod } 4 =0 $ the relation is symmetric?
– Doesbaddel
Nov 27 '18 at 19:03
add a comment |
For the symmetry, you just need to realize that $-0 = 0$. Transivity is obvious since $(a-b)+(b-c) = a-c$ (so mod should be the same). It might help to realize that $a$ and $b$ are related iff $a$ and $b$ have the same remainder when divided by $4$ which makes three properties pretty obvious
– user160738
Nov 27 '18 at 15:36
Yeah, I've already finished r and t, just needed some help for s. Thanks, that helped!
– Doesbaddel
Nov 27 '18 at 18:47
Can I write: $(a-b) text{ mod } 4=0 iff {mid a-b mid} text{ mod } 4=0$. Therefore, $(b-a) text{ mod } 4=0$ must be ${mid b-a mid} text{ mod } 4=0$ and because ${mid a-b mid} = {mid b-a mid} text{ mod } 4 =0 $ the relation is symmetric?
– Doesbaddel
Nov 27 '18 at 19:03
For the symmetry, you just need to realize that $-0 = 0$. Transivity is obvious since $(a-b)+(b-c) = a-c$ (so mod should be the same). It might help to realize that $a$ and $b$ are related iff $a$ and $b$ have the same remainder when divided by $4$ which makes three properties pretty obvious
– user160738
Nov 27 '18 at 15:36
For the symmetry, you just need to realize that $-0 = 0$. Transivity is obvious since $(a-b)+(b-c) = a-c$ (so mod should be the same). It might help to realize that $a$ and $b$ are related iff $a$ and $b$ have the same remainder when divided by $4$ which makes three properties pretty obvious
– user160738
Nov 27 '18 at 15:36
Yeah, I've already finished r and t, just needed some help for s. Thanks, that helped!
– Doesbaddel
Nov 27 '18 at 18:47
Yeah, I've already finished r and t, just needed some help for s. Thanks, that helped!
– Doesbaddel
Nov 27 '18 at 18:47
Can I write: $(a-b) text{ mod } 4=0 iff {mid a-b mid} text{ mod } 4=0$. Therefore, $(b-a) text{ mod } 4=0$ must be ${mid b-a mid} text{ mod } 4=0$ and because ${mid a-b mid} = {mid b-a mid} text{ mod } 4 =0 $ the relation is symmetric?
– Doesbaddel
Nov 27 '18 at 19:03
Can I write: $(a-b) text{ mod } 4=0 iff {mid a-b mid} text{ mod } 4=0$. Therefore, $(b-a) text{ mod } 4=0$ must be ${mid b-a mid} text{ mod } 4=0$ and because ${mid a-b mid} = {mid b-a mid} text{ mod } 4 =0 $ the relation is symmetric?
– Doesbaddel
Nov 27 '18 at 19:03
add a comment |
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For the symmetry, you just need to realize that $-0 = 0$. Transivity is obvious since $(a-b)+(b-c) = a-c$ (so mod should be the same). It might help to realize that $a$ and $b$ are related iff $a$ and $b$ have the same remainder when divided by $4$ which makes three properties pretty obvious
– user160738
Nov 27 '18 at 15:36
Yeah, I've already finished r and t, just needed some help for s. Thanks, that helped!
– Doesbaddel
Nov 27 '18 at 18:47
Can I write: $(a-b) text{ mod } 4=0 iff {mid a-b mid} text{ mod } 4=0$. Therefore, $(b-a) text{ mod } 4=0$ must be ${mid b-a mid} text{ mod } 4=0$ and because ${mid a-b mid} = {mid b-a mid} text{ mod } 4 =0 $ the relation is symmetric?
– Doesbaddel
Nov 27 '18 at 19:03