Prove, that $M={(a,b)mid a,b in mathbb{N_0} land (a-b) text{ mod } 4 = 0} $ is a equivalence relation












0















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?










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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 at 15:36










  • Yeah, I've already finished r and t, just needed some help for s. Thanks, that helped!
    – Doesbaddel
    Nov 27 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 at 19:03


















0















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?










share|cite|improve this question






















  • 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 at 15:36










  • Yeah, I've already finished r and t, just needed some help for s. Thanks, that helped!
    – Doesbaddel
    Nov 27 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 at 19:03
















0












0








0








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?










share|cite|improve this question














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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 27 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 at 15:36










  • Yeah, I've already finished r and t, just needed some help for s. Thanks, that helped!
    – Doesbaddel
    Nov 27 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 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 at 15:36










  • Yeah, I've already finished r and t, just needed some help for s. Thanks, that helped!
    – Doesbaddel
    Nov 27 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 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 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 at 15:36












Yeah, I've already finished r and t, just needed some help for s. Thanks, that helped!
– Doesbaddel
Nov 27 at 18:47




Yeah, I've already finished r and t, just needed some help for s. Thanks, that helped!
– Doesbaddel
Nov 27 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 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 at 19:03

















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