A question from 1989 leningrad mathematical olympiad












4














Prove that we cannot define an binary operation $*$ on the set of integers Z satisfy all of the three properties below simultaneously:



For any $A∈Z,B∈Z,C∈Z:$



1.$A*B=-(B*A)$



2.$(A*B)*C=A*(B*C)$ (Associative Law)



3.For every $Ain Z$ there exist $B∈Z,C∈Z$ such that $A=B*C$



I have got stucked for three days on this questions.Anyway,I will show some result and idea I had:



1.for any X∈Z,we have $X*X=-(X*X)$.So we have $X*X=0$



2.for any X∈Z,we have $X*0=-(0*X)=X*(X*X)=(X*X)*X=0*X$.So we have $X*0=0*X=0$



3.Now For any $X∈Z$($X≠0$).We define the orbit of X--$Ox$ to be the set $Ox$={S|$∃Y∈Z$ such that $X*Y=S$},and the stabilizer of X--$Fx$ to be the set $Fx$={T|$T*X=X*T=0$}.



My goal is to prove that actually $Ox=Fx$.And therefore since $X∈Fx$,so $X∈Ox$,and we reach a contradiction since $X∉Ox$(otherwise if $∃Y∈Z$ such that $X*Y=X$,then $(X*Y)*Y=X*(Y*Y)=X*0=X*Y=X=0$)



It is easy to see that $Ox⊆Fx$,since for any $S∈Ox$,we have $X*S=X*(X*Y)=(X*X)*Y=0*Y=0$,so $S∈Fx$



However,for the other direction,I cannot deduce out,which I need help.



I think the backgroud of this question is the orbit&stabilizer theorem in the course abstract algebra.So I have a strong intuition that I am on the right track.










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  • 1




    Not sure this is clear. Does $AB$ mean $Astar B$? If so, how can you deduce, say, that $Xstar X=-Xstar Ximplies Xstar X=0$? All we know is that $Xstar X=(-1star X)star X$. If $AB$ doesn't mean $Astar B$ I think you need to clarify when you are using the new operation and when you are using ordinary multiplication.
    – lulu
    1 hour ago












  • @AndrewArmstrong So you mean they are all supposed to be the new operaton? That there is no ordinary multiplication involved in any of these laws?
    – Bram28
    1 hour ago












  • In that case, do you agree with me that it is not obvious that $Xstar X=-Xstar Ximplies Xstar X=0$? At least that requires a proper proof. here, by the way, is a good tutorial on formatting for this site. I am using "star" to get $star$).
    – lulu
    1 hour ago






  • 1




    @lulu In his first requirement he has parentheses around the product so $Xstar X=-(Xstar X)$. In ordinary integer arithmetic I interpret $-X$ as the additive inverse of $X$ and it is a thing to prove that $-X=(-1)*X$,.i.e. the additive inverse of an integer is the additive inverse of $1$ times the integer.
    – John Douma
    1 hour ago






  • 1




    @JohnDouma if $-A$ is intended to denote the additive inverse of the integer $A$ in the usual sense, then $A=-A$ does indeed imply that $A=0$. And it is entirely possible that this is what is intended.
    – lulu
    1 hour ago
















4














Prove that we cannot define an binary operation $*$ on the set of integers Z satisfy all of the three properties below simultaneously:



For any $A∈Z,B∈Z,C∈Z:$



1.$A*B=-(B*A)$



2.$(A*B)*C=A*(B*C)$ (Associative Law)



3.For every $Ain Z$ there exist $B∈Z,C∈Z$ such that $A=B*C$



I have got stucked for three days on this questions.Anyway,I will show some result and idea I had:



1.for any X∈Z,we have $X*X=-(X*X)$.So we have $X*X=0$



2.for any X∈Z,we have $X*0=-(0*X)=X*(X*X)=(X*X)*X=0*X$.So we have $X*0=0*X=0$



3.Now For any $X∈Z$($X≠0$).We define the orbit of X--$Ox$ to be the set $Ox$={S|$∃Y∈Z$ such that $X*Y=S$},and the stabilizer of X--$Fx$ to be the set $Fx$={T|$T*X=X*T=0$}.



My goal is to prove that actually $Ox=Fx$.And therefore since $X∈Fx$,so $X∈Ox$,and we reach a contradiction since $X∉Ox$(otherwise if $∃Y∈Z$ such that $X*Y=X$,then $(X*Y)*Y=X*(Y*Y)=X*0=X*Y=X=0$)



It is easy to see that $Ox⊆Fx$,since for any $S∈Ox$,we have $X*S=X*(X*Y)=(X*X)*Y=0*Y=0$,so $S∈Fx$



However,for the other direction,I cannot deduce out,which I need help.



I think the backgroud of this question is the orbit&stabilizer theorem in the course abstract algebra.So I have a strong intuition that I am on the right track.










share|cite|improve this question




















  • 1




    Not sure this is clear. Does $AB$ mean $Astar B$? If so, how can you deduce, say, that $Xstar X=-Xstar Ximplies Xstar X=0$? All we know is that $Xstar X=(-1star X)star X$. If $AB$ doesn't mean $Astar B$ I think you need to clarify when you are using the new operation and when you are using ordinary multiplication.
    – lulu
    1 hour ago












  • @AndrewArmstrong So you mean they are all supposed to be the new operaton? That there is no ordinary multiplication involved in any of these laws?
    – Bram28
    1 hour ago












  • In that case, do you agree with me that it is not obvious that $Xstar X=-Xstar Ximplies Xstar X=0$? At least that requires a proper proof. here, by the way, is a good tutorial on formatting for this site. I am using "star" to get $star$).
    – lulu
    1 hour ago






  • 1




    @lulu In his first requirement he has parentheses around the product so $Xstar X=-(Xstar X)$. In ordinary integer arithmetic I interpret $-X$ as the additive inverse of $X$ and it is a thing to prove that $-X=(-1)*X$,.i.e. the additive inverse of an integer is the additive inverse of $1$ times the integer.
    – John Douma
    1 hour ago






  • 1




    @JohnDouma if $-A$ is intended to denote the additive inverse of the integer $A$ in the usual sense, then $A=-A$ does indeed imply that $A=0$. And it is entirely possible that this is what is intended.
    – lulu
    1 hour ago














4












4








4







Prove that we cannot define an binary operation $*$ on the set of integers Z satisfy all of the three properties below simultaneously:



For any $A∈Z,B∈Z,C∈Z:$



1.$A*B=-(B*A)$



2.$(A*B)*C=A*(B*C)$ (Associative Law)



3.For every $Ain Z$ there exist $B∈Z,C∈Z$ such that $A=B*C$



I have got stucked for three days on this questions.Anyway,I will show some result and idea I had:



1.for any X∈Z,we have $X*X=-(X*X)$.So we have $X*X=0$



2.for any X∈Z,we have $X*0=-(0*X)=X*(X*X)=(X*X)*X=0*X$.So we have $X*0=0*X=0$



3.Now For any $X∈Z$($X≠0$).We define the orbit of X--$Ox$ to be the set $Ox$={S|$∃Y∈Z$ such that $X*Y=S$},and the stabilizer of X--$Fx$ to be the set $Fx$={T|$T*X=X*T=0$}.



My goal is to prove that actually $Ox=Fx$.And therefore since $X∈Fx$,so $X∈Ox$,and we reach a contradiction since $X∉Ox$(otherwise if $∃Y∈Z$ such that $X*Y=X$,then $(X*Y)*Y=X*(Y*Y)=X*0=X*Y=X=0$)



It is easy to see that $Ox⊆Fx$,since for any $S∈Ox$,we have $X*S=X*(X*Y)=(X*X)*Y=0*Y=0$,so $S∈Fx$



However,for the other direction,I cannot deduce out,which I need help.



I think the backgroud of this question is the orbit&stabilizer theorem in the course abstract algebra.So I have a strong intuition that I am on the right track.










share|cite|improve this question















Prove that we cannot define an binary operation $*$ on the set of integers Z satisfy all of the three properties below simultaneously:



For any $A∈Z,B∈Z,C∈Z:$



1.$A*B=-(B*A)$



2.$(A*B)*C=A*(B*C)$ (Associative Law)



3.For every $Ain Z$ there exist $B∈Z,C∈Z$ such that $A=B*C$



I have got stucked for three days on this questions.Anyway,I will show some result and idea I had:



1.for any X∈Z,we have $X*X=-(X*X)$.So we have $X*X=0$



2.for any X∈Z,we have $X*0=-(0*X)=X*(X*X)=(X*X)*X=0*X$.So we have $X*0=0*X=0$



3.Now For any $X∈Z$($X≠0$).We define the orbit of X--$Ox$ to be the set $Ox$={S|$∃Y∈Z$ such that $X*Y=S$},and the stabilizer of X--$Fx$ to be the set $Fx$={T|$T*X=X*T=0$}.



My goal is to prove that actually $Ox=Fx$.And therefore since $X∈Fx$,so $X∈Ox$,and we reach a contradiction since $X∉Ox$(otherwise if $∃Y∈Z$ such that $X*Y=X$,then $(X*Y)*Y=X*(Y*Y)=X*0=X*Y=X=0$)



It is easy to see that $Ox⊆Fx$,since for any $S∈Ox$,we have $X*S=X*(X*Y)=(X*X)*Y=0*Y=0$,so $S∈Fx$



However,for the other direction,I cannot deduce out,which I need help.



I think the backgroud of this question is the orbit&stabilizer theorem in the course abstract algebra.So I have a strong intuition that I am on the right track.







abstract-algebra contest-math binary-operations






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share|cite|improve this question













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share|cite|improve this question








edited 49 mins ago

























asked 2 hours ago









Andrew Armstrong

935




935








  • 1




    Not sure this is clear. Does $AB$ mean $Astar B$? If so, how can you deduce, say, that $Xstar X=-Xstar Ximplies Xstar X=0$? All we know is that $Xstar X=(-1star X)star X$. If $AB$ doesn't mean $Astar B$ I think you need to clarify when you are using the new operation and when you are using ordinary multiplication.
    – lulu
    1 hour ago












  • @AndrewArmstrong So you mean they are all supposed to be the new operaton? That there is no ordinary multiplication involved in any of these laws?
    – Bram28
    1 hour ago












  • In that case, do you agree with me that it is not obvious that $Xstar X=-Xstar Ximplies Xstar X=0$? At least that requires a proper proof. here, by the way, is a good tutorial on formatting for this site. I am using "star" to get $star$).
    – lulu
    1 hour ago






  • 1




    @lulu In his first requirement he has parentheses around the product so $Xstar X=-(Xstar X)$. In ordinary integer arithmetic I interpret $-X$ as the additive inverse of $X$ and it is a thing to prove that $-X=(-1)*X$,.i.e. the additive inverse of an integer is the additive inverse of $1$ times the integer.
    – John Douma
    1 hour ago






  • 1




    @JohnDouma if $-A$ is intended to denote the additive inverse of the integer $A$ in the usual sense, then $A=-A$ does indeed imply that $A=0$. And it is entirely possible that this is what is intended.
    – lulu
    1 hour ago














  • 1




    Not sure this is clear. Does $AB$ mean $Astar B$? If so, how can you deduce, say, that $Xstar X=-Xstar Ximplies Xstar X=0$? All we know is that $Xstar X=(-1star X)star X$. If $AB$ doesn't mean $Astar B$ I think you need to clarify when you are using the new operation and when you are using ordinary multiplication.
    – lulu
    1 hour ago












  • @AndrewArmstrong So you mean they are all supposed to be the new operaton? That there is no ordinary multiplication involved in any of these laws?
    – Bram28
    1 hour ago












  • In that case, do you agree with me that it is not obvious that $Xstar X=-Xstar Ximplies Xstar X=0$? At least that requires a proper proof. here, by the way, is a good tutorial on formatting for this site. I am using "star" to get $star$).
    – lulu
    1 hour ago






  • 1




    @lulu In his first requirement he has parentheses around the product so $Xstar X=-(Xstar X)$. In ordinary integer arithmetic I interpret $-X$ as the additive inverse of $X$ and it is a thing to prove that $-X=(-1)*X$,.i.e. the additive inverse of an integer is the additive inverse of $1$ times the integer.
    – John Douma
    1 hour ago






  • 1




    @JohnDouma if $-A$ is intended to denote the additive inverse of the integer $A$ in the usual sense, then $A=-A$ does indeed imply that $A=0$. And it is entirely possible that this is what is intended.
    – lulu
    1 hour ago








1




1




Not sure this is clear. Does $AB$ mean $Astar B$? If so, how can you deduce, say, that $Xstar X=-Xstar Ximplies Xstar X=0$? All we know is that $Xstar X=(-1star X)star X$. If $AB$ doesn't mean $Astar B$ I think you need to clarify when you are using the new operation and when you are using ordinary multiplication.
– lulu
1 hour ago






Not sure this is clear. Does $AB$ mean $Astar B$? If so, how can you deduce, say, that $Xstar X=-Xstar Ximplies Xstar X=0$? All we know is that $Xstar X=(-1star X)star X$. If $AB$ doesn't mean $Astar B$ I think you need to clarify when you are using the new operation and when you are using ordinary multiplication.
– lulu
1 hour ago














@AndrewArmstrong So you mean they are all supposed to be the new operaton? That there is no ordinary multiplication involved in any of these laws?
– Bram28
1 hour ago






@AndrewArmstrong So you mean they are all supposed to be the new operaton? That there is no ordinary multiplication involved in any of these laws?
– Bram28
1 hour ago














In that case, do you agree with me that it is not obvious that $Xstar X=-Xstar Ximplies Xstar X=0$? At least that requires a proper proof. here, by the way, is a good tutorial on formatting for this site. I am using "star" to get $star$).
– lulu
1 hour ago




In that case, do you agree with me that it is not obvious that $Xstar X=-Xstar Ximplies Xstar X=0$? At least that requires a proper proof. here, by the way, is a good tutorial on formatting for this site. I am using "star" to get $star$).
– lulu
1 hour ago




1




1




@lulu In his first requirement he has parentheses around the product so $Xstar X=-(Xstar X)$. In ordinary integer arithmetic I interpret $-X$ as the additive inverse of $X$ and it is a thing to prove that $-X=(-1)*X$,.i.e. the additive inverse of an integer is the additive inverse of $1$ times the integer.
– John Douma
1 hour ago




@lulu In his first requirement he has parentheses around the product so $Xstar X=-(Xstar X)$. In ordinary integer arithmetic I interpret $-X$ as the additive inverse of $X$ and it is a thing to prove that $-X=(-1)*X$,.i.e. the additive inverse of an integer is the additive inverse of $1$ times the integer.
– John Douma
1 hour ago




1




1




@JohnDouma if $-A$ is intended to denote the additive inverse of the integer $A$ in the usual sense, then $A=-A$ does indeed imply that $A=0$. And it is entirely possible that this is what is intended.
– lulu
1 hour ago




@JohnDouma if $-A$ is intended to denote the additive inverse of the integer $A$ in the usual sense, then $A=-A$ does indeed imply that $A=0$. And it is entirely possible that this is what is intended.
– lulu
1 hour ago










1 Answer
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Note the associativity condition implies that parentheses are redundant, so something like $a star b star c star d$ is uniquely-defined without parentheses.



Observe that, by axiom $3$, $1 = c star d$ for some $c,d in mathbb{Z}$. A repeated application of axiom $3$ gives us that $d= e star f$ for some integers $e,f$.



Hence we get that $1 = c star e star f$. Observe now that $-1 = f star c star e$ by the first axiom. A repeated application of the first axiom gives that $1 = e star f star c$. A third application gives $-1 = c star e star f$. But this is a contradiction, since we know that $1 = c star e star f$.






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    Note the associativity condition implies that parentheses are redundant, so something like $a star b star c star d$ is uniquely-defined without parentheses.



    Observe that, by axiom $3$, $1 = c star d$ for some $c,d in mathbb{Z}$. A repeated application of axiom $3$ gives us that $d= e star f$ for some integers $e,f$.



    Hence we get that $1 = c star e star f$. Observe now that $-1 = f star c star e$ by the first axiom. A repeated application of the first axiom gives that $1 = e star f star c$. A third application gives $-1 = c star e star f$. But this is a contradiction, since we know that $1 = c star e star f$.






    share|cite|improve this answer


























      3














      Note the associativity condition implies that parentheses are redundant, so something like $a star b star c star d$ is uniquely-defined without parentheses.



      Observe that, by axiom $3$, $1 = c star d$ for some $c,d in mathbb{Z}$. A repeated application of axiom $3$ gives us that $d= e star f$ for some integers $e,f$.



      Hence we get that $1 = c star e star f$. Observe now that $-1 = f star c star e$ by the first axiom. A repeated application of the first axiom gives that $1 = e star f star c$. A third application gives $-1 = c star e star f$. But this is a contradiction, since we know that $1 = c star e star f$.






      share|cite|improve this answer
























        3












        3








        3






        Note the associativity condition implies that parentheses are redundant, so something like $a star b star c star d$ is uniquely-defined without parentheses.



        Observe that, by axiom $3$, $1 = c star d$ for some $c,d in mathbb{Z}$. A repeated application of axiom $3$ gives us that $d= e star f$ for some integers $e,f$.



        Hence we get that $1 = c star e star f$. Observe now that $-1 = f star c star e$ by the first axiom. A repeated application of the first axiom gives that $1 = e star f star c$. A third application gives $-1 = c star e star f$. But this is a contradiction, since we know that $1 = c star e star f$.






        share|cite|improve this answer












        Note the associativity condition implies that parentheses are redundant, so something like $a star b star c star d$ is uniquely-defined without parentheses.



        Observe that, by axiom $3$, $1 = c star d$ for some $c,d in mathbb{Z}$. A repeated application of axiom $3$ gives us that $d= e star f$ for some integers $e,f$.



        Hence we get that $1 = c star e star f$. Observe now that $-1 = f star c star e$ by the first axiom. A repeated application of the first axiom gives that $1 = e star f star c$. A third application gives $-1 = c star e star f$. But this is a contradiction, since we know that $1 = c star e star f$.







        share|cite|improve this answer












        share|cite|improve this answer



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        answered 41 mins ago









        MathematicsStudent1122

        7,89122263




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