A question from 1989 leningrad mathematical olympiad
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
asked 2 hours ago
Andrew Armstrong
935
|
show 6 more comments
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
asked 2 hours ago
Andrew Armstrong
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
|
show 6 more comments
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
asked 2 hours ago
Andrew Armstrong
935
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
abstract-algebra contest-math binary-operations
asked 2 hours ago
Andrew Armstrong
935
asked 2 hours ago
Andrew Armstrong
935
edited 49 mins ago
asked 2 hours ago
Andrew Armstrong
935
asked 2 hours ago
Andrew Armstrong
935
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
|
show 6 more comments
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
|
show 6 more comments
1 Answer
1
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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 $1 = c star e star f$.
answered 41 mins ago
MathematicsStudent1122
7,89122263
add a comment |
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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 $1 = c star e star f$.
answered 41 mins ago
MathematicsStudent1122
7,89122263
add a comment |
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 $1 = c star e star f$.
answered 41 mins ago
MathematicsStudent1122
7,89122263
add a comment |
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 $1 = c star e star f$.
answered 41 mins ago
MathematicsStudent1122
7,89122263
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 $1 = c star e star f$.
answered 41 mins ago
MathematicsStudent1122
7,89122263
answered 41 mins ago
MathematicsStudent1122
7,89122263
answered 41 mins ago
MathematicsStudent1122
7,89122263
answered 41 mins ago
MathematicsStudent1122
7,89122263
7,89122263
add a comment |
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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