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 yesterday
Andrew Armstrong
1538
|
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 yesterday
Andrew Armstrong
1538
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
23 hours 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
23 hours 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
23 hours 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
23 hours 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
22 hours 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 yesterday
Andrew Armstrong
1538
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 yesterday
Andrew Armstrong
1538
asked yesterday
Andrew Armstrong
1538
edited 22 hours ago
asked yesterday
Andrew Armstrong
1538
asked yesterday
Andrew Armstrong
1538
asked yesterday
Andrew Armstrong
1538
1538
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
23 hours 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
23 hours 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
23 hours 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
23 hours 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
22 hours 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
23 hours 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
23 hours 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
23 hours 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
23 hours 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
22 hours 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
23 hours 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
23 hours 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
23 hours 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
23 hours 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
23 hours 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
23 hours 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
23 hours 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
23 hours 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
22 hours 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
22 hours 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 22 hours ago
MathematicsStudent1122
8,09122364
More generally, in any abstract algebra with binary $ star $ and unary $ - $ satisfying the 3 properties in the question, $ - $ must be the identity. This can be proven by supposing $x ≠ -x$ for some $x$ and deriving a contradiction as above.
– Hans Lub
3 hours ago
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 22 hours ago
MathematicsStudent1122
8,09122364
More generally, in any abstract algebra with binary $ star $ and unary $ - $ satisfying the 3 properties in the question, $ - $ must be the identity. This can be proven by supposing $x ≠ -x$ for some $x$ and deriving a contradiction as above.
– Hans Lub
3 hours ago
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 22 hours ago
MathematicsStudent1122
8,09122364
More generally, in any abstract algebra with binary $ star $ and unary $ - $ satisfying the 3 properties in the question, $ - $ must be the identity. This can be proven by supposing $x ≠ -x$ for some $x$ and deriving a contradiction as above.
– Hans Lub
3 hours ago
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 22 hours ago
MathematicsStudent1122
8,09122364
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 22 hours ago
MathematicsStudent1122
8,09122364
answered 22 hours ago
MathematicsStudent1122
8,09122364
answered 22 hours ago
MathematicsStudent1122
8,09122364
answered 22 hours ago
MathematicsStudent1122
8,09122364
8,09122364
More generally, in any abstract algebra with binary $ star $ and unary $ - $ satisfying the 3 properties in the question, $ - $ must be the identity. This can be proven by supposing $x ≠ -x$ for some $x$ and deriving a contradiction as above.
– Hans Lub
3 hours ago
add a comment |
More generally, in any abstract algebra with binary $ star $ and unary $ - $ satisfying the 3 properties in the question, $ - $ must be the identity. This can be proven by supposing $x ≠ -x$ for some $x$ and deriving a contradiction as above.
– Hans Lub
3 hours ago
More generally, in any abstract algebra with binary $ star $ and unary $ - $ satisfying the 3 properties in the question, $ - $ must be the identity. This can be proven by supposing $x ≠ -x$ for some $x$ and deriving a contradiction as above.
– Hans Lub
3 hours ago
More generally, in any abstract algebra with binary $ star $ and unary $ - $ satisfying the 3 properties in the question, $ - $ must be the identity. This can be proven by supposing $x ≠ -x$ for some $x$ and deriving a contradiction as above.
– Hans Lub
3 hours ago
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
23 hours 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
23 hours 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
23 hours 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
23 hours 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
22 hours ago