A question from 1989 leningrad mathematical olympiad
Clash Royale CLAN TAG#URR8PPP
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 $Ox$=$∃Y∈Z$ such that $X*Y=S$,and the stabilizer of X--$Fx$ to be the set $Fx$=T.
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
|
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 $Ox$=$∃Y∈Z$ such that $X*Y=S$,and the stabilizer of X--$Fx$ to be the set $Fx$=T.
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
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
Dec 25 '18 at 14:00
@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
Dec 25 '18 at 14:07
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
Dec 25 '18 at 14:07
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
Dec 25 '18 at 14:41
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
Dec 25 '18 at 14:55
|
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 $Ox$=$∃Y∈Z$ such that $X*Y=S$,and the stabilizer of X--$Fx$ to be the set $Fx$=T.
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
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 $Ox$=$∃Y∈Z$ such that $X*Y=S$,and the stabilizer of X--$Fx$ to be the set $Fx$=T.
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
edited Dec 25 '18 at 15:05
asked Dec 25 '18 at 13:38
Andrew Armstrong
1688
1688
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
Dec 25 '18 at 14:00
@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
Dec 25 '18 at 14:07
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
Dec 25 '18 at 14:07
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
Dec 25 '18 at 14:41
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
Dec 25 '18 at 14:55
|
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
Dec 25 '18 at 14:00
@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
Dec 25 '18 at 14:07
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
Dec 25 '18 at 14:07
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
Dec 25 '18 at 14:41
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
Dec 25 '18 at 14:55
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
Dec 25 '18 at 14:00
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
Dec 25 '18 at 14:00
@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
Dec 25 '18 at 14:07
@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
Dec 25 '18 at 14:07
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
Dec 25 '18 at 14:07
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
Dec 25 '18 at 14:07
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
Dec 25 '18 at 14:41
@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
Dec 25 '18 at 14:41
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
Dec 25 '18 at 14:55
@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
Dec 25 '18 at 14:55
|
show 6 more comments
1 Answer
1
active
oldest
votes
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 $c,d in mathbbZ$. 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 $1 = c star e star f$.
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
Dec 26 '18 at 10:14
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052110%2fa-question-from-1989-leningrad-mathematical-olympiad%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $c,d in mathbbZ$. 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 $1 = c star e star f$.
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
Dec 26 '18 at 10:14
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 $c,d in mathbbZ$. 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 $1 = c star e star f$.
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
Dec 26 '18 at 10:14
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 $c,d in mathbbZ$. 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 $1 = c star e star f$.
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 $c,d in mathbbZ$. 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 Dec 25 '18 at 15:13
MathematicsStudent1122
8,55122366
8,55122366
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
Dec 26 '18 at 10:14
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
Dec 26 '18 at 10:14
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
Dec 26 '18 at 10:14
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
Dec 26 '18 at 10:14
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052110%2fa-question-from-1989-leningrad-mathematical-olympiad%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
Dec 25 '18 at 14:00
@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
Dec 25 '18 at 14:07
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
Dec 25 '18 at 14:07
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
Dec 25 '18 at 14:41
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
Dec 25 '18 at 14:55