Mixing
Introduction to Modern Algebra
$begingroup$
Define an operation $ * $ on $ Z $ by $ x*y = 1 + xy $. Prove or disprove whether we have an identity element and/or inverse elements.
I am in Introduction to Modern Algebra class and I do not know how to start this problem. I can't find any example problems that are similar as well. Any help would be much appreciated!
abstract-algebra
edited Jan 21 at 3:09
user549397
1,5061418
asked Jan 21 at 2:37
Katelyn EldredgeKatelyn Eldredge
111
$endgroup$
add a comment |
$begingroup$
Define an operation $ * $ on $ Z $ by $ x*y = 1 + xy $. Prove or disprove whether we have an identity element and/or inverse elements.
I am in Introduction to Modern Algebra class and I do not know how to start this problem. I can't find any example problems that are similar as well. Any help would be much appreciated!
abstract-algebra
edited Jan 21 at 3:09
user549397
1,5061418
asked Jan 21 at 2:37
Katelyn EldredgeKatelyn Eldredge
111
$endgroup$
4
$begingroup$
Check the definition of identity elements.
$endgroup$
– xbh
Jan 21 at 2:38
2
$begingroup$
If there is an identity element then $0=0*i=1$ which is impossible. No identity, no inverses.
$endgroup$
– Ben W
Jan 21 at 2:53
$begingroup$
Please choose an informative title. That means something other than subject tags.
$endgroup$
– rschwieb
Jan 21 at 3:08
add a comment |
$begingroup$
Define an operation $ * $ on $ Z $ by $ x*y = 1 + xy $. Prove or disprove whether we have an identity element and/or inverse elements.
I am in Introduction to Modern Algebra class and I do not know how to start this problem. I can't find any example problems that are similar as well. Any help would be much appreciated!
abstract-algebra
edited Jan 21 at 3:09
user549397
1,5061418
asked Jan 21 at 2:37
Katelyn EldredgeKatelyn Eldredge
111
$endgroup$
Define an operation $ * $ on $ Z $ by $ x*y = 1 + xy $. Prove or disprove whether we have an identity element and/or inverse elements.
I am in Introduction to Modern Algebra class and I do not know how to start this problem. I can't find any example problems that are similar as well. Any help would be much appreciated!
abstract-algebra
abstract-algebra
edited Jan 21 at 3:09
user549397
1,5061418
asked Jan 21 at 2:37
Katelyn EldredgeKatelyn Eldredge
111
edited Jan 21 at 3:09
user549397
1,5061418
asked Jan 21 at 2:37
Katelyn EldredgeKatelyn Eldredge
111
edited Jan 21 at 3:09
user549397
1,5061418
edited Jan 21 at 3:09
user549397
1,5061418
edited Jan 21 at 3:09
user549397
1,5061418
1,5061418
asked Jan 21 at 2:37
Katelyn EldredgeKatelyn Eldredge
111
asked Jan 21 at 2:37
Katelyn EldredgeKatelyn Eldredge
111
asked Jan 21 at 2:37
Katelyn EldredgeKatelyn Eldredge
111
111
4
$begingroup$
Check the definition of identity elements.
$endgroup$
– xbh
Jan 21 at 2:38
2
$begingroup$
If there is an identity element then $0=0*i=1$ which is impossible. No identity, no inverses.
$endgroup$
– Ben W
Jan 21 at 2:53
$begingroup$
Please choose an informative title. That means something other than subject tags.
$endgroup$
– rschwieb
Jan 21 at 3:08
add a comment |
4
$begingroup$
Check the definition of identity elements.
$endgroup$
– xbh
Jan 21 at 2:38
2
$begingroup$
If there is an identity element then $0=0*i=1$ which is impossible. No identity, no inverses.
$endgroup$
– Ben W
Jan 21 at 2:53
$begingroup$
Please choose an informative title. That means something other than subject tags.
$endgroup$
– rschwieb
Jan 21 at 3:08
4
4
$begingroup$
Check the definition of identity elements.
$endgroup$
– xbh
Jan 21 at 2:38
$begingroup$
Check the definition of identity elements.
$endgroup$
– xbh
Jan 21 at 2:38
2
2
$begingroup$
If there is an identity element then $0=0*i=1$ which is impossible. No identity, no inverses.
$endgroup$
– Ben W
Jan 21 at 2:53
$begingroup$
If there is an identity element then $0=0*i=1$ which is impossible. No identity, no inverses.
$endgroup$
– Ben W
Jan 21 at 2:53
$begingroup$
Please choose an informative title. That means something other than subject tags.
$endgroup$
– rschwieb
Jan 21 at 3:08
$begingroup$
Please choose an informative title. That means something other than subject tags.
$endgroup$
– rschwieb
Jan 21 at 3:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We need an element $e$ such that $x=x*e=e*x,,forall x$. When $x=0$, we get $0=0*e=1+0e=1$ , a contradiction.
Since we don't have $e$, we don't have inverses either.
answered Jan 21 at 2:58
Chris CusterChris Custer
14.1k3827
$endgroup$
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%2f3081422%2fintroduction-to-modern-algebra%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
$begingroup$
We need an element $e$ such that $x=x*e=e*x,,forall x$. When $x=0$, we get $0=0*e=1+0e=1$ , a contradiction.
Since we don't have $e$, we don't have inverses either.
answered Jan 21 at 2:58
Chris CusterChris Custer
14.1k3827
$endgroup$
add a comment |
$begingroup$
We need an element $e$ such that $x=x*e=e*x,,forall x$. When $x=0$, we get $0=0*e=1+0e=1$ , a contradiction.
Since we don't have $e$, we don't have inverses either.
answered Jan 21 at 2:58
Chris CusterChris Custer
14.1k3827
$endgroup$
add a comment |
$begingroup$
We need an element $e$ such that $x=x*e=e*x,,forall x$. When $x=0$, we get $0=0*e=1+0e=1$ , a contradiction.
Since we don't have $e$, we don't have inverses either.
answered Jan 21 at 2:58
Chris CusterChris Custer
14.1k3827
$endgroup$
We need an element $e$ such that $x=x*e=e*x,,forall x$. When $x=0$, we get $0=0*e=1+0e=1$ , a contradiction.
Since we don't have $e$, we don't have inverses either.
answered Jan 21 at 2:58
Chris CusterChris Custer
14.1k3827
answered Jan 21 at 2:58
Chris CusterChris Custer
14.1k3827
answered Jan 21 at 2:58
Chris CusterChris Custer
14.1k3827
answered Jan 21 at 2:58
Chris CusterChris Custer
14.1k3827
14.1k3827
add a comment |
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.
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%2f3081422%2fintroduction-to-modern-algebra%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
4
$begingroup$
Check the definition of identity elements.
$endgroup$
– xbh
Jan 21 at 2:38
2
$begingroup$
If there is an identity element then $0=0*i=1$ which is impossible. No identity, no inverses.
$endgroup$
– Ben W
Jan 21 at 2:53
$begingroup$
Please choose an informative title. That means something other than subject tags.
$endgroup$
– rschwieb
Jan 21 at 3:08