Let $K$ be the ring of all real functions and let $f in K$. Suppose that f is not a zero-divisor. Prove that...
$begingroup$
Let K be the ring of all real functions and let $f in K$. Suppose that $f$ is
not a zero-divisor. Prove that $f in U(K)$.
ring-theory
edited Dec 8 '18 at 22:36
Robert Lewis
46.6k23067
asked Dec 8 '18 at 22:29
Полина СергеевнаПолина Сергеевна
63
$endgroup$
closed as off-topic by Dando18, Davide Giraudo, José Carlos Santos, KReiser, Shailesh Dec 9 '18 at 3:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dando18, Davide Giraudo, José Carlos Santos, KReiser, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let K be the ring of all real functions and let $f in K$. Suppose that $f$ is
not a zero-divisor. Prove that $f in U(K)$.
ring-theory
edited Dec 8 '18 at 22:36
Robert Lewis
46.6k23067
asked Dec 8 '18 at 22:29
Полина СергеевнаПолина Сергеевна
63
$endgroup$
closed as off-topic by Dando18, Davide Giraudo, José Carlos Santos, KReiser, Shailesh Dec 9 '18 at 3:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dando18, Davide Giraudo, José Carlos Santos, KReiser, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let K be the ring of all real functions and let $f in K$. Suppose that $f$ is
not a zero-divisor. Prove that $f in U(K)$.
ring-theory
edited Dec 8 '18 at 22:36
Robert Lewis
46.6k23067
asked Dec 8 '18 at 22:29
Полина СергеевнаПолина Сергеевна
63
$endgroup$
Let K be the ring of all real functions and let $f in K$. Suppose that $f$ is
not a zero-divisor. Prove that $f in U(K)$.
ring-theory
ring-theory
edited Dec 8 '18 at 22:36
Robert Lewis
46.6k23067
asked Dec 8 '18 at 22:29
Полина СергеевнаПолина Сергеевна
63
edited Dec 8 '18 at 22:36
Robert Lewis
46.6k23067
asked Dec 8 '18 at 22:29
Полина СергеевнаПолина Сергеевна
63
edited Dec 8 '18 at 22:36
Robert Lewis
46.6k23067
edited Dec 8 '18 at 22:36
Robert Lewis
46.6k23067
edited Dec 8 '18 at 22:36
Robert Lewis
46.6k23067
46.6k23067
asked Dec 8 '18 at 22:29
Полина СергеевнаПолина Сергеевна
63
asked Dec 8 '18 at 22:29
Полина СергеевнаПолина Сергеевна
63
asked Dec 8 '18 at 22:29
Полина СергеевнаПолина Сергеевна
63
63
closed as off-topic by Dando18, Davide Giraudo, José Carlos Santos, KReiser, Shailesh Dec 9 '18 at 3:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dando18, Davide Giraudo, José Carlos Santos, KReiser, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Dando18, Davide Giraudo, José Carlos Santos, KReiser, Shailesh Dec 9 '18 at 3:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dando18, Davide Giraudo, José Carlos Santos, KReiser, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
It is enough to show that $E={x:f(x)=0}$ is empty. Suppose $E$ is not empty $f(x)=0$, define $g(x)=1, g(y)=0, yneq x, fg=0$. Contradiction $h(x)={1over{f(x)}}$ is the inverse of $f$.
answered Dec 8 '18 at 22:32
Tsemo AristideTsemo Aristide
58.1k11445
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is enough to show that $E={x:f(x)=0}$ is empty. Suppose $E$ is not empty $f(x)=0$, define $g(x)=1, g(y)=0, yneq x, fg=0$. Contradiction $h(x)={1over{f(x)}}$ is the inverse of $f$.
answered Dec 8 '18 at 22:32
Tsemo AristideTsemo Aristide
58.1k11445
$endgroup$
add a comment |
$begingroup$
It is enough to show that $E={x:f(x)=0}$ is empty. Suppose $E$ is not empty $f(x)=0$, define $g(x)=1, g(y)=0, yneq x, fg=0$. Contradiction $h(x)={1over{f(x)}}$ is the inverse of $f$.
answered Dec 8 '18 at 22:32
Tsemo AristideTsemo Aristide
58.1k11445
$endgroup$
add a comment |
$begingroup$
It is enough to show that $E={x:f(x)=0}$ is empty. Suppose $E$ is not empty $f(x)=0$, define $g(x)=1, g(y)=0, yneq x, fg=0$. Contradiction $h(x)={1over{f(x)}}$ is the inverse of $f$.
answered Dec 8 '18 at 22:32
Tsemo AristideTsemo Aristide
58.1k11445
$endgroup$
It is enough to show that $E={x:f(x)=0}$ is empty. Suppose $E$ is not empty $f(x)=0$, define $g(x)=1, g(y)=0, yneq x, fg=0$. Contradiction $h(x)={1over{f(x)}}$ is the inverse of $f$.
answered Dec 8 '18 at 22:32
Tsemo AristideTsemo Aristide
58.1k11445
answered Dec 8 '18 at 22:32
Tsemo AristideTsemo Aristide
58.1k11445
answered Dec 8 '18 at 22:32
Tsemo AristideTsemo Aristide
58.1k11445
answered Dec 8 '18 at 22:32
Tsemo AristideTsemo Aristide
58.1k11445
58.1k11445
add a comment |
add a comment |
