A ring without identity: deny the hypothesis
Let $R$ a ring without identity. So, $$(Rquadtext{does not have the identity})Rightarrow (nexists;ein R;text{such that};ea=a;forall ain R)[text{It's correct?}].$$
Now, if I found an element $ein R$ and an element $ain R$ such that $ea=a$, can I conclude that this contradicts the hypothesis according to which $R$ not have identity?
abstract-algebra ring-theory
edited Dec 3 '18 at 14:39
José Carlos Santos
150k22122223
asked Dec 3 '18 at 14:23
Jack J.
4421419
add a comment |
Let $R$ a ring without identity. So, $$(Rquadtext{does not have the identity})Rightarrow (nexists;ein R;text{such that};ea=a;forall ain R)[text{It's correct?}].$$
Now, if I found an element $ein R$ and an element $ain R$ such that $ea=a$, can I conclude that this contradicts the hypothesis according to which $R$ not have identity?
abstract-algebra ring-theory
edited Dec 3 '18 at 14:39
José Carlos Santos
150k22122223
asked Dec 3 '18 at 14:23
Jack J.
4421419
For $e$ to be an identity we need $ea = a$ for all $a$, not just one single $a$. So you can‘t conclude what you said.
– Lukas Kofler
Dec 3 '18 at 14:25
add a comment |
Let $R$ a ring without identity. So, $$(Rquadtext{does not have the identity})Rightarrow (nexists;ein R;text{such that};ea=a;forall ain R)[text{It's correct?}].$$
Now, if I found an element $ein R$ and an element $ain R$ such that $ea=a$, can I conclude that this contradicts the hypothesis according to which $R$ not have identity?
abstract-algebra ring-theory
edited Dec 3 '18 at 14:39
José Carlos Santos
150k22122223
asked Dec 3 '18 at 14:23
Jack J.
4421419
Let $R$ a ring without identity. So, $$(Rquadtext{does not have the identity})Rightarrow (nexists;ein R;text{such that};ea=a;forall ain R)[text{It's correct?}].$$
Now, if I found an element $ein R$ and an element $ain R$ such that $ea=a$, can I conclude that this contradicts the hypothesis according to which $R$ not have identity?
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Dec 3 '18 at 14:39
José Carlos Santos
150k22122223
asked Dec 3 '18 at 14:23
Jack J.
4421419
edited Dec 3 '18 at 14:39
José Carlos Santos
150k22122223
asked Dec 3 '18 at 14:23
Jack J.
4421419
edited Dec 3 '18 at 14:39
José Carlos Santos
150k22122223
edited Dec 3 '18 at 14:39
José Carlos Santos
150k22122223
edited Dec 3 '18 at 14:39
José Carlos Santos
150k22122223
150k22122223
asked Dec 3 '18 at 14:23
Jack J.
4421419
asked Dec 3 '18 at 14:23
Jack J.
4421419
asked Dec 3 '18 at 14:23
Jack J.
4421419
4421419
For $e$ to be an identity we need $ea = a$ for all $a$, not just one single $a$. So you can‘t conclude what you said.
– Lukas Kofler
Dec 3 '18 at 14:25
add a comment |
For $e$ to be an identity we need $ea = a$ for all $a$, not just one single $a$. So you can‘t conclude what you said.
– Lukas Kofler
Dec 3 '18 at 14:25
For $e$ to be an identity we need $ea = a$ for all $a$, not just one single $a$. So you can‘t conclude what you said.
– Lukas Kofler
Dec 3 '18 at 14:25
For $e$ to be an identity we need $ea = a$ for all $a$, not just one single $a$. So you can‘t conclude what you said.
– Lukas Kofler
Dec 3 '18 at 14:25
add a comment |
2 Answers
2
active
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votes
First of all, we write quantifiers before statements they are referring. The notation
$$notexists ein R text{ such that }ea=aforall ain R$$
is invalid, simply because it is unclear. We don't know if it means
$$forall ain R:notexists ein R text{ such that }ea=a$$
or
$$notexists ein R: forall ain R: text{ such that }ea=a.$$
The two statements above are not equivalent, and are in fact very much different. The second statement is much stronger.
Second of all, the negation of the statement "$R$ has an identity" is the negation of the statement
$$exists ein R: forall ain R: ae=e$$
and the negation of this statement is the second of the two statements above.
Your argument that if you find some $e,a$ such that $ea=a$, is not enough to prove that $R$ has no identity, since even in rings without an identity, we can set $a=e=0$ and have $ea=a$.
answered Dec 3 '18 at 14:34
5xum
89.5k393161
Thanks for this. You said what I wanted to, and probably better.
– Lubin
Dec 3 '18 at 21:24
add a comment |
No, you cannot. In any ring (with or without identity) such elements exist: just take $e=a=0$.
answered Dec 3 '18 at 14:25
José Carlos Santos
150k22122223
add a comment |
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2 Answers
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oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
First of all, we write quantifiers before statements they are referring. The notation
$$notexists ein R text{ such that }ea=aforall ain R$$
is invalid, simply because it is unclear. We don't know if it means
$$forall ain R:notexists ein R text{ such that }ea=a$$
or
$$notexists ein R: forall ain R: text{ such that }ea=a.$$
The two statements above are not equivalent, and are in fact very much different. The second statement is much stronger.
Second of all, the negation of the statement "$R$ has an identity" is the negation of the statement
$$exists ein R: forall ain R: ae=e$$
and the negation of this statement is the second of the two statements above.
Your argument that if you find some $e,a$ such that $ea=a$, is not enough to prove that $R$ has no identity, since even in rings without an identity, we can set $a=e=0$ and have $ea=a$.
answered Dec 3 '18 at 14:34
5xum
89.5k393161
Thanks for this. You said what I wanted to, and probably better.
– Lubin
Dec 3 '18 at 21:24
add a comment |
First of all, we write quantifiers before statements they are referring. The notation
$$notexists ein R text{ such that }ea=aforall ain R$$
is invalid, simply because it is unclear. We don't know if it means
$$forall ain R:notexists ein R text{ such that }ea=a$$
or
$$notexists ein R: forall ain R: text{ such that }ea=a.$$
The two statements above are not equivalent, and are in fact very much different. The second statement is much stronger.
Second of all, the negation of the statement "$R$ has an identity" is the negation of the statement
$$exists ein R: forall ain R: ae=e$$
and the negation of this statement is the second of the two statements above.
Your argument that if you find some $e,a$ such that $ea=a$, is not enough to prove that $R$ has no identity, since even in rings without an identity, we can set $a=e=0$ and have $ea=a$.
answered Dec 3 '18 at 14:34
5xum
89.5k393161
Thanks for this. You said what I wanted to, and probably better.
– Lubin
Dec 3 '18 at 21:24
add a comment |
First of all, we write quantifiers before statements they are referring. The notation
$$notexists ein R text{ such that }ea=aforall ain R$$
is invalid, simply because it is unclear. We don't know if it means
$$forall ain R:notexists ein R text{ such that }ea=a$$
or
$$notexists ein R: forall ain R: text{ such that }ea=a.$$
The two statements above are not equivalent, and are in fact very much different. The second statement is much stronger.
Second of all, the negation of the statement "$R$ has an identity" is the negation of the statement
$$exists ein R: forall ain R: ae=e$$
and the negation of this statement is the second of the two statements above.
Your argument that if you find some $e,a$ such that $ea=a$, is not enough to prove that $R$ has no identity, since even in rings without an identity, we can set $a=e=0$ and have $ea=a$.
answered Dec 3 '18 at 14:34
5xum
89.5k393161
First of all, we write quantifiers before statements they are referring. The notation
$$notexists ein R text{ such that }ea=aforall ain R$$
is invalid, simply because it is unclear. We don't know if it means
$$forall ain R:notexists ein R text{ such that }ea=a$$
or
$$notexists ein R: forall ain R: text{ such that }ea=a.$$
The two statements above are not equivalent, and are in fact very much different. The second statement is much stronger.
Second of all, the negation of the statement "$R$ has an identity" is the negation of the statement
$$exists ein R: forall ain R: ae=e$$
and the negation of this statement is the second of the two statements above.
Your argument that if you find some $e,a$ such that $ea=a$, is not enough to prove that $R$ has no identity, since even in rings without an identity, we can set $a=e=0$ and have $ea=a$.
answered Dec 3 '18 at 14:34
5xum
89.5k393161
answered Dec 3 '18 at 14:34
5xum
89.5k393161
answered Dec 3 '18 at 14:34
5xum
89.5k393161
answered Dec 3 '18 at 14:34
5xum
89.5k393161
89.5k393161
Thanks for this. You said what I wanted to, and probably better.
– Lubin
Dec 3 '18 at 21:24
add a comment |
Thanks for this. You said what I wanted to, and probably better.
– Lubin
Dec 3 '18 at 21:24
Thanks for this. You said what I wanted to, and probably better.
– Lubin
Dec 3 '18 at 21:24
Thanks for this. You said what I wanted to, and probably better.
– Lubin
Dec 3 '18 at 21:24
add a comment |
No, you cannot. In any ring (with or without identity) such elements exist: just take $e=a=0$.
answered Dec 3 '18 at 14:25
José Carlos Santos
150k22122223
add a comment |
No, you cannot. In any ring (with or without identity) such elements exist: just take $e=a=0$.
answered Dec 3 '18 at 14:25
José Carlos Santos
150k22122223
add a comment |
No, you cannot. In any ring (with or without identity) such elements exist: just take $e=a=0$.
answered Dec 3 '18 at 14:25
José Carlos Santos
150k22122223
No, you cannot. In any ring (with or without identity) such elements exist: just take $e=a=0$.
answered Dec 3 '18 at 14:25
José Carlos Santos
150k22122223
answered Dec 3 '18 at 14:25
José Carlos Santos
150k22122223
answered Dec 3 '18 at 14:25
José Carlos Santos
150k22122223
answered Dec 3 '18 at 14:25
José Carlos Santos
150k22122223
150k22122223
add a comment |
add a comment |
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For $e$ to be an identity we need $ea = a$ for all $a$, not just one single $a$. So you can‘t conclude what you said.
– Lukas Kofler
Dec 3 '18 at 14:25