Mixing
Meadows with usual real division
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A "meadow" is a commutative ring with a multiplicative identity element and a total multiplicative inverse operation satisfying the two equations $(x^-1)^-1 = x$ and $x times (x times x^-1) = x$.
Is there a meadow in which the inverses for all reals (except zero) are the usual inverses of the field of real numbers?
References would be appreciated. Thanks.
abstract-algebra reference-request
asked Jul 31 at 15:10
YeatsL
815
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up vote
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A "meadow" is a commutative ring with a multiplicative identity element and a total multiplicative inverse operation satisfying the two equations $(x^-1)^-1 = x$ and $x times (x times x^-1) = x$.
Is there a meadow in which the inverses for all reals (except zero) are the usual inverses of the field of real numbers?
References would be appreciated. Thanks.
abstract-algebra reference-request
asked Jul 31 at 15:10
YeatsL
815
Yes, of course. Define $0^-1:=0$ on $Bbb R$.
– Berci
Jul 31 at 15:14
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A "meadow" is a commutative ring with a multiplicative identity element and a total multiplicative inverse operation satisfying the two equations $(x^-1)^-1 = x$ and $x times (x times x^-1) = x$.
Is there a meadow in which the inverses for all reals (except zero) are the usual inverses of the field of real numbers?
References would be appreciated. Thanks.
abstract-algebra reference-request
asked Jul 31 at 15:10
YeatsL
815
A "meadow" is a commutative ring with a multiplicative identity element and a total multiplicative inverse operation satisfying the two equations $(x^-1)^-1 = x$ and $x times (x times x^-1) = x$.
Is there a meadow in which the inverses for all reals (except zero) are the usual inverses of the field of real numbers?
References would be appreciated. Thanks.
abstract-algebra reference-request
asked Jul 31 at 15:10
YeatsL
815
asked Jul 31 at 15:10
YeatsL
815
asked Jul 31 at 15:10
YeatsL
815
asked Jul 31 at 15:10
YeatsL
815
815
Yes, of course. Define $0^-1:=0$ on $Bbb R$.
– Berci
Jul 31 at 15:14
add a comment |Â
Yes, of course. Define $0^-1:=0$ on $Bbb R$.
– Berci
Jul 31 at 15:14
Yes, of course. Define $0^-1:=0$ on $Bbb R$.
– Berci
Jul 31 at 15:14
Yes, of course. Define $0^-1:=0$ on $Bbb R$.
– Berci
Jul 31 at 15:14
add a comment |Â
1 Answer
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Quoting from the abstract of the paper which apparently introduced meadows:
We study a new axiomatic concept for number systems with division that uses only equations: a meadow is a commutative ring with a total inverse operator satisfying two equations which imply that the inverse of zero is zero. All fields and products of fields can be viewed as meadows.
In other words, since $mathbbR$ is a field you just need to extend it with $0^-1 = 0$.
answered Jul 31 at 15:27
Peter Taylor
7,67012239
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Quoting from the abstract of the paper which apparently introduced meadows:
We study a new axiomatic concept for number systems with division that uses only equations: a meadow is a commutative ring with a total inverse operator satisfying two equations which imply that the inverse of zero is zero. All fields and products of fields can be viewed as meadows.
In other words, since $mathbbR$ is a field you just need to extend it with $0^-1 = 0$.
answered Jul 31 at 15:27
Peter Taylor
7,67012239
add a comment |Â
up vote
2
down vote
accepted
Quoting from the abstract of the paper which apparently introduced meadows:
We study a new axiomatic concept for number systems with division that uses only equations: a meadow is a commutative ring with a total inverse operator satisfying two equations which imply that the inverse of zero is zero. All fields and products of fields can be viewed as meadows.
In other words, since $mathbbR$ is a field you just need to extend it with $0^-1 = 0$.
answered Jul 31 at 15:27
Peter Taylor
7,67012239
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Quoting from the abstract of the paper which apparently introduced meadows:
We study a new axiomatic concept for number systems with division that uses only equations: a meadow is a commutative ring with a total inverse operator satisfying two equations which imply that the inverse of zero is zero. All fields and products of fields can be viewed as meadows.
In other words, since $mathbbR$ is a field you just need to extend it with $0^-1 = 0$.
answered Jul 31 at 15:27
Peter Taylor
7,67012239
Quoting from the abstract of the paper which apparently introduced meadows:
We study a new axiomatic concept for number systems with division that uses only equations: a meadow is a commutative ring with a total inverse operator satisfying two equations which imply that the inverse of zero is zero. All fields and products of fields can be viewed as meadows.
In other words, since $mathbbR$ is a field you just need to extend it with $0^-1 = 0$.
answered Jul 31 at 15:27
Peter Taylor
7,67012239
answered Jul 31 at 15:27
Peter Taylor
7,67012239
answered Jul 31 at 15:27
Peter Taylor
7,67012239
answered Jul 31 at 15:27
Peter Taylor
7,67012239
7,67012239
add a comment |Â
add a comment |Â
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Yes, of course. Define $0^-1:=0$ on $Bbb R$.
– Berci
Jul 31 at 15:14