In ring theory, is every real number divisible by any other nonzero real number, precisely because nonzero...












0














In ring theory, is every real number divisible by any other nonzero real number, precisely because nonzero division is closed in $mathbb{R}$?



And so in general this idea of "closure under division" can be used to describe divisibility in a ring? E.g. one might say:



$mathbb{R}^*$ is closed under division, hence each one of its elements is divisible by any other one. (Which for example, in $mathbb{Z}$, remains false.)










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    That's like saying an integer plus an integer is an integer, "because" addition is closed in $Bbb Z$. It doesn't explain anything, it's just two ways of saying the same thing.
    – TonyK
    Dec 26 at 2:49


















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In ring theory, is every real number divisible by any other nonzero real number, precisely because nonzero division is closed in $mathbb{R}$?



And so in general this idea of "closure under division" can be used to describe divisibility in a ring? E.g. one might say:



$mathbb{R}^*$ is closed under division, hence each one of its elements is divisible by any other one. (Which for example, in $mathbb{Z}$, remains false.)










share|cite|improve this question


















  • 4




    That's like saying an integer plus an integer is an integer, "because" addition is closed in $Bbb Z$. It doesn't explain anything, it's just two ways of saying the same thing.
    – TonyK
    Dec 26 at 2:49
















0












0








0







In ring theory, is every real number divisible by any other nonzero real number, precisely because nonzero division is closed in $mathbb{R}$?



And so in general this idea of "closure under division" can be used to describe divisibility in a ring? E.g. one might say:



$mathbb{R}^*$ is closed under division, hence each one of its elements is divisible by any other one. (Which for example, in $mathbb{Z}$, remains false.)










share|cite|improve this question













In ring theory, is every real number divisible by any other nonzero real number, precisely because nonzero division is closed in $mathbb{R}$?



And so in general this idea of "closure under division" can be used to describe divisibility in a ring? E.g. one might say:



$mathbb{R}^*$ is closed under division, hence each one of its elements is divisible by any other one. (Which for example, in $mathbb{Z}$, remains false.)







abstract-algebra ring-theory






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asked Dec 26 at 2:25









Stephen

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  • 4




    That's like saying an integer plus an integer is an integer, "because" addition is closed in $Bbb Z$. It doesn't explain anything, it's just two ways of saying the same thing.
    – TonyK
    Dec 26 at 2:49
















  • 4




    That's like saying an integer plus an integer is an integer, "because" addition is closed in $Bbb Z$. It doesn't explain anything, it's just two ways of saying the same thing.
    – TonyK
    Dec 26 at 2:49










4




4




That's like saying an integer plus an integer is an integer, "because" addition is closed in $Bbb Z$. It doesn't explain anything, it's just two ways of saying the same thing.
– TonyK
Dec 26 at 2:49






That's like saying an integer plus an integer is an integer, "because" addition is closed in $Bbb Z$. It doesn't explain anything, it's just two ways of saying the same thing.
– TonyK
Dec 26 at 2:49












3 Answers
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The answer to the first question is unequivocally yes. This is why it's important that divisibility (and hence primality) is only meaningful relative to a base ring.



Mainly answered to say that since the previous answers weren't being sufficiently definite about it. That said:



As to the second question, one would normally just say that $mathbb{R}$ is a field. An element that every element is divisible by is called a unit, so you equivalently say that every nonzero element is a unit. The reverse condition, being divisible by every nonzero element, doesn't have a name I'm aware of, but is a nontrivial condition in certain rings.



People are objecting to the term "closure" on the grounds that it doesn't make sense except in the context of an operation defined on a larger set. This is a valid objection which also applies to a common way you'll see group theory taught: it's fine to say "a subgroup is a subset closed under the group operations," but it's not okay to say that "a group is a set closed under multiplication such that..." because what would it even mean for it not to be closed?






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    I don't think it is wise to talk about the division operator being closed in the ring minus the $0$ element, but rather that every nonzero element has a inverse element such that $xcdot x^{-1}=1$. Every real number "is divisible" by every nonzero real number because every nonzero real number has a multiplicative inverse.



    I want to make this seemingly pedantic distinction because in a more general ring it is not clear what "division" should be for elements that don't already have a defined inverse. The best we can do is build the field of fractions, but again it is all defined in terms of the multiplication operation and the idea of what a multiplicative inverse is.






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      Well, yes, it is true that if $dinmathbb R setminus {0}$ and $ain mathbb R$, then $d|a$, since this means
      $$exists kin mathbb R, ;a=dk,$$
      (and is enough to take $k=frac ad$). But this is equivalent to have a multiplicative inverse for every non-zero element of the ring, which implies that this is in fact a field (or a division ring, if it were not commutative).



      I would not say that $mathbb R^*$ is "closed under division", since for a ring $R$ the set $R^*$ is by definition the set of elements of $R$ with an inverse, so this is kind of the same. For instance, $$mathbb Z^*={-1,1},$$ and so $mathbb Z^*$ is also "closed under division", since
      $$1|1, quad 1|-1, quad -1|1 quad text{and}quad -1|-1.$$






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        3 Answers
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        3 Answers
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        The answer to the first question is unequivocally yes. This is why it's important that divisibility (and hence primality) is only meaningful relative to a base ring.



        Mainly answered to say that since the previous answers weren't being sufficiently definite about it. That said:



        As to the second question, one would normally just say that $mathbb{R}$ is a field. An element that every element is divisible by is called a unit, so you equivalently say that every nonzero element is a unit. The reverse condition, being divisible by every nonzero element, doesn't have a name I'm aware of, but is a nontrivial condition in certain rings.



        People are objecting to the term "closure" on the grounds that it doesn't make sense except in the context of an operation defined on a larger set. This is a valid objection which also applies to a common way you'll see group theory taught: it's fine to say "a subgroup is a subset closed under the group operations," but it's not okay to say that "a group is a set closed under multiplication such that..." because what would it even mean for it not to be closed?






        share|cite|improve this answer




























          1














          The answer to the first question is unequivocally yes. This is why it's important that divisibility (and hence primality) is only meaningful relative to a base ring.



          Mainly answered to say that since the previous answers weren't being sufficiently definite about it. That said:



          As to the second question, one would normally just say that $mathbb{R}$ is a field. An element that every element is divisible by is called a unit, so you equivalently say that every nonzero element is a unit. The reverse condition, being divisible by every nonzero element, doesn't have a name I'm aware of, but is a nontrivial condition in certain rings.



          People are objecting to the term "closure" on the grounds that it doesn't make sense except in the context of an operation defined on a larger set. This is a valid objection which also applies to a common way you'll see group theory taught: it's fine to say "a subgroup is a subset closed under the group operations," but it's not okay to say that "a group is a set closed under multiplication such that..." because what would it even mean for it not to be closed?






          share|cite|improve this answer


























            1












            1








            1






            The answer to the first question is unequivocally yes. This is why it's important that divisibility (and hence primality) is only meaningful relative to a base ring.



            Mainly answered to say that since the previous answers weren't being sufficiently definite about it. That said:



            As to the second question, one would normally just say that $mathbb{R}$ is a field. An element that every element is divisible by is called a unit, so you equivalently say that every nonzero element is a unit. The reverse condition, being divisible by every nonzero element, doesn't have a name I'm aware of, but is a nontrivial condition in certain rings.



            People are objecting to the term "closure" on the grounds that it doesn't make sense except in the context of an operation defined on a larger set. This is a valid objection which also applies to a common way you'll see group theory taught: it's fine to say "a subgroup is a subset closed under the group operations," but it's not okay to say that "a group is a set closed under multiplication such that..." because what would it even mean for it not to be closed?






            share|cite|improve this answer














            The answer to the first question is unequivocally yes. This is why it's important that divisibility (and hence primality) is only meaningful relative to a base ring.



            Mainly answered to say that since the previous answers weren't being sufficiently definite about it. That said:



            As to the second question, one would normally just say that $mathbb{R}$ is a field. An element that every element is divisible by is called a unit, so you equivalently say that every nonzero element is a unit. The reverse condition, being divisible by every nonzero element, doesn't have a name I'm aware of, but is a nontrivial condition in certain rings.



            People are objecting to the term "closure" on the grounds that it doesn't make sense except in the context of an operation defined on a larger set. This is a valid objection which also applies to a common way you'll see group theory taught: it's fine to say "a subgroup is a subset closed under the group operations," but it's not okay to say that "a group is a set closed under multiplication such that..." because what would it even mean for it not to be closed?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 26 at 3:18

























            answered Dec 26 at 3:03









            Daniel McLaury

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            15.5k32977























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                I don't think it is wise to talk about the division operator being closed in the ring minus the $0$ element, but rather that every nonzero element has a inverse element such that $xcdot x^{-1}=1$. Every real number "is divisible" by every nonzero real number because every nonzero real number has a multiplicative inverse.



                I want to make this seemingly pedantic distinction because in a more general ring it is not clear what "division" should be for elements that don't already have a defined inverse. The best we can do is build the field of fractions, but again it is all defined in terms of the multiplication operation and the idea of what a multiplicative inverse is.






                share|cite|improve this answer








                New contributor




                ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                  1














                  I don't think it is wise to talk about the division operator being closed in the ring minus the $0$ element, but rather that every nonzero element has a inverse element such that $xcdot x^{-1}=1$. Every real number "is divisible" by every nonzero real number because every nonzero real number has a multiplicative inverse.



                  I want to make this seemingly pedantic distinction because in a more general ring it is not clear what "division" should be for elements that don't already have a defined inverse. The best we can do is build the field of fractions, but again it is all defined in terms of the multiplication operation and the idea of what a multiplicative inverse is.






                  share|cite|improve this answer








                  New contributor




                  ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





















                    1












                    1








                    1






                    I don't think it is wise to talk about the division operator being closed in the ring minus the $0$ element, but rather that every nonzero element has a inverse element such that $xcdot x^{-1}=1$. Every real number "is divisible" by every nonzero real number because every nonzero real number has a multiplicative inverse.



                    I want to make this seemingly pedantic distinction because in a more general ring it is not clear what "division" should be for elements that don't already have a defined inverse. The best we can do is build the field of fractions, but again it is all defined in terms of the multiplication operation and the idea of what a multiplicative inverse is.






                    share|cite|improve this answer








                    New contributor




                    ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    I don't think it is wise to talk about the division operator being closed in the ring minus the $0$ element, but rather that every nonzero element has a inverse element such that $xcdot x^{-1}=1$. Every real number "is divisible" by every nonzero real number because every nonzero real number has a multiplicative inverse.



                    I want to make this seemingly pedantic distinction because in a more general ring it is not clear what "division" should be for elements that don't already have a defined inverse. The best we can do is build the field of fractions, but again it is all defined in terms of the multiplication operation and the idea of what a multiplicative inverse is.







                    share|cite|improve this answer








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                    ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                    share|cite|improve this answer



                    share|cite|improve this answer






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                    answered Dec 26 at 2:30









                    ImNotTheGuy

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                        0














                        Well, yes, it is true that if $dinmathbb R setminus {0}$ and $ain mathbb R$, then $d|a$, since this means
                        $$exists kin mathbb R, ;a=dk,$$
                        (and is enough to take $k=frac ad$). But this is equivalent to have a multiplicative inverse for every non-zero element of the ring, which implies that this is in fact a field (or a division ring, if it were not commutative).



                        I would not say that $mathbb R^*$ is "closed under division", since for a ring $R$ the set $R^*$ is by definition the set of elements of $R$ with an inverse, so this is kind of the same. For instance, $$mathbb Z^*={-1,1},$$ and so $mathbb Z^*$ is also "closed under division", since
                        $$1|1, quad 1|-1, quad -1|1 quad text{and}quad -1|-1.$$






                        share|cite|improve this answer




























                          0














                          Well, yes, it is true that if $dinmathbb R setminus {0}$ and $ain mathbb R$, then $d|a$, since this means
                          $$exists kin mathbb R, ;a=dk,$$
                          (and is enough to take $k=frac ad$). But this is equivalent to have a multiplicative inverse for every non-zero element of the ring, which implies that this is in fact a field (or a division ring, if it were not commutative).



                          I would not say that $mathbb R^*$ is "closed under division", since for a ring $R$ the set $R^*$ is by definition the set of elements of $R$ with an inverse, so this is kind of the same. For instance, $$mathbb Z^*={-1,1},$$ and so $mathbb Z^*$ is also "closed under division", since
                          $$1|1, quad 1|-1, quad -1|1 quad text{and}quad -1|-1.$$






                          share|cite|improve this answer


























                            0












                            0








                            0






                            Well, yes, it is true that if $dinmathbb R setminus {0}$ and $ain mathbb R$, then $d|a$, since this means
                            $$exists kin mathbb R, ;a=dk,$$
                            (and is enough to take $k=frac ad$). But this is equivalent to have a multiplicative inverse for every non-zero element of the ring, which implies that this is in fact a field (or a division ring, if it were not commutative).



                            I would not say that $mathbb R^*$ is "closed under division", since for a ring $R$ the set $R^*$ is by definition the set of elements of $R$ with an inverse, so this is kind of the same. For instance, $$mathbb Z^*={-1,1},$$ and so $mathbb Z^*$ is also "closed under division", since
                            $$1|1, quad 1|-1, quad -1|1 quad text{and}quad -1|-1.$$






                            share|cite|improve this answer














                            Well, yes, it is true that if $dinmathbb R setminus {0}$ and $ain mathbb R$, then $d|a$, since this means
                            $$exists kin mathbb R, ;a=dk,$$
                            (and is enough to take $k=frac ad$). But this is equivalent to have a multiplicative inverse for every non-zero element of the ring, which implies that this is in fact a field (or a division ring, if it were not commutative).



                            I would not say that $mathbb R^*$ is "closed under division", since for a ring $R$ the set $R^*$ is by definition the set of elements of $R$ with an inverse, so this is kind of the same. For instance, $$mathbb Z^*={-1,1},$$ and so $mathbb Z^*$ is also "closed under division", since
                            $$1|1, quad 1|-1, quad -1|1 quad text{and}quad -1|-1.$$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 26 at 2:50

























                            answered Dec 26 at 2:41









                            Alejandro Nasif Salum

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