Why are real and complex numbers consider rings when irrationals aren't? [closed]
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Irrational numbers are not closed under multiplication, then why are real and complex numbers considered rings? Aren't real and complex numbers supposed to have irrational numbers?
*Made the mistake of thinking if subsets were to be invalid, the higher sets would be invalid too. Thanks to @Zachary Selk for clarify it!
abstract-algebra ring-theory
asked Nov 16 at 5:49
John
12
closed as unclear what you're asking by Bungo, Morgan Rodgers, José Carlos Santos, amWhy, rschwieb Nov 16 at 12:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
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Irrational numbers are not closed under multiplication, then why are real and complex numbers considered rings? Aren't real and complex numbers supposed to have irrational numbers?
*Made the mistake of thinking if subsets were to be invalid, the higher sets would be invalid too. Thanks to @Zachary Selk for clarify it!
abstract-algebra ring-theory
asked Nov 16 at 5:49
John
12
closed as unclear what you're asking by Bungo, Morgan Rodgers, José Carlos Santos, amWhy, rschwieb Nov 16 at 12:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
Irrationals are not closed under addition/multiplication.
– Lord Shark the Unknown
Nov 16 at 5:51
3
You seem to be arguing that if $A$ is not a ring and $Asubset B$ then $B$ is not a ring. This is not correct.
– MPW
Nov 16 at 6:24
Using the operations of the real numbers, the product of two irrationals isn't necessarily irrational, the sum of two irrationals isn't necessarily irrational, and there is no irrational additive identity, so the set of irrational numbers wildly fails the criteria for being a subring of $mathbb R$. Of course, you could make them into a ring with some weird product, but that isn't apparently what you're talking about.
– rschwieb
Nov 16 at 19:52
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Irrational numbers are not closed under multiplication, then why are real and complex numbers considered rings? Aren't real and complex numbers supposed to have irrational numbers?
*Made the mistake of thinking if subsets were to be invalid, the higher sets would be invalid too. Thanks to @Zachary Selk for clarify it!
abstract-algebra ring-theory
asked Nov 16 at 5:49
John
12
Irrational numbers are not closed under multiplication, then why are real and complex numbers considered rings? Aren't real and complex numbers supposed to have irrational numbers?
*Made the mistake of thinking if subsets were to be invalid, the higher sets would be invalid too. Thanks to @Zachary Selk for clarify it!
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Nov 16 at 5:49
John
12
asked Nov 16 at 5:49
John
12
edited Nov 17 at 5:02
asked Nov 16 at 5:49
John
12
asked Nov 16 at 5:49
John
12
asked Nov 16 at 5:49
John
12
12
closed as unclear what you're asking by Bungo, Morgan Rodgers, José Carlos Santos, amWhy, rschwieb Nov 16 at 12:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Bungo, Morgan Rodgers, José Carlos Santos, amWhy, rschwieb Nov 16 at 12:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
Irrationals are not closed under addition/multiplication.
– Lord Shark the Unknown
Nov 16 at 5:51
3
You seem to be arguing that if $A$ is not a ring and $Asubset B$ then $B$ is not a ring. This is not correct.
– MPW
Nov 16 at 6:24
Using the operations of the real numbers, the product of two irrationals isn't necessarily irrational, the sum of two irrationals isn't necessarily irrational, and there is no irrational additive identity, so the set of irrational numbers wildly fails the criteria for being a subring of $mathbb R$. Of course, you could make them into a ring with some weird product, but that isn't apparently what you're talking about.
– rschwieb
Nov 16 at 19:52
add a comment |
Irrationals are not closed under addition/multiplication.
– Lord Shark the Unknown
Nov 16 at 5:51
3
You seem to be arguing that if $A$ is not a ring and $Asubset B$ then $B$ is not a ring. This is not correct.
– MPW
Nov 16 at 6:24
Using the operations of the real numbers, the product of two irrationals isn't necessarily irrational, the sum of two irrationals isn't necessarily irrational, and there is no irrational additive identity, so the set of irrational numbers wildly fails the criteria for being a subring of $mathbb R$. Of course, you could make them into a ring with some weird product, but that isn't apparently what you're talking about.
– rschwieb
Nov 16 at 19:52
Irrationals are not closed under addition/multiplication.
– Lord Shark the Unknown
Nov 16 at 5:51
Irrationals are not closed under addition/multiplication.
– Lord Shark the Unknown
Nov 16 at 5:51
3
3
You seem to be arguing that if $A$ is not a ring and $Asubset B$ then $B$ is not a ring. This is not correct.
– MPW
Nov 16 at 6:24
You seem to be arguing that if $A$ is not a ring and $Asubset B$ then $B$ is not a ring. This is not correct.
– MPW
Nov 16 at 6:24
Using the operations of the real numbers, the product of two irrationals isn't necessarily irrational, the sum of two irrationals isn't necessarily irrational, and there is no irrational additive identity, so the set of irrational numbers wildly fails the criteria for being a subring of $mathbb R$. Of course, you could make them into a ring with some weird product, but that isn't apparently what you're talking about.
– rschwieb
Nov 16 at 19:52
Using the operations of the real numbers, the product of two irrationals isn't necessarily irrational, the sum of two irrationals isn't necessarily irrational, and there is no irrational additive identity, so the set of irrational numbers wildly fails the criteria for being a subring of $mathbb R$. Of course, you could make them into a ring with some weird product, but that isn't apparently what you're talking about.
– rschwieb
Nov 16 at 19:52
add a comment |
3 Answers
3
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up vote
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The irrationals are not closed under multiplication, but the reals and complex numbers are. Why should the irrationals not being closed matter to whether the reals and complex numbers are rings? ${2}$ is not closed under multiplication (or addition), but it is a subset of the integers, which form a ring.
answered Nov 16 at 5:51
Ross Millikan
287k23195364
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up vote
1
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Yes, real numbers do indeed contain irrational numbers. However, the only thing irrationals can "escape to" are rationals, i.e. reals. So reals are a ring.
As a matter of fact, every nonzero ring contains a subset that is not a ring.
answered Nov 16 at 5:55
Zachary Selk
412211
Yea, I made the mistake of thinking if a subset is invalid, the higher set would be too. Sort of a bad question on my part but thanks for the simple explanation. This answers my question best, thanks.
– John
Nov 17 at 5:00
add a comment |
up vote
0
down vote
Well, the irrationals are not closed under multiplication, e.g., $sqrt 2 cdot sqrt 2 = 2$. And what would be the zero and unit elements?
However, what is interesting is that one can define a ring structure on the set ${Bbb I}$ of irrational numbers by using (any) bijection $phi:{Bbb I}rightarrow{Bbb R}$.
answered Nov 16 at 6:33
Wuestenfux
2,4791410
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The irrationals are not closed under multiplication, but the reals and complex numbers are. Why should the irrationals not being closed matter to whether the reals and complex numbers are rings? ${2}$ is not closed under multiplication (or addition), but it is a subset of the integers, which form a ring.
answered Nov 16 at 5:51
Ross Millikan
287k23195364
add a comment |
up vote
3
down vote
The irrationals are not closed under multiplication, but the reals and complex numbers are. Why should the irrationals not being closed matter to whether the reals and complex numbers are rings? ${2}$ is not closed under multiplication (or addition), but it is a subset of the integers, which form a ring.
answered Nov 16 at 5:51
Ross Millikan
287k23195364
add a comment |
up vote
3
down vote
up vote
3
down vote
The irrationals are not closed under multiplication, but the reals and complex numbers are. Why should the irrationals not being closed matter to whether the reals and complex numbers are rings? ${2}$ is not closed under multiplication (or addition), but it is a subset of the integers, which form a ring.
answered Nov 16 at 5:51
Ross Millikan
287k23195364
The irrationals are not closed under multiplication, but the reals and complex numbers are. Why should the irrationals not being closed matter to whether the reals and complex numbers are rings? ${2}$ is not closed under multiplication (or addition), but it is a subset of the integers, which form a ring.
answered Nov 16 at 5:51
Ross Millikan
287k23195364
answered Nov 16 at 5:51
Ross Millikan
287k23195364
answered Nov 16 at 5:51
Ross Millikan
287k23195364
answered Nov 16 at 5:51
Ross Millikan
287k23195364
287k23195364
add a comment |
add a comment |
up vote
1
down vote
Yes, real numbers do indeed contain irrational numbers. However, the only thing irrationals can "escape to" are rationals, i.e. reals. So reals are a ring.
As a matter of fact, every nonzero ring contains a subset that is not a ring.
answered Nov 16 at 5:55
Zachary Selk
412211
Yea, I made the mistake of thinking if a subset is invalid, the higher set would be too. Sort of a bad question on my part but thanks for the simple explanation. This answers my question best, thanks.
– John
Nov 17 at 5:00
add a comment |
up vote
1
down vote
Yes, real numbers do indeed contain irrational numbers. However, the only thing irrationals can "escape to" are rationals, i.e. reals. So reals are a ring.
As a matter of fact, every nonzero ring contains a subset that is not a ring.
answered Nov 16 at 5:55
Zachary Selk
412211
Yea, I made the mistake of thinking if a subset is invalid, the higher set would be too. Sort of a bad question on my part but thanks for the simple explanation. This answers my question best, thanks.
– John
Nov 17 at 5:00
add a comment |
up vote
1
down vote
up vote
1
down vote
Yes, real numbers do indeed contain irrational numbers. However, the only thing irrationals can "escape to" are rationals, i.e. reals. So reals are a ring.
As a matter of fact, every nonzero ring contains a subset that is not a ring.
answered Nov 16 at 5:55
Zachary Selk
412211
Yes, real numbers do indeed contain irrational numbers. However, the only thing irrationals can "escape to" are rationals, i.e. reals. So reals are a ring.
As a matter of fact, every nonzero ring contains a subset that is not a ring.
answered Nov 16 at 5:55
Zachary Selk
412211
answered Nov 16 at 5:55
Zachary Selk
412211
answered Nov 16 at 5:55
Zachary Selk
412211
answered Nov 16 at 5:55
Zachary Selk
412211
412211
Yea, I made the mistake of thinking if a subset is invalid, the higher set would be too. Sort of a bad question on my part but thanks for the simple explanation. This answers my question best, thanks.
– John
Nov 17 at 5:00
add a comment |
Yea, I made the mistake of thinking if a subset is invalid, the higher set would be too. Sort of a bad question on my part but thanks for the simple explanation. This answers my question best, thanks.
– John
Nov 17 at 5:00
Yea, I made the mistake of thinking if a subset is invalid, the higher set would be too. Sort of a bad question on my part but thanks for the simple explanation. This answers my question best, thanks.
– John
Nov 17 at 5:00
Yea, I made the mistake of thinking if a subset is invalid, the higher set would be too. Sort of a bad question on my part but thanks for the simple explanation. This answers my question best, thanks.
– John
Nov 17 at 5:00
add a comment |
up vote
0
down vote
Well, the irrationals are not closed under multiplication, e.g., $sqrt 2 cdot sqrt 2 = 2$. And what would be the zero and unit elements?
However, what is interesting is that one can define a ring structure on the set ${Bbb I}$ of irrational numbers by using (any) bijection $phi:{Bbb I}rightarrow{Bbb R}$.
answered Nov 16 at 6:33
Wuestenfux
2,4791410
add a comment |
up vote
0
down vote
Well, the irrationals are not closed under multiplication, e.g., $sqrt 2 cdot sqrt 2 = 2$. And what would be the zero and unit elements?
However, what is interesting is that one can define a ring structure on the set ${Bbb I}$ of irrational numbers by using (any) bijection $phi:{Bbb I}rightarrow{Bbb R}$.
answered Nov 16 at 6:33
Wuestenfux
2,4791410
add a comment |
up vote
0
down vote
up vote
0
down vote
Well, the irrationals are not closed under multiplication, e.g., $sqrt 2 cdot sqrt 2 = 2$. And what would be the zero and unit elements?
However, what is interesting is that one can define a ring structure on the set ${Bbb I}$ of irrational numbers by using (any) bijection $phi:{Bbb I}rightarrow{Bbb R}$.
answered Nov 16 at 6:33
Wuestenfux
2,4791410
Well, the irrationals are not closed under multiplication, e.g., $sqrt 2 cdot sqrt 2 = 2$. And what would be the zero and unit elements?
However, what is interesting is that one can define a ring structure on the set ${Bbb I}$ of irrational numbers by using (any) bijection $phi:{Bbb I}rightarrow{Bbb R}$.
answered Nov 16 at 6:33
Wuestenfux
2,4791410
edited Nov 16 at 8:47
answered Nov 16 at 6:33
Wuestenfux
2,4791410
answered Nov 16 at 6:33
Wuestenfux
2,4791410
answered Nov 16 at 6:33
Wuestenfux
2,4791410
2,4791410
add a comment |
add a comment |
Irrationals are not closed under addition/multiplication.
– Lord Shark the Unknown
Nov 16 at 5:51
3
You seem to be arguing that if $A$ is not a ring and $Asubset B$ then $B$ is not a ring. This is not correct.
– MPW
Nov 16 at 6:24
Using the operations of the real numbers, the product of two irrationals isn't necessarily irrational, the sum of two irrationals isn't necessarily irrational, and there is no irrational additive identity, so the set of irrational numbers wildly fails the criteria for being a subring of $mathbb R$. Of course, you could make them into a ring with some weird product, but that isn't apparently what you're talking about.
– rschwieb
Nov 16 at 19:52