Mixing
How do I interpret the formating of these given structures and prove they are not isomorphic?
$begingroup$
$(Bbb{Z}^+,cdot)$ and $(Bbb{Q}^+,cdot)$
First and foremost, I am not looking for a direct answer just a method of interpreting what these "structures" are, and some techniques I can use to prove they are not isomorphic. I have recently started Abstract Algebra I and keep hearing Cardinality and the form $f(ab)=f(a)cdot f(b)$ while looking throughout the internet, but I haven't heard this covered in class.
An attempt at interpretation:
I assume the first structure is a set of all number that can be represented as a multiplication of two positive integers (meaning all positive integers), and the second is the same with positive rational numbers (meaning all positive rational numbers).
An attempt at proving they are not isomorphic:
Because isomorphic means "to have the same shape" I assume this means that the two sets need to cover the same range of numbers. Because the second set can output non-integer values while the first set can't this would prove their isomorphic.
If I am misinterpreting what isomorphic means or if there is a better way to go about proving it I would greatly appreciate any input. Also, I'd like to apologize if I messed up the tags or formatting, I tried to follow the guidelines to the best of my ability; a first-time poster.
abstract-algebra elementary-set-theory
asked Jan 20 at 14:30
NoalineNoaline
62
$endgroup$
add a comment |
$begingroup$
$(Bbb{Z}^+,cdot)$ and $(Bbb{Q}^+,cdot)$
First and foremost, I am not looking for a direct answer just a method of interpreting what these "structures" are, and some techniques I can use to prove they are not isomorphic. I have recently started Abstract Algebra I and keep hearing Cardinality and the form $f(ab)=f(a)cdot f(b)$ while looking throughout the internet, but I haven't heard this covered in class.
An attempt at interpretation:
I assume the first structure is a set of all number that can be represented as a multiplication of two positive integers (meaning all positive integers), and the second is the same with positive rational numbers (meaning all positive rational numbers).
An attempt at proving they are not isomorphic:
Because isomorphic means "to have the same shape" I assume this means that the two sets need to cover the same range of numbers. Because the second set can output non-integer values while the first set can't this would prove their isomorphic.
If I am misinterpreting what isomorphic means or if there is a better way to go about proving it I would greatly appreciate any input. Also, I'd like to apologize if I messed up the tags or formatting, I tried to follow the guidelines to the best of my ability; a first-time poster.
abstract-algebra elementary-set-theory
asked Jan 20 at 14:30
NoalineNoaline
62
$endgroup$
$begingroup$
$mathbb{Z}^+$ with product is not a group. $2$ doesn't have an inverse.
$endgroup$
– Yanko
Jan 20 at 14:35
$begingroup$
I'm not sure I follow. Why is the first not a group (what would it be classified as), and how is it known that the second group has no inverse and what would that mean about being isomorphic with group 1. Sorry for so many questions, I am really trying to get a grasp on this.
$endgroup$
– Noaline
Jan 20 at 14:41
1
$begingroup$
You should read about the definition of a group here en.wikipedia.org/wiki/… The first is not a group because $2$ has no inverse as $2a=1$ implies that $a=frac{1}{2}$ which is not an integer. Group without inverse is sometimes called a semi-group and it can't be isomorphic to a group.
$endgroup$
– Yanko
Jan 20 at 14:46
1
$begingroup$
Oh, so groups can't be isomorphic with non-groups, and there are certain criteria for being a group that must be checked? If my understanding is correct, thank you very much, I had spent all night last night trying to figure this out.
$endgroup$
– Noaline
Jan 20 at 14:51
add a comment |
$begingroup$
$(Bbb{Z}^+,cdot)$ and $(Bbb{Q}^+,cdot)$
First and foremost, I am not looking for a direct answer just a method of interpreting what these "structures" are, and some techniques I can use to prove they are not isomorphic. I have recently started Abstract Algebra I and keep hearing Cardinality and the form $f(ab)=f(a)cdot f(b)$ while looking throughout the internet, but I haven't heard this covered in class.
An attempt at interpretation:
I assume the first structure is a set of all number that can be represented as a multiplication of two positive integers (meaning all positive integers), and the second is the same with positive rational numbers (meaning all positive rational numbers).
An attempt at proving they are not isomorphic:
Because isomorphic means "to have the same shape" I assume this means that the two sets need to cover the same range of numbers. Because the second set can output non-integer values while the first set can't this would prove their isomorphic.
If I am misinterpreting what isomorphic means or if there is a better way to go about proving it I would greatly appreciate any input. Also, I'd like to apologize if I messed up the tags or formatting, I tried to follow the guidelines to the best of my ability; a first-time poster.
abstract-algebra elementary-set-theory
asked Jan 20 at 14:30
NoalineNoaline
62
$endgroup$
$(Bbb{Z}^+,cdot)$ and $(Bbb{Q}^+,cdot)$
First and foremost, I am not looking for a direct answer just a method of interpreting what these "structures" are, and some techniques I can use to prove they are not isomorphic. I have recently started Abstract Algebra I and keep hearing Cardinality and the form $f(ab)=f(a)cdot f(b)$ while looking throughout the internet, but I haven't heard this covered in class.
An attempt at interpretation:
I assume the first structure is a set of all number that can be represented as a multiplication of two positive integers (meaning all positive integers), and the second is the same with positive rational numbers (meaning all positive rational numbers).
An attempt at proving they are not isomorphic:
Because isomorphic means "to have the same shape" I assume this means that the two sets need to cover the same range of numbers. Because the second set can output non-integer values while the first set can't this would prove their isomorphic.
If I am misinterpreting what isomorphic means or if there is a better way to go about proving it I would greatly appreciate any input. Also, I'd like to apologize if I messed up the tags or formatting, I tried to follow the guidelines to the best of my ability; a first-time poster.
abstract-algebra elementary-set-theory
abstract-algebra elementary-set-theory
asked Jan 20 at 14:30
NoalineNoaline
62
asked Jan 20 at 14:30
NoalineNoaline
62
asked Jan 20 at 14:30
NoalineNoaline
62
asked Jan 20 at 14:30
NoalineNoaline
62
asked Jan 20 at 14:30
NoalineNoaline
62
62
$begingroup$
$mathbb{Z}^+$ with product is not a group. $2$ doesn't have an inverse.
$endgroup$
– Yanko
Jan 20 at 14:35
$begingroup$
I'm not sure I follow. Why is the first not a group (what would it be classified as), and how is it known that the second group has no inverse and what would that mean about being isomorphic with group 1. Sorry for so many questions, I am really trying to get a grasp on this.
$endgroup$
– Noaline
Jan 20 at 14:41
1
$begingroup$
You should read about the definition of a group here en.wikipedia.org/wiki/… The first is not a group because $2$ has no inverse as $2a=1$ implies that $a=frac{1}{2}$ which is not an integer. Group without inverse is sometimes called a semi-group and it can't be isomorphic to a group.
$endgroup$
– Yanko
Jan 20 at 14:46
1
$begingroup$
Oh, so groups can't be isomorphic with non-groups, and there are certain criteria for being a group that must be checked? If my understanding is correct, thank you very much, I had spent all night last night trying to figure this out.
$endgroup$
– Noaline
Jan 20 at 14:51
add a comment |
$begingroup$
$mathbb{Z}^+$ with product is not a group. $2$ doesn't have an inverse.
$endgroup$
– Yanko
Jan 20 at 14:35
$begingroup$
I'm not sure I follow. Why is the first not a group (what would it be classified as), and how is it known that the second group has no inverse and what would that mean about being isomorphic with group 1. Sorry for so many questions, I am really trying to get a grasp on this.
$endgroup$
– Noaline
Jan 20 at 14:41
1
$begingroup$
You should read about the definition of a group here en.wikipedia.org/wiki/… The first is not a group because $2$ has no inverse as $2a=1$ implies that $a=frac{1}{2}$ which is not an integer. Group without inverse is sometimes called a semi-group and it can't be isomorphic to a group.
$endgroup$
– Yanko
Jan 20 at 14:46
1
$begingroup$
Oh, so groups can't be isomorphic with non-groups, and there are certain criteria for being a group that must be checked? If my understanding is correct, thank you very much, I had spent all night last night trying to figure this out.
$endgroup$
– Noaline
Jan 20 at 14:51
$begingroup$
$mathbb{Z}^+$ with product is not a group. $2$ doesn't have an inverse.
$endgroup$
– Yanko
Jan 20 at 14:35
$begingroup$
$mathbb{Z}^+$ with product is not a group. $2$ doesn't have an inverse.
$endgroup$
– Yanko
Jan 20 at 14:35
$begingroup$
I'm not sure I follow. Why is the first not a group (what would it be classified as), and how is it known that the second group has no inverse and what would that mean about being isomorphic with group 1. Sorry for so many questions, I am really trying to get a grasp on this.
$endgroup$
– Noaline
Jan 20 at 14:41
$begingroup$
I'm not sure I follow. Why is the first not a group (what would it be classified as), and how is it known that the second group has no inverse and what would that mean about being isomorphic with group 1. Sorry for so many questions, I am really trying to get a grasp on this.
$endgroup$
– Noaline
Jan 20 at 14:41
1
1
$begingroup$
You should read about the definition of a group here en.wikipedia.org/wiki/… The first is not a group because $2$ has no inverse as $2a=1$ implies that $a=frac{1}{2}$ which is not an integer. Group without inverse is sometimes called a semi-group and it can't be isomorphic to a group.
$endgroup$
– Yanko
Jan 20 at 14:46
$begingroup$
You should read about the definition of a group here en.wikipedia.org/wiki/… The first is not a group because $2$ has no inverse as $2a=1$ implies that $a=frac{1}{2}$ which is not an integer. Group without inverse is sometimes called a semi-group and it can't be isomorphic to a group.
$endgroup$
– Yanko
Jan 20 at 14:46
1
1
$begingroup$
Oh, so groups can't be isomorphic with non-groups, and there are certain criteria for being a group that must be checked? If my understanding is correct, thank you very much, I had spent all night last night trying to figure this out.
$endgroup$
– Noaline
Jan 20 at 14:51
$begingroup$
Oh, so groups can't be isomorphic with non-groups, and there are certain criteria for being a group that must be checked? If my understanding is correct, thank you very much, I had spent all night last night trying to figure this out.
$endgroup$
– Noaline
Jan 20 at 14:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To say that two structures are or are not isomorphic, you need to specify what you are saying they are(n't) isomorphic as. In this case, asking about isomorphism of monoids would be reasonable, since both $mathbb{Z}^+$ and $mathbb{Q}^+$ are monoids under multiplication.
Fortunately in this case, any reasonable notion of homomorphism $(mathbb{Q}^+, {cdot}) to (mathbb{Z}^+, {cdot})$ must satisfy $f(ab) = f(a)f(b)$ for all $a,b in mathbb{Q}^+$. We can prove that no such homomorphism is an isomorphism (be it of monoids or otherwise), since it is not even bijective.
Indeed, first note that we must have
$$f(1) = f(1 cdot 1) = f(1) cdot f(1) quad Rightarrow quad f(1) = 1$$
We were able to cancel the $f(1)$ since $f(1) > 0$. But then we also have
$$1 = f(1) = fleft( frac{1}{2} cdot 2 right) = fleft(frac{1}{2}right) cdot f(2)$$
and hence $f(2) = 1$ since that is the only positive integral factor of $1$.
But then $f(1)=f(2)$, so $f$ is not a bijection, so $f$ is not an isomorphism.
Footnote: what we actually proved is that $mathbb{Z}^+$ and $mathbb{Q}^+$ are not isomorphic as magmas under multiplication. Since every monoid, or even semigroup, has an underlying magma, it follows that they are not isomorphic as semigroups or as magmas.
answered Jan 20 at 14:55
Clive NewsteadClive Newstead
51.9k474136
$endgroup$
$begingroup$
Alright, this seems like a much more formal approach, assuming bijective then reaching a contradiction, which I been seeing a lot of. However, my only question with this kind of explanation is what if the operations paired with the sets are different, such as one is multiplication and the other is addition. Is the problem set up in the same way? Also, we haven't been introduced to terminology such as "magmus" or "monoinds;" are these concepts important to know in the early stages of Abstract Algebra.
$endgroup$
– Noaline
Jan 20 at 15:03
$begingroup$
@Noaline: If you're asking about whether there exist binary operations on $mathbb{Z}^+$ and $mathbb{Q}^+$ such that they are isomorphic (as, say, monoids), then the answer is 'yes'. For example, fix a bijection $f : mathbb{Z}^+ to mathbb{Q}^+$ and define $star$ on $mathbb{Z}^+$ by $a star b = f(a)f(b)$. Then $f$ trivially extends to an isomorphism $(mathbb{Z}^+, star) to (mathbb{Q}^+, times)$.
$endgroup$
– Clive Newstead
Jan 20 at 15:10
$begingroup$
...but I'm not sure I can be very helpful. You're asking a very vague question and it sounds like you need more of an algebra background to understand any answer that could be given to such a question.
$endgroup$
– Clive Newstead
Jan 20 at 15:11
$begingroup$
@CliveNewstead: I don't see how the map $star$ you define is on $Bbb Z^+$: how do we know if $f(a)f(b)inBbb Z^+$? The only thing we can say is that $star$ is a map $Bbb Z^+timesBbb Z^+toBbb Q^+$
$endgroup$
– learner
Jan 20 at 15:15
$begingroup$
@CliveNewstead Alright, thank you. I have only recently put my pedal to the metal when it comes to studies recently so my linear algebra and calculus classes have not been very beneficial for me. Are there any resources you know of that could help me get a better grasp on the topic?
$endgroup$
– Noaline
Jan 20 at 15:15
|
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1 Answer
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$begingroup$
To say that two structures are or are not isomorphic, you need to specify what you are saying they are(n't) isomorphic as. In this case, asking about isomorphism of monoids would be reasonable, since both $mathbb{Z}^+$ and $mathbb{Q}^+$ are monoids under multiplication.
Fortunately in this case, any reasonable notion of homomorphism $(mathbb{Q}^+, {cdot}) to (mathbb{Z}^+, {cdot})$ must satisfy $f(ab) = f(a)f(b)$ for all $a,b in mathbb{Q}^+$. We can prove that no such homomorphism is an isomorphism (be it of monoids or otherwise), since it is not even bijective.
Indeed, first note that we must have
$$f(1) = f(1 cdot 1) = f(1) cdot f(1) quad Rightarrow quad f(1) = 1$$
We were able to cancel the $f(1)$ since $f(1) > 0$. But then we also have
$$1 = f(1) = fleft( frac{1}{2} cdot 2 right) = fleft(frac{1}{2}right) cdot f(2)$$
and hence $f(2) = 1$ since that is the only positive integral factor of $1$.
But then $f(1)=f(2)$, so $f$ is not a bijection, so $f$ is not an isomorphism.
Footnote: what we actually proved is that $mathbb{Z}^+$ and $mathbb{Q}^+$ are not isomorphic as magmas under multiplication. Since every monoid, or even semigroup, has an underlying magma, it follows that they are not isomorphic as semigroups or as magmas.
answered Jan 20 at 14:55
Clive NewsteadClive Newstead
51.9k474136
$endgroup$
$begingroup$
Alright, this seems like a much more formal approach, assuming bijective then reaching a contradiction, which I been seeing a lot of. However, my only question with this kind of explanation is what if the operations paired with the sets are different, such as one is multiplication and the other is addition. Is the problem set up in the same way? Also, we haven't been introduced to terminology such as "magmus" or "monoinds;" are these concepts important to know in the early stages of Abstract Algebra.
$endgroup$
– Noaline
Jan 20 at 15:03
$begingroup$
@Noaline: If you're asking about whether there exist binary operations on $mathbb{Z}^+$ and $mathbb{Q}^+$ such that they are isomorphic (as, say, monoids), then the answer is 'yes'. For example, fix a bijection $f : mathbb{Z}^+ to mathbb{Q}^+$ and define $star$ on $mathbb{Z}^+$ by $a star b = f(a)f(b)$. Then $f$ trivially extends to an isomorphism $(mathbb{Z}^+, star) to (mathbb{Q}^+, times)$.
$endgroup$
– Clive Newstead
Jan 20 at 15:10
$begingroup$
...but I'm not sure I can be very helpful. You're asking a very vague question and it sounds like you need more of an algebra background to understand any answer that could be given to such a question.
$endgroup$
– Clive Newstead
Jan 20 at 15:11
$begingroup$
@CliveNewstead: I don't see how the map $star$ you define is on $Bbb Z^+$: how do we know if $f(a)f(b)inBbb Z^+$? The only thing we can say is that $star$ is a map $Bbb Z^+timesBbb Z^+toBbb Q^+$
$endgroup$
– learner
Jan 20 at 15:15
$begingroup$
@CliveNewstead Alright, thank you. I have only recently put my pedal to the metal when it comes to studies recently so my linear algebra and calculus classes have not been very beneficial for me. Are there any resources you know of that could help me get a better grasp on the topic?
$endgroup$
– Noaline
Jan 20 at 15:15
|
show 1 more comment
$begingroup$
To say that two structures are or are not isomorphic, you need to specify what you are saying they are(n't) isomorphic as. In this case, asking about isomorphism of monoids would be reasonable, since both $mathbb{Z}^+$ and $mathbb{Q}^+$ are monoids under multiplication.
Fortunately in this case, any reasonable notion of homomorphism $(mathbb{Q}^+, {cdot}) to (mathbb{Z}^+, {cdot})$ must satisfy $f(ab) = f(a)f(b)$ for all $a,b in mathbb{Q}^+$. We can prove that no such homomorphism is an isomorphism (be it of monoids or otherwise), since it is not even bijective.
Indeed, first note that we must have
$$f(1) = f(1 cdot 1) = f(1) cdot f(1) quad Rightarrow quad f(1) = 1$$
We were able to cancel the $f(1)$ since $f(1) > 0$. But then we also have
$$1 = f(1) = fleft( frac{1}{2} cdot 2 right) = fleft(frac{1}{2}right) cdot f(2)$$
and hence $f(2) = 1$ since that is the only positive integral factor of $1$.
But then $f(1)=f(2)$, so $f$ is not a bijection, so $f$ is not an isomorphism.
Footnote: what we actually proved is that $mathbb{Z}^+$ and $mathbb{Q}^+$ are not isomorphic as magmas under multiplication. Since every monoid, or even semigroup, has an underlying magma, it follows that they are not isomorphic as semigroups or as magmas.
answered Jan 20 at 14:55
Clive NewsteadClive Newstead
51.9k474136
$endgroup$
$begingroup$
Alright, this seems like a much more formal approach, assuming bijective then reaching a contradiction, which I been seeing a lot of. However, my only question with this kind of explanation is what if the operations paired with the sets are different, such as one is multiplication and the other is addition. Is the problem set up in the same way? Also, we haven't been introduced to terminology such as "magmus" or "monoinds;" are these concepts important to know in the early stages of Abstract Algebra.
$endgroup$
– Noaline
Jan 20 at 15:03
$begingroup$
@Noaline: If you're asking about whether there exist binary operations on $mathbb{Z}^+$ and $mathbb{Q}^+$ such that they are isomorphic (as, say, monoids), then the answer is 'yes'. For example, fix a bijection $f : mathbb{Z}^+ to mathbb{Q}^+$ and define $star$ on $mathbb{Z}^+$ by $a star b = f(a)f(b)$. Then $f$ trivially extends to an isomorphism $(mathbb{Z}^+, star) to (mathbb{Q}^+, times)$.
$endgroup$
– Clive Newstead
Jan 20 at 15:10
$begingroup$
...but I'm not sure I can be very helpful. You're asking a very vague question and it sounds like you need more of an algebra background to understand any answer that could be given to such a question.
$endgroup$
– Clive Newstead
Jan 20 at 15:11
$begingroup$
@CliveNewstead: I don't see how the map $star$ you define is on $Bbb Z^+$: how do we know if $f(a)f(b)inBbb Z^+$? The only thing we can say is that $star$ is a map $Bbb Z^+timesBbb Z^+toBbb Q^+$
$endgroup$
– learner
Jan 20 at 15:15
$begingroup$
@CliveNewstead Alright, thank you. I have only recently put my pedal to the metal when it comes to studies recently so my linear algebra and calculus classes have not been very beneficial for me. Are there any resources you know of that could help me get a better grasp on the topic?
$endgroup$
– Noaline
Jan 20 at 15:15
|
show 1 more comment
$begingroup$
To say that two structures are or are not isomorphic, you need to specify what you are saying they are(n't) isomorphic as. In this case, asking about isomorphism of monoids would be reasonable, since both $mathbb{Z}^+$ and $mathbb{Q}^+$ are monoids under multiplication.
Fortunately in this case, any reasonable notion of homomorphism $(mathbb{Q}^+, {cdot}) to (mathbb{Z}^+, {cdot})$ must satisfy $f(ab) = f(a)f(b)$ for all $a,b in mathbb{Q}^+$. We can prove that no such homomorphism is an isomorphism (be it of monoids or otherwise), since it is not even bijective.
Indeed, first note that we must have
$$f(1) = f(1 cdot 1) = f(1) cdot f(1) quad Rightarrow quad f(1) = 1$$
We were able to cancel the $f(1)$ since $f(1) > 0$. But then we also have
$$1 = f(1) = fleft( frac{1}{2} cdot 2 right) = fleft(frac{1}{2}right) cdot f(2)$$
and hence $f(2) = 1$ since that is the only positive integral factor of $1$.
But then $f(1)=f(2)$, so $f$ is not a bijection, so $f$ is not an isomorphism.
Footnote: what we actually proved is that $mathbb{Z}^+$ and $mathbb{Q}^+$ are not isomorphic as magmas under multiplication. Since every monoid, or even semigroup, has an underlying magma, it follows that they are not isomorphic as semigroups or as magmas.
answered Jan 20 at 14:55
Clive NewsteadClive Newstead
51.9k474136
$endgroup$
To say that two structures are or are not isomorphic, you need to specify what you are saying they are(n't) isomorphic as. In this case, asking about isomorphism of monoids would be reasonable, since both $mathbb{Z}^+$ and $mathbb{Q}^+$ are monoids under multiplication.
Fortunately in this case, any reasonable notion of homomorphism $(mathbb{Q}^+, {cdot}) to (mathbb{Z}^+, {cdot})$ must satisfy $f(ab) = f(a)f(b)$ for all $a,b in mathbb{Q}^+$. We can prove that no such homomorphism is an isomorphism (be it of monoids or otherwise), since it is not even bijective.
Indeed, first note that we must have
$$f(1) = f(1 cdot 1) = f(1) cdot f(1) quad Rightarrow quad f(1) = 1$$
We were able to cancel the $f(1)$ since $f(1) > 0$. But then we also have
$$1 = f(1) = fleft( frac{1}{2} cdot 2 right) = fleft(frac{1}{2}right) cdot f(2)$$
and hence $f(2) = 1$ since that is the only positive integral factor of $1$.
But then $f(1)=f(2)$, so $f$ is not a bijection, so $f$ is not an isomorphism.
Footnote: what we actually proved is that $mathbb{Z}^+$ and $mathbb{Q}^+$ are not isomorphic as magmas under multiplication. Since every monoid, or even semigroup, has an underlying magma, it follows that they are not isomorphic as semigroups or as magmas.
answered Jan 20 at 14:55
Clive NewsteadClive Newstead
51.9k474136
answered Jan 20 at 14:55
Clive NewsteadClive Newstead
51.9k474136
answered Jan 20 at 14:55
Clive NewsteadClive Newstead
51.9k474136
answered Jan 20 at 14:55
Clive NewsteadClive Newstead
51.9k474136
51.9k474136
$begingroup$
Alright, this seems like a much more formal approach, assuming bijective then reaching a contradiction, which I been seeing a lot of. However, my only question with this kind of explanation is what if the operations paired with the sets are different, such as one is multiplication and the other is addition. Is the problem set up in the same way? Also, we haven't been introduced to terminology such as "magmus" or "monoinds;" are these concepts important to know in the early stages of Abstract Algebra.
$endgroup$
– Noaline
Jan 20 at 15:03
$begingroup$
@Noaline: If you're asking about whether there exist binary operations on $mathbb{Z}^+$ and $mathbb{Q}^+$ such that they are isomorphic (as, say, monoids), then the answer is 'yes'. For example, fix a bijection $f : mathbb{Z}^+ to mathbb{Q}^+$ and define $star$ on $mathbb{Z}^+$ by $a star b = f(a)f(b)$. Then $f$ trivially extends to an isomorphism $(mathbb{Z}^+, star) to (mathbb{Q}^+, times)$.
$endgroup$
– Clive Newstead
Jan 20 at 15:10
$begingroup$
...but I'm not sure I can be very helpful. You're asking a very vague question and it sounds like you need more of an algebra background to understand any answer that could be given to such a question.
$endgroup$
– Clive Newstead
Jan 20 at 15:11
$begingroup$
@CliveNewstead: I don't see how the map $star$ you define is on $Bbb Z^+$: how do we know if $f(a)f(b)inBbb Z^+$? The only thing we can say is that $star$ is a map $Bbb Z^+timesBbb Z^+toBbb Q^+$
$endgroup$
– learner
Jan 20 at 15:15
$begingroup$
@CliveNewstead Alright, thank you. I have only recently put my pedal to the metal when it comes to studies recently so my linear algebra and calculus classes have not been very beneficial for me. Are there any resources you know of that could help me get a better grasp on the topic?
$endgroup$
– Noaline
Jan 20 at 15:15
|
show 1 more comment
$begingroup$
Alright, this seems like a much more formal approach, assuming bijective then reaching a contradiction, which I been seeing a lot of. However, my only question with this kind of explanation is what if the operations paired with the sets are different, such as one is multiplication and the other is addition. Is the problem set up in the same way? Also, we haven't been introduced to terminology such as "magmus" or "monoinds;" are these concepts important to know in the early stages of Abstract Algebra.
$endgroup$
– Noaline
Jan 20 at 15:03
$begingroup$
@Noaline: If you're asking about whether there exist binary operations on $mathbb{Z}^+$ and $mathbb{Q}^+$ such that they are isomorphic (as, say, monoids), then the answer is 'yes'. For example, fix a bijection $f : mathbb{Z}^+ to mathbb{Q}^+$ and define $star$ on $mathbb{Z}^+$ by $a star b = f(a)f(b)$. Then $f$ trivially extends to an isomorphism $(mathbb{Z}^+, star) to (mathbb{Q}^+, times)$.
$endgroup$
– Clive Newstead
Jan 20 at 15:10
$begingroup$
...but I'm not sure I can be very helpful. You're asking a very vague question and it sounds like you need more of an algebra background to understand any answer that could be given to such a question.
$endgroup$
– Clive Newstead
Jan 20 at 15:11
$begingroup$
@CliveNewstead: I don't see how the map $star$ you define is on $Bbb Z^+$: how do we know if $f(a)f(b)inBbb Z^+$? The only thing we can say is that $star$ is a map $Bbb Z^+timesBbb Z^+toBbb Q^+$
$endgroup$
– learner
Jan 20 at 15:15
$begingroup$
@CliveNewstead Alright, thank you. I have only recently put my pedal to the metal when it comes to studies recently so my linear algebra and calculus classes have not been very beneficial for me. Are there any resources you know of that could help me get a better grasp on the topic?
$endgroup$
– Noaline
Jan 20 at 15:15
$begingroup$
Alright, this seems like a much more formal approach, assuming bijective then reaching a contradiction, which I been seeing a lot of. However, my only question with this kind of explanation is what if the operations paired with the sets are different, such as one is multiplication and the other is addition. Is the problem set up in the same way? Also, we haven't been introduced to terminology such as "magmus" or "monoinds;" are these concepts important to know in the early stages of Abstract Algebra.
$endgroup$
– Noaline
Jan 20 at 15:03
$begingroup$
Alright, this seems like a much more formal approach, assuming bijective then reaching a contradiction, which I been seeing a lot of. However, my only question with this kind of explanation is what if the operations paired with the sets are different, such as one is multiplication and the other is addition. Is the problem set up in the same way? Also, we haven't been introduced to terminology such as "magmus" or "monoinds;" are these concepts important to know in the early stages of Abstract Algebra.
$endgroup$
– Noaline
Jan 20 at 15:03
$begingroup$
@Noaline: If you're asking about whether there exist binary operations on $mathbb{Z}^+$ and $mathbb{Q}^+$ such that they are isomorphic (as, say, monoids), then the answer is 'yes'. For example, fix a bijection $f : mathbb{Z}^+ to mathbb{Q}^+$ and define $star$ on $mathbb{Z}^+$ by $a star b = f(a)f(b)$. Then $f$ trivially extends to an isomorphism $(mathbb{Z}^+, star) to (mathbb{Q}^+, times)$.
$endgroup$
– Clive Newstead
Jan 20 at 15:10
$begingroup$
@Noaline: If you're asking about whether there exist binary operations on $mathbb{Z}^+$ and $mathbb{Q}^+$ such that they are isomorphic (as, say, monoids), then the answer is 'yes'. For example, fix a bijection $f : mathbb{Z}^+ to mathbb{Q}^+$ and define $star$ on $mathbb{Z}^+$ by $a star b = f(a)f(b)$. Then $f$ trivially extends to an isomorphism $(mathbb{Z}^+, star) to (mathbb{Q}^+, times)$.
$endgroup$
– Clive Newstead
Jan 20 at 15:10
$begingroup$
...but I'm not sure I can be very helpful. You're asking a very vague question and it sounds like you need more of an algebra background to understand any answer that could be given to such a question.
$endgroup$
– Clive Newstead
Jan 20 at 15:11
$begingroup$
...but I'm not sure I can be very helpful. You're asking a very vague question and it sounds like you need more of an algebra background to understand any answer that could be given to such a question.
$endgroup$
– Clive Newstead
Jan 20 at 15:11
$begingroup$
@CliveNewstead: I don't see how the map $star$ you define is on $Bbb Z^+$: how do we know if $f(a)f(b)inBbb Z^+$? The only thing we can say is that $star$ is a map $Bbb Z^+timesBbb Z^+toBbb Q^+$
$endgroup$
– learner
Jan 20 at 15:15
$begingroup$
@CliveNewstead: I don't see how the map $star$ you define is on $Bbb Z^+$: how do we know if $f(a)f(b)inBbb Z^+$? The only thing we can say is that $star$ is a map $Bbb Z^+timesBbb Z^+toBbb Q^+$
$endgroup$
– learner
Jan 20 at 15:15
$begingroup$
@CliveNewstead Alright, thank you. I have only recently put my pedal to the metal when it comes to studies recently so my linear algebra and calculus classes have not been very beneficial for me. Are there any resources you know of that could help me get a better grasp on the topic?
$endgroup$
– Noaline
Jan 20 at 15:15
$begingroup$
@CliveNewstead Alright, thank you. I have only recently put my pedal to the metal when it comes to studies recently so my linear algebra and calculus classes have not been very beneficial for me. Are there any resources you know of that could help me get a better grasp on the topic?
$endgroup$
– Noaline
Jan 20 at 15:15
|
show 1 more comment
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$begingroup$
$mathbb{Z}^+$ with product is not a group. $2$ doesn't have an inverse.
$endgroup$
– Yanko
Jan 20 at 14:35
$begingroup$
I'm not sure I follow. Why is the first not a group (what would it be classified as), and how is it known that the second group has no inverse and what would that mean about being isomorphic with group 1. Sorry for so many questions, I am really trying to get a grasp on this.
$endgroup$
– Noaline
Jan 20 at 14:41
1
$begingroup$
You should read about the definition of a group here en.wikipedia.org/wiki/… The first is not a group because $2$ has no inverse as $2a=1$ implies that $a=frac{1}{2}$ which is not an integer. Group without inverse is sometimes called a semi-group and it can't be isomorphic to a group.
$endgroup$
– Yanko
Jan 20 at 14:46
1
$begingroup$
Oh, so groups can't be isomorphic with non-groups, and there are certain criteria for being a group that must be checked? If my understanding is correct, thank you very much, I had spent all night last night trying to figure this out.
$endgroup$
– Noaline
Jan 20 at 14:51