Mixing
Notions of basis and span in a magma
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
Suppose that $C$ is a set with closure under the binary operation $+$. $(C,+)$ is therefore a magma.
I am trying to figure out if notions of basis, or span make sense in a magma.
Spanning set (?)
Suppose that I can find a subset of C: $C_s subset C$, which has a finite number of elements: $C_s = c_s_1, ldots, c_s_n$, so that every element of $C$ can be expressed as a "sum" of the elements of $C_s$.
Since my "sum" is not associative or commutative, I can have the following:
$c_i = ((c_s_4 + c_s_1) + c_s_4) + c_s_2$
- Is $C_s$ necessarily closed under +?
Since my $+$ is not associative or commutative, I could potentially have an infinite number of operands correct?
2.1 subsidiary question: is there a clean/short way to write a non-associative non-commutative binary operation on a possibly infinite number of elements? Something like $sum_k^n ...$ but which clearly shows that there can be repeating elements (like $c_s_4$ above) and that the "sum" is not commutative/associative?
Basis
Let us now assume that this subset $C_s$ is made of "linearly independent" elements, so that I could ultimately define a basis in my magma.
- How would I properly define this "linear independence" property in my magma with a non-associative and non-commutative +?
I am familiar with the definition of a basis in a vector space E on K where:
$forall x in E, exists lambda_1, ldots, lambda_p in K: x = lambda_1e_1 + ldots + lambda_pe_p$
and
$lambda_1e_1 + ldots + lambda_pe_p = 0 implies lambda_1 = 0, ldots, lambda_p = 0$
but I have problems connecting this with the magma
linear-algebra magma
asked yesterday
Hazan Tayeb
356
add a comment |Â
up vote
3
down vote
favorite
Suppose that $C$ is a set with closure under the binary operation $+$. $(C,+)$ is therefore a magma.
I am trying to figure out if notions of basis, or span make sense in a magma.
Spanning set (?)
Suppose that I can find a subset of C: $C_s subset C$, which has a finite number of elements: $C_s = c_s_1, ldots, c_s_n$, so that every element of $C$ can be expressed as a "sum" of the elements of $C_s$.
Since my "sum" is not associative or commutative, I can have the following:
$c_i = ((c_s_4 + c_s_1) + c_s_4) + c_s_2$
- Is $C_s$ necessarily closed under +?
Since my $+$ is not associative or commutative, I could potentially have an infinite number of operands correct?
2.1 subsidiary question: is there a clean/short way to write a non-associative non-commutative binary operation on a possibly infinite number of elements? Something like $sum_k^n ...$ but which clearly shows that there can be repeating elements (like $c_s_4$ above) and that the "sum" is not commutative/associative?
Basis
Let us now assume that this subset $C_s$ is made of "linearly independent" elements, so that I could ultimately define a basis in my magma.
- How would I properly define this "linear independence" property in my magma with a non-associative and non-commutative +?
I am familiar with the definition of a basis in a vector space E on K where:
$forall x in E, exists lambda_1, ldots, lambda_p in K: x = lambda_1e_1 + ldots + lambda_pe_p$
and
$lambda_1e_1 + ldots + lambda_pe_p = 0 implies lambda_1 = 0, ldots, lambda_p = 0$
but I have problems connecting this with the magma
linear-algebra magma
asked yesterday
Hazan Tayeb
356
1
In writing $c_s_4 + c_s_1 + c_s_4 + c_s_2$, where do you put the brackets? Are you writing left-associatively, in that it can be interpreted as $$((c_s_4 + c_s_1) + c_s_4) + c_s_2?$$ You should be careful with non-associative operations.
– Theo Bendit
yesterday
Indeed. I will make it explicit in the question.
– Hazan Tayeb
yesterday
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose that $C$ is a set with closure under the binary operation $+$. $(C,+)$ is therefore a magma.
I am trying to figure out if notions of basis, or span make sense in a magma.
Spanning set (?)
Suppose that I can find a subset of C: $C_s subset C$, which has a finite number of elements: $C_s = c_s_1, ldots, c_s_n$, so that every element of $C$ can be expressed as a "sum" of the elements of $C_s$.
Since my "sum" is not associative or commutative, I can have the following:
$c_i = ((c_s_4 + c_s_1) + c_s_4) + c_s_2$
- Is $C_s$ necessarily closed under +?
Since my $+$ is not associative or commutative, I could potentially have an infinite number of operands correct?
2.1 subsidiary question: is there a clean/short way to write a non-associative non-commutative binary operation on a possibly infinite number of elements? Something like $sum_k^n ...$ but which clearly shows that there can be repeating elements (like $c_s_4$ above) and that the "sum" is not commutative/associative?
Basis
Let us now assume that this subset $C_s$ is made of "linearly independent" elements, so that I could ultimately define a basis in my magma.
- How would I properly define this "linear independence" property in my magma with a non-associative and non-commutative +?
I am familiar with the definition of a basis in a vector space E on K where:
$forall x in E, exists lambda_1, ldots, lambda_p in K: x = lambda_1e_1 + ldots + lambda_pe_p$
and
$lambda_1e_1 + ldots + lambda_pe_p = 0 implies lambda_1 = 0, ldots, lambda_p = 0$
but I have problems connecting this with the magma
linear-algebra magma
asked yesterday
Hazan Tayeb
356
Suppose that $C$ is a set with closure under the binary operation $+$. $(C,+)$ is therefore a magma.
I am trying to figure out if notions of basis, or span make sense in a magma.
Spanning set (?)
Suppose that I can find a subset of C: $C_s subset C$, which has a finite number of elements: $C_s = c_s_1, ldots, c_s_n$, so that every element of $C$ can be expressed as a "sum" of the elements of $C_s$.
Since my "sum" is not associative or commutative, I can have the following:
$c_i = ((c_s_4 + c_s_1) + c_s_4) + c_s_2$
- Is $C_s$ necessarily closed under +?
Since my $+$ is not associative or commutative, I could potentially have an infinite number of operands correct?
2.1 subsidiary question: is there a clean/short way to write a non-associative non-commutative binary operation on a possibly infinite number of elements? Something like $sum_k^n ...$ but which clearly shows that there can be repeating elements (like $c_s_4$ above) and that the "sum" is not commutative/associative?
Basis
Let us now assume that this subset $C_s$ is made of "linearly independent" elements, so that I could ultimately define a basis in my magma.
- How would I properly define this "linear independence" property in my magma with a non-associative and non-commutative +?
I am familiar with the definition of a basis in a vector space E on K where:
$forall x in E, exists lambda_1, ldots, lambda_p in K: x = lambda_1e_1 + ldots + lambda_pe_p$
and
$lambda_1e_1 + ldots + lambda_pe_p = 0 implies lambda_1 = 0, ldots, lambda_p = 0$
but I have problems connecting this with the magma
linear-algebra magma
linear-algebra magma
asked yesterday
Hazan Tayeb
356
asked yesterday
Hazan Tayeb
356
edited yesterday
asked yesterday
Hazan Tayeb
356
asked yesterday
Hazan Tayeb
356
asked yesterday
Hazan Tayeb
356
356
1
In writing $c_s_4 + c_s_1 + c_s_4 + c_s_2$, where do you put the brackets? Are you writing left-associatively, in that it can be interpreted as $$((c_s_4 + c_s_1) + c_s_4) + c_s_2?$$ You should be careful with non-associative operations.
– Theo Bendit
yesterday
Indeed. I will make it explicit in the question.
– Hazan Tayeb
yesterday
add a comment |Â
1
In writing $c_s_4 + c_s_1 + c_s_4 + c_s_2$, where do you put the brackets? Are you writing left-associatively, in that it can be interpreted as $$((c_s_4 + c_s_1) + c_s_4) + c_s_2?$$ You should be careful with non-associative operations.
– Theo Bendit
yesterday
Indeed. I will make it explicit in the question.
– Hazan Tayeb
yesterday
1
1
In writing $c_s_4 + c_s_1 + c_s_4 + c_s_2$, where do you put the brackets? Are you writing left-associatively, in that it can be interpreted as $$((c_s_4 + c_s_1) + c_s_4) + c_s_2?$$ You should be careful with non-associative operations.
– Theo Bendit
yesterday
In writing $c_s_4 + c_s_1 + c_s_4 + c_s_2$, where do you put the brackets? Are you writing left-associatively, in that it can be interpreted as $$((c_s_4 + c_s_1) + c_s_4) + c_s_2?$$ You should be careful with non-associative operations.
– Theo Bendit
yesterday
Indeed. I will make it explicit in the question.
– Hazan Tayeb
yesterday
Indeed. I will make it explicit in the question.
– Hazan Tayeb
yesterday
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Suppose that I can find a subset of C: $C_s subset C$, which has a finite number of elements: $C_s = c_s_1, ldots, c_s_n$, so that every element of $C$ can be expressed as a "sum" of the elements of $C_s$.
Making "sum" precise here would be to say that every $cin C$ can be obtained from the elements of $C_s$ by finitely many $+$ operations. You would then call $C_s$ a generating set of $(C,+)$
- Is $C_s$ necessarily closed under +?
No it is not, consider the magma $(mathbb Z_>0, +)$ with generating set $1$.
- Since my $+$ is not associative or commutative, I could potentially have an infinite number of operands correct?
No, you can only have finitely many operands. Just like in any structure without a notion of limits, you only have finite products, finite sums, finite linear combinations, etc.
Of course, you can have arbitrarily many operands in a finite (non-associative) sum.
2.1 subsidiary question: is there a clean/short way to write a non-associative non-commutative binary operation on a possibly infinite number of elements? Something like $sum_k^n ...$ but which clearly shows that there can be repeating elements (like $c_s_4$ above) and that the "sum" is not commutative/associative?
You can draw finite binary rooted trees with leaves labelled by elements of the magma. You have to encode the order of operations (the brackets) one way or another.
- How would I properly define this "linear independence" property in my magma with a non-associative and non-commutative +?
The closest you can get is saying a subset $Isubseteq C$ is independent if every $cin I$ can not be generated from $Isetminusc$. This is an analogy with a subset of a vector space being linearly independent iff no vector can be written as a linear combination of the others.
You could then define a "basis" to be an independent generating set, but do not expect any nice properties from this.
answered yesterday
Christoph
10.9k1240
add a comment |Â
up vote
2
down vote
If $+$ is any operation on a set $C$ and $A$ is a subset of $C$, there exists the smallest subset $B$ of $C$ such that
- $Asubseteq B$
- $B$ is closed under $+$
Such a set $B$ can be described as the intersection of all subsets of $C$ that satisfy properties 1 and 2 above or in a recursive fashion. Define $A_0=A$; assuming you have defined $A_n$, define
$$
A_n+1=a+b:a,bin A_n
$$
Then set $B=bigcup_nge0A_n$ and verify $B$ is closed under $+$ and is contained in every subset of $C$ that contains $A$ and is closed under $+$, so it's the same as the one defined above.
Let's denote by $langle Arangle$ the set $B$, as determined in one of the two equivalent ways above. We say that $A$ is a set of generators for $C$ if $langle Arangle=C$.
If you're looking for uniqueness properties of sets of generators similar to bases in vector spaces, then you're out of luck. It's quite easy to show that there are magmas with finite sets of generators which are minimal (in the sense that removing one element leads to a set which doesn't generate the magma) but have different cardinalities. A simple example is the set $mathbbZ/6mathbbZ$ with the standard addition (modulo $6$): then $2,3$ is a minimal set of generators, but also $1$ is a minimal set of generators.
answered yesterday
egreg
166k1180189
In your definition of $A_n+1$ you want to allow both $a$ and $c$ to be picked from $A_n$ I think. Otherwise, do you get $(a+a)+(a+a)$ from $A=a$?
– Christoph
yesterday
In your example of $mathbb Z$ with addition, $1$ generates only the positive integers and $2,3$ generates only the integers $2,3,4,5,ldots$
– Christoph
yesterday
@Christoph You're right on both issues, I fixed the answer.
– egreg
yesterday
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Suppose that I can find a subset of C: $C_s subset C$, which has a finite number of elements: $C_s = c_s_1, ldots, c_s_n$, so that every element of $C$ can be expressed as a "sum" of the elements of $C_s$.
Making "sum" precise here would be to say that every $cin C$ can be obtained from the elements of $C_s$ by finitely many $+$ operations. You would then call $C_s$ a generating set of $(C,+)$
- Is $C_s$ necessarily closed under +?
No it is not, consider the magma $(mathbb Z_>0, +)$ with generating set $1$.
- Since my $+$ is not associative or commutative, I could potentially have an infinite number of operands correct?
No, you can only have finitely many operands. Just like in any structure without a notion of limits, you only have finite products, finite sums, finite linear combinations, etc.
Of course, you can have arbitrarily many operands in a finite (non-associative) sum.
2.1 subsidiary question: is there a clean/short way to write a non-associative non-commutative binary operation on a possibly infinite number of elements? Something like $sum_k^n ...$ but which clearly shows that there can be repeating elements (like $c_s_4$ above) and that the "sum" is not commutative/associative?
You can draw finite binary rooted trees with leaves labelled by elements of the magma. You have to encode the order of operations (the brackets) one way or another.
- How would I properly define this "linear independence" property in my magma with a non-associative and non-commutative +?
The closest you can get is saying a subset $Isubseteq C$ is independent if every $cin I$ can not be generated from $Isetminusc$. This is an analogy with a subset of a vector space being linearly independent iff no vector can be written as a linear combination of the others.
You could then define a "basis" to be an independent generating set, but do not expect any nice properties from this.
answered yesterday
Christoph
10.9k1240
add a comment |Â
up vote
3
down vote
accepted
Suppose that I can find a subset of C: $C_s subset C$, which has a finite number of elements: $C_s = c_s_1, ldots, c_s_n$, so that every element of $C$ can be expressed as a "sum" of the elements of $C_s$.
Making "sum" precise here would be to say that every $cin C$ can be obtained from the elements of $C_s$ by finitely many $+$ operations. You would then call $C_s$ a generating set of $(C,+)$
- Is $C_s$ necessarily closed under +?
No it is not, consider the magma $(mathbb Z_>0, +)$ with generating set $1$.
- Since my $+$ is not associative or commutative, I could potentially have an infinite number of operands correct?
No, you can only have finitely many operands. Just like in any structure without a notion of limits, you only have finite products, finite sums, finite linear combinations, etc.
Of course, you can have arbitrarily many operands in a finite (non-associative) sum.
2.1 subsidiary question: is there a clean/short way to write a non-associative non-commutative binary operation on a possibly infinite number of elements? Something like $sum_k^n ...$ but which clearly shows that there can be repeating elements (like $c_s_4$ above) and that the "sum" is not commutative/associative?
You can draw finite binary rooted trees with leaves labelled by elements of the magma. You have to encode the order of operations (the brackets) one way or another.
- How would I properly define this "linear independence" property in my magma with a non-associative and non-commutative +?
The closest you can get is saying a subset $Isubseteq C$ is independent if every $cin I$ can not be generated from $Isetminusc$. This is an analogy with a subset of a vector space being linearly independent iff no vector can be written as a linear combination of the others.
You could then define a "basis" to be an independent generating set, but do not expect any nice properties from this.
answered yesterday
Christoph
10.9k1240
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Suppose that I can find a subset of C: $C_s subset C$, which has a finite number of elements: $C_s = c_s_1, ldots, c_s_n$, so that every element of $C$ can be expressed as a "sum" of the elements of $C_s$.
Making "sum" precise here would be to say that every $cin C$ can be obtained from the elements of $C_s$ by finitely many $+$ operations. You would then call $C_s$ a generating set of $(C,+)$
- Is $C_s$ necessarily closed under +?
No it is not, consider the magma $(mathbb Z_>0, +)$ with generating set $1$.
- Since my $+$ is not associative or commutative, I could potentially have an infinite number of operands correct?
No, you can only have finitely many operands. Just like in any structure without a notion of limits, you only have finite products, finite sums, finite linear combinations, etc.
Of course, you can have arbitrarily many operands in a finite (non-associative) sum.
2.1 subsidiary question: is there a clean/short way to write a non-associative non-commutative binary operation on a possibly infinite number of elements? Something like $sum_k^n ...$ but which clearly shows that there can be repeating elements (like $c_s_4$ above) and that the "sum" is not commutative/associative?
You can draw finite binary rooted trees with leaves labelled by elements of the magma. You have to encode the order of operations (the brackets) one way or another.
- How would I properly define this "linear independence" property in my magma with a non-associative and non-commutative +?
The closest you can get is saying a subset $Isubseteq C$ is independent if every $cin I$ can not be generated from $Isetminusc$. This is an analogy with a subset of a vector space being linearly independent iff no vector can be written as a linear combination of the others.
You could then define a "basis" to be an independent generating set, but do not expect any nice properties from this.
answered yesterday
Christoph
10.9k1240
Suppose that I can find a subset of C: $C_s subset C$, which has a finite number of elements: $C_s = c_s_1, ldots, c_s_n$, so that every element of $C$ can be expressed as a "sum" of the elements of $C_s$.
Making "sum" precise here would be to say that every $cin C$ can be obtained from the elements of $C_s$ by finitely many $+$ operations. You would then call $C_s$ a generating set of $(C,+)$
- Is $C_s$ necessarily closed under +?
No it is not, consider the magma $(mathbb Z_>0, +)$ with generating set $1$.
- Since my $+$ is not associative or commutative, I could potentially have an infinite number of operands correct?
No, you can only have finitely many operands. Just like in any structure without a notion of limits, you only have finite products, finite sums, finite linear combinations, etc.
Of course, you can have arbitrarily many operands in a finite (non-associative) sum.
2.1 subsidiary question: is there a clean/short way to write a non-associative non-commutative binary operation on a possibly infinite number of elements? Something like $sum_k^n ...$ but which clearly shows that there can be repeating elements (like $c_s_4$ above) and that the "sum" is not commutative/associative?
You can draw finite binary rooted trees with leaves labelled by elements of the magma. You have to encode the order of operations (the brackets) one way or another.
- How would I properly define this "linear independence" property in my magma with a non-associative and non-commutative +?
The closest you can get is saying a subset $Isubseteq C$ is independent if every $cin I$ can not be generated from $Isetminusc$. This is an analogy with a subset of a vector space being linearly independent iff no vector can be written as a linear combination of the others.
You could then define a "basis" to be an independent generating set, but do not expect any nice properties from this.
answered yesterday
Christoph
10.9k1240
edited yesterday
answered yesterday
Christoph
10.9k1240
answered yesterday
Christoph
10.9k1240
answered yesterday
Christoph
10.9k1240
10.9k1240
add a comment |Â
add a comment |Â
up vote
2
down vote
If $+$ is any operation on a set $C$ and $A$ is a subset of $C$, there exists the smallest subset $B$ of $C$ such that
- $Asubseteq B$
- $B$ is closed under $+$
Such a set $B$ can be described as the intersection of all subsets of $C$ that satisfy properties 1 and 2 above or in a recursive fashion. Define $A_0=A$; assuming you have defined $A_n$, define
$$
A_n+1=a+b:a,bin A_n
$$
Then set $B=bigcup_nge0A_n$ and verify $B$ is closed under $+$ and is contained in every subset of $C$ that contains $A$ and is closed under $+$, so it's the same as the one defined above.
Let's denote by $langle Arangle$ the set $B$, as determined in one of the two equivalent ways above. We say that $A$ is a set of generators for $C$ if $langle Arangle=C$.
If you're looking for uniqueness properties of sets of generators similar to bases in vector spaces, then you're out of luck. It's quite easy to show that there are magmas with finite sets of generators which are minimal (in the sense that removing one element leads to a set which doesn't generate the magma) but have different cardinalities. A simple example is the set $mathbbZ/6mathbbZ$ with the standard addition (modulo $6$): then $2,3$ is a minimal set of generators, but also $1$ is a minimal set of generators.
answered yesterday
egreg
166k1180189
In your definition of $A_n+1$ you want to allow both $a$ and $c$ to be picked from $A_n$ I think. Otherwise, do you get $(a+a)+(a+a)$ from $A=a$?
– Christoph
yesterday
In your example of $mathbb Z$ with addition, $1$ generates only the positive integers and $2,3$ generates only the integers $2,3,4,5,ldots$
– Christoph
yesterday
@Christoph You're right on both issues, I fixed the answer.
– egreg
yesterday
add a comment |Â
up vote
2
down vote
If $+$ is any operation on a set $C$ and $A$ is a subset of $C$, there exists the smallest subset $B$ of $C$ such that
- $Asubseteq B$
- $B$ is closed under $+$
Such a set $B$ can be described as the intersection of all subsets of $C$ that satisfy properties 1 and 2 above or in a recursive fashion. Define $A_0=A$; assuming you have defined $A_n$, define
$$
A_n+1=a+b:a,bin A_n
$$
Then set $B=bigcup_nge0A_n$ and verify $B$ is closed under $+$ and is contained in every subset of $C$ that contains $A$ and is closed under $+$, so it's the same as the one defined above.
Let's denote by $langle Arangle$ the set $B$, as determined in one of the two equivalent ways above. We say that $A$ is a set of generators for $C$ if $langle Arangle=C$.
If you're looking for uniqueness properties of sets of generators similar to bases in vector spaces, then you're out of luck. It's quite easy to show that there are magmas with finite sets of generators which are minimal (in the sense that removing one element leads to a set which doesn't generate the magma) but have different cardinalities. A simple example is the set $mathbbZ/6mathbbZ$ with the standard addition (modulo $6$): then $2,3$ is a minimal set of generators, but also $1$ is a minimal set of generators.
answered yesterday
egreg
166k1180189
In your definition of $A_n+1$ you want to allow both $a$ and $c$ to be picked from $A_n$ I think. Otherwise, do you get $(a+a)+(a+a)$ from $A=a$?
– Christoph
yesterday
In your example of $mathbb Z$ with addition, $1$ generates only the positive integers and $2,3$ generates only the integers $2,3,4,5,ldots$
– Christoph
yesterday
@Christoph You're right on both issues, I fixed the answer.
– egreg
yesterday
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If $+$ is any operation on a set $C$ and $A$ is a subset of $C$, there exists the smallest subset $B$ of $C$ such that
- $Asubseteq B$
- $B$ is closed under $+$
Such a set $B$ can be described as the intersection of all subsets of $C$ that satisfy properties 1 and 2 above or in a recursive fashion. Define $A_0=A$; assuming you have defined $A_n$, define
$$
A_n+1=a+b:a,bin A_n
$$
Then set $B=bigcup_nge0A_n$ and verify $B$ is closed under $+$ and is contained in every subset of $C$ that contains $A$ and is closed under $+$, so it's the same as the one defined above.
Let's denote by $langle Arangle$ the set $B$, as determined in one of the two equivalent ways above. We say that $A$ is a set of generators for $C$ if $langle Arangle=C$.
If you're looking for uniqueness properties of sets of generators similar to bases in vector spaces, then you're out of luck. It's quite easy to show that there are magmas with finite sets of generators which are minimal (in the sense that removing one element leads to a set which doesn't generate the magma) but have different cardinalities. A simple example is the set $mathbbZ/6mathbbZ$ with the standard addition (modulo $6$): then $2,3$ is a minimal set of generators, but also $1$ is a minimal set of generators.
answered yesterday
egreg
166k1180189
If $+$ is any operation on a set $C$ and $A$ is a subset of $C$, there exists the smallest subset $B$ of $C$ such that
- $Asubseteq B$
- $B$ is closed under $+$
Such a set $B$ can be described as the intersection of all subsets of $C$ that satisfy properties 1 and 2 above or in a recursive fashion. Define $A_0=A$; assuming you have defined $A_n$, define
$$
A_n+1=a+b:a,bin A_n
$$
Then set $B=bigcup_nge0A_n$ and verify $B$ is closed under $+$ and is contained in every subset of $C$ that contains $A$ and is closed under $+$, so it's the same as the one defined above.
Let's denote by $langle Arangle$ the set $B$, as determined in one of the two equivalent ways above. We say that $A$ is a set of generators for $C$ if $langle Arangle=C$.
If you're looking for uniqueness properties of sets of generators similar to bases in vector spaces, then you're out of luck. It's quite easy to show that there are magmas with finite sets of generators which are minimal (in the sense that removing one element leads to a set which doesn't generate the magma) but have different cardinalities. A simple example is the set $mathbbZ/6mathbbZ$ with the standard addition (modulo $6$): then $2,3$ is a minimal set of generators, but also $1$ is a minimal set of generators.
answered yesterday
egreg
166k1180189
edited yesterday
answered yesterday
egreg
166k1180189
answered yesterday
egreg
166k1180189
answered yesterday
egreg
166k1180189
166k1180189
In your definition of $A_n+1$ you want to allow both $a$ and $c$ to be picked from $A_n$ I think. Otherwise, do you get $(a+a)+(a+a)$ from $A=a$?
– Christoph
yesterday
In your example of $mathbb Z$ with addition, $1$ generates only the positive integers and $2,3$ generates only the integers $2,3,4,5,ldots$
– Christoph
yesterday
@Christoph You're right on both issues, I fixed the answer.
– egreg
yesterday
add a comment |Â
In your definition of $A_n+1$ you want to allow both $a$ and $c$ to be picked from $A_n$ I think. Otherwise, do you get $(a+a)+(a+a)$ from $A=a$?
– Christoph
yesterday
In your example of $mathbb Z$ with addition, $1$ generates only the positive integers and $2,3$ generates only the integers $2,3,4,5,ldots$
– Christoph
yesterday
@Christoph You're right on both issues, I fixed the answer.
– egreg
yesterday
In your definition of $A_n+1$ you want to allow both $a$ and $c$ to be picked from $A_n$ I think. Otherwise, do you get $(a+a)+(a+a)$ from $A=a$?
– Christoph
yesterday
In your definition of $A_n+1$ you want to allow both $a$ and $c$ to be picked from $A_n$ I think. Otherwise, do you get $(a+a)+(a+a)$ from $A=a$?
– Christoph
yesterday
In your example of $mathbb Z$ with addition, $1$ generates only the positive integers and $2,3$ generates only the integers $2,3,4,5,ldots$
– Christoph
yesterday
In your example of $mathbb Z$ with addition, $1$ generates only the positive integers and $2,3$ generates only the integers $2,3,4,5,ldots$
– Christoph
yesterday
@Christoph You're right on both issues, I fixed the answer.
– egreg
yesterday
@Christoph You're right on both issues, I fixed the answer.
– egreg
yesterday
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2914138%2fnotions-of-basis-and-span-in-a-magma%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
In writing $c_s_4 + c_s_1 + c_s_4 + c_s_2$, where do you put the brackets? Are you writing left-associatively, in that it can be interpreted as $$((c_s_4 + c_s_1) + c_s_4) + c_s_2?$$ You should be careful with non-associative operations.
– Theo Bendit
yesterday
Indeed. I will make it explicit in the question.
– Hazan Tayeb
yesterday