A question about associativity in monoids.
$begingroup$
Lang's "Algebra" (on pg. 4) says the following:
Let $G$ be a monoid. Then $Pi{x_i}$ is defined as $(x_2x_2dots)x_n$.
This probably means $Pi x_i=(((x_1x_2)x_3)dots)x_n$.
He then says
We then have the following rule $Pi_a^bx_i.Pi_{b+1}^c x_j=Pi_a^cx_k$. This ensures that parenthesis can be placed in any way.
I don't understand this. A monoid anyway has associative property. Shouldn't the fact that parentheses can be placed in any way have followed from the first statement itself?
Thanks in advance!
abstract-algebra
asked Feb 26 '14 at 9:42
fierydemonfierydemon
4,42222156
$endgroup$
add a comment |
$begingroup$
Lang's "Algebra" (on pg. 4) says the following:
Let $G$ be a monoid. Then $Pi{x_i}$ is defined as $(x_2x_2dots)x_n$.
This probably means $Pi x_i=(((x_1x_2)x_3)dots)x_n$.
He then says
We then have the following rule $Pi_a^bx_i.Pi_{b+1}^c x_j=Pi_a^cx_k$. This ensures that parenthesis can be placed in any way.
I don't understand this. A monoid anyway has associative property. Shouldn't the fact that parentheses can be placed in any way have followed from the first statement itself?
Thanks in advance!
abstract-algebra
asked Feb 26 '14 at 9:42
fierydemonfierydemon
4,42222156
$endgroup$
$begingroup$
He goes on to say "The proof is easy by induction, and we shall leave it as an exercise". It follows indeed, but I think this rule may be seen as an intermediate step in proving the fact the parentheses can be placed in any way.
$endgroup$
– Marcin Łoś
Feb 26 '14 at 9:47
$begingroup$
@MarcinŁoś- The proof that I have in mind does not involve the second step at all!
$endgroup$
– fierydemon
Feb 26 '14 at 9:54
$begingroup$
The associativity rule is for precisely 3 elements. You must prove the version for any finite number of elements using induction. Anything else is hand-waving and amounts to assuming what you intend to prove.
$endgroup$
– MPW
Feb 26 '14 at 9:59
$begingroup$
@MPW- Say we have $a_1a_2a_3a_4$. The first statement defines it as $(((a_1a_2)a_3)a_4)$. Due to the associativity of the elements of a monoid, you can arrange the parentheses however you like (for more than 3 elements too).
$endgroup$
– fierydemon
Feb 26 '14 at 10:03
$begingroup$
@AyushKhaitan That is true, but it requires proof, which is Lang's point. Though one may have hardwired intuitive beliefs why this is true, they need to be translated into a formal proof to be rigorous.
$endgroup$
– Bill Dubuque
Feb 26 '14 at 17:25
add a comment |
$begingroup$
Lang's "Algebra" (on pg. 4) says the following:
Let $G$ be a monoid. Then $Pi{x_i}$ is defined as $(x_2x_2dots)x_n$.
This probably means $Pi x_i=(((x_1x_2)x_3)dots)x_n$.
He then says
We then have the following rule $Pi_a^bx_i.Pi_{b+1}^c x_j=Pi_a^cx_k$. This ensures that parenthesis can be placed in any way.
I don't understand this. A monoid anyway has associative property. Shouldn't the fact that parentheses can be placed in any way have followed from the first statement itself?
Thanks in advance!
abstract-algebra
asked Feb 26 '14 at 9:42
fierydemonfierydemon
4,42222156
$endgroup$
Lang's "Algebra" (on pg. 4) says the following:
Let $G$ be a monoid. Then $Pi{x_i}$ is defined as $(x_2x_2dots)x_n$.
This probably means $Pi x_i=(((x_1x_2)x_3)dots)x_n$.
He then says
We then have the following rule $Pi_a^bx_i.Pi_{b+1}^c x_j=Pi_a^cx_k$. This ensures that parenthesis can be placed in any way.
I don't understand this. A monoid anyway has associative property. Shouldn't the fact that parentheses can be placed in any way have followed from the first statement itself?
Thanks in advance!
abstract-algebra
abstract-algebra
asked Feb 26 '14 at 9:42
fierydemonfierydemon
4,42222156
asked Feb 26 '14 at 9:42
fierydemonfierydemon
4,42222156
edited Feb 26 '14 at 9:55
fierydemon
asked Feb 26 '14 at 9:42
fierydemonfierydemon
4,42222156
asked Feb 26 '14 at 9:42
fierydemonfierydemon
4,42222156
asked Feb 26 '14 at 9:42
fierydemonfierydemon
4,42222156
4,42222156
$begingroup$
He goes on to say "The proof is easy by induction, and we shall leave it as an exercise". It follows indeed, but I think this rule may be seen as an intermediate step in proving the fact the parentheses can be placed in any way.
$endgroup$
– Marcin Łoś
Feb 26 '14 at 9:47
$begingroup$
@MarcinŁoś- The proof that I have in mind does not involve the second step at all!
$endgroup$
– fierydemon
Feb 26 '14 at 9:54
$begingroup$
The associativity rule is for precisely 3 elements. You must prove the version for any finite number of elements using induction. Anything else is hand-waving and amounts to assuming what you intend to prove.
$endgroup$
– MPW
Feb 26 '14 at 9:59
$begingroup$
@MPW- Say we have $a_1a_2a_3a_4$. The first statement defines it as $(((a_1a_2)a_3)a_4)$. Due to the associativity of the elements of a monoid, you can arrange the parentheses however you like (for more than 3 elements too).
$endgroup$
– fierydemon
Feb 26 '14 at 10:03
$begingroup$
@AyushKhaitan That is true, but it requires proof, which is Lang's point. Though one may have hardwired intuitive beliefs why this is true, they need to be translated into a formal proof to be rigorous.
$endgroup$
– Bill Dubuque
Feb 26 '14 at 17:25
add a comment |
$begingroup$
He goes on to say "The proof is easy by induction, and we shall leave it as an exercise". It follows indeed, but I think this rule may be seen as an intermediate step in proving the fact the parentheses can be placed in any way.
$endgroup$
– Marcin Łoś
Feb 26 '14 at 9:47
$begingroup$
@MarcinŁoś- The proof that I have in mind does not involve the second step at all!
$endgroup$
– fierydemon
Feb 26 '14 at 9:54
$begingroup$
The associativity rule is for precisely 3 elements. You must prove the version for any finite number of elements using induction. Anything else is hand-waving and amounts to assuming what you intend to prove.
$endgroup$
– MPW
Feb 26 '14 at 9:59
$begingroup$
@MPW- Say we have $a_1a_2a_3a_4$. The first statement defines it as $(((a_1a_2)a_3)a_4)$. Due to the associativity of the elements of a monoid, you can arrange the parentheses however you like (for more than 3 elements too).
$endgroup$
– fierydemon
Feb 26 '14 at 10:03
$begingroup$
@AyushKhaitan That is true, but it requires proof, which is Lang's point. Though one may have hardwired intuitive beliefs why this is true, they need to be translated into a formal proof to be rigorous.
$endgroup$
– Bill Dubuque
Feb 26 '14 at 17:25
$begingroup$
He goes on to say "The proof is easy by induction, and we shall leave it as an exercise". It follows indeed, but I think this rule may be seen as an intermediate step in proving the fact the parentheses can be placed in any way.
$endgroup$
– Marcin Łoś
Feb 26 '14 at 9:47
$begingroup$
He goes on to say "The proof is easy by induction, and we shall leave it as an exercise". It follows indeed, but I think this rule may be seen as an intermediate step in proving the fact the parentheses can be placed in any way.
$endgroup$
– Marcin Łoś
Feb 26 '14 at 9:47
$begingroup$
@MarcinŁoś- The proof that I have in mind does not involve the second step at all!
$endgroup$
– fierydemon
Feb 26 '14 at 9:54
$begingroup$
@MarcinŁoś- The proof that I have in mind does not involve the second step at all!
$endgroup$
– fierydemon
Feb 26 '14 at 9:54
$begingroup$
The associativity rule is for precisely 3 elements. You must prove the version for any finite number of elements using induction. Anything else is hand-waving and amounts to assuming what you intend to prove.
$endgroup$
– MPW
Feb 26 '14 at 9:59
$begingroup$
The associativity rule is for precisely 3 elements. You must prove the version for any finite number of elements using induction. Anything else is hand-waving and amounts to assuming what you intend to prove.
$endgroup$
– MPW
Feb 26 '14 at 9:59
$begingroup$
@MPW- Say we have $a_1a_2a_3a_4$. The first statement defines it as $(((a_1a_2)a_3)a_4)$. Due to the associativity of the elements of a monoid, you can arrange the parentheses however you like (for more than 3 elements too).
$endgroup$
– fierydemon
Feb 26 '14 at 10:03
$begingroup$
@MPW- Say we have $a_1a_2a_3a_4$. The first statement defines it as $(((a_1a_2)a_3)a_4)$. Due to the associativity of the elements of a monoid, you can arrange the parentheses however you like (for more than 3 elements too).
$endgroup$
– fierydemon
Feb 26 '14 at 10:03
$begingroup$
@AyushKhaitan That is true, but it requires proof, which is Lang's point. Though one may have hardwired intuitive beliefs why this is true, they need to be translated into a formal proof to be rigorous.
$endgroup$
– Bill Dubuque
Feb 26 '14 at 17:25
$begingroup$
@AyushKhaitan That is true, but it requires proof, which is Lang's point. Though one may have hardwired intuitive beliefs why this is true, they need to be translated into a formal proof to be rigorous.
$endgroup$
– Bill Dubuque
Feb 26 '14 at 17:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Lang's point is that these intuitive normalizations using the assoicative law require formal proof. Below is one way to do so from Bergman's superb Companion to Lang's Algebra, from pp. 5-7 of the Introduction, and Notes to Chapter I. I highly recommend that anyone reading Lang's Algebra have Bergman's notes close at hand. I suggest that you attempt to prove the Lemma's below before consulting their proofs in Bergman's notes.
answered Feb 26 '14 at 17:42
Bill DubuqueBill Dubuque
211k29192645
$endgroup$
$begingroup$
Thanks Bill, that was most helpful.
$endgroup$
– fierydemon
Feb 27 '14 at 1:51
add a comment |
$begingroup$
I am just getting into this book, and came up with the following proof, which I thought was simple enough. I am curious what you all would think....
Let $G$ be a monoid. Fix an integer $m ge 1$. Suppose $n=1$.
Then we have
$$(Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{n} x_{q+m})$$
$$ = (x_1 * ... * x_m) * x_{m+1}$$
$$ = Pi_{p=1}^{m+1} x_p.$$
Suppose $n=k$ for some positive integer $k$. Assume this inductive hypothesis:
$$(Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{k} x_{q+m}) = (Pi_{p=1}^{m+k} x_p)$$
Then we have
$$(Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{k+1} x_{q+m})$$
$= (Pi_{p=1}^{m} x_p) * bigl((x_{1+m} * ... * x_{k+m}) * x_{k+1+m}bigr)$ by definition of product
$= bigl((Pi_{p=1}^{m} x_p) * (x_{1+m} * ... * x_{k+m})bigr) * x_{k+1+m}$ by associativity of the law of composition of $G$.
$= bigl((Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{k} x_{q+m})bigr) * x_{k+1+m}$ by definition of product
$= (Pi_{p=1}^{m+k} x_p) * x_{k+1+m}$ by our inductive hypothesis
$= Pi_{p=1}^{k+1+m} x_p$ by our definition of product.
By the principle of mathematical induction, that proves the point for all $n$.
answered Dec 14 '18 at 6:04
AddisonAddison
416315
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Lang's point is that these intuitive normalizations using the assoicative law require formal proof. Below is one way to do so from Bergman's superb Companion to Lang's Algebra, from pp. 5-7 of the Introduction, and Notes to Chapter I. I highly recommend that anyone reading Lang's Algebra have Bergman's notes close at hand. I suggest that you attempt to prove the Lemma's below before consulting their proofs in Bergman's notes.
answered Feb 26 '14 at 17:42
Bill DubuqueBill Dubuque
211k29192645
$endgroup$
$begingroup$
Thanks Bill, that was most helpful.
$endgroup$
– fierydemon
Feb 27 '14 at 1:51
add a comment |
$begingroup$
Lang's point is that these intuitive normalizations using the assoicative law require formal proof. Below is one way to do so from Bergman's superb Companion to Lang's Algebra, from pp. 5-7 of the Introduction, and Notes to Chapter I. I highly recommend that anyone reading Lang's Algebra have Bergman's notes close at hand. I suggest that you attempt to prove the Lemma's below before consulting their proofs in Bergman's notes.
answered Feb 26 '14 at 17:42
Bill DubuqueBill Dubuque
211k29192645
$endgroup$
$begingroup$
Thanks Bill, that was most helpful.
$endgroup$
– fierydemon
Feb 27 '14 at 1:51
add a comment |
$begingroup$
Lang's point is that these intuitive normalizations using the assoicative law require formal proof. Below is one way to do so from Bergman's superb Companion to Lang's Algebra, from pp. 5-7 of the Introduction, and Notes to Chapter I. I highly recommend that anyone reading Lang's Algebra have Bergman's notes close at hand. I suggest that you attempt to prove the Lemma's below before consulting their proofs in Bergman's notes.
answered Feb 26 '14 at 17:42
Bill DubuqueBill Dubuque
211k29192645
$endgroup$
Lang's point is that these intuitive normalizations using the assoicative law require formal proof. Below is one way to do so from Bergman's superb Companion to Lang's Algebra, from pp. 5-7 of the Introduction, and Notes to Chapter I. I highly recommend that anyone reading Lang's Algebra have Bergman's notes close at hand. I suggest that you attempt to prove the Lemma's below before consulting their proofs in Bergman's notes.
answered Feb 26 '14 at 17:42
Bill DubuqueBill Dubuque
211k29192645
answered Feb 26 '14 at 17:42
Bill DubuqueBill Dubuque
211k29192645
answered Feb 26 '14 at 17:42
Bill DubuqueBill Dubuque
211k29192645
answered Feb 26 '14 at 17:42
Bill DubuqueBill Dubuque
211k29192645
211k29192645
$begingroup$
Thanks Bill, that was most helpful.
$endgroup$
– fierydemon
Feb 27 '14 at 1:51
add a comment |
$begingroup$
Thanks Bill, that was most helpful.
$endgroup$
– fierydemon
Feb 27 '14 at 1:51
$begingroup$
Thanks Bill, that was most helpful.
$endgroup$
– fierydemon
Feb 27 '14 at 1:51
$begingroup$
Thanks Bill, that was most helpful.
$endgroup$
– fierydemon
Feb 27 '14 at 1:51
add a comment |
$begingroup$
I am just getting into this book, and came up with the following proof, which I thought was simple enough. I am curious what you all would think....
Let $G$ be a monoid. Fix an integer $m ge 1$. Suppose $n=1$.
Then we have
$$(Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{n} x_{q+m})$$
$$ = (x_1 * ... * x_m) * x_{m+1}$$
$$ = Pi_{p=1}^{m+1} x_p.$$
Suppose $n=k$ for some positive integer $k$. Assume this inductive hypothesis:
$$(Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{k} x_{q+m}) = (Pi_{p=1}^{m+k} x_p)$$
Then we have
$$(Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{k+1} x_{q+m})$$
$= (Pi_{p=1}^{m} x_p) * bigl((x_{1+m} * ... * x_{k+m}) * x_{k+1+m}bigr)$ by definition of product
$= bigl((Pi_{p=1}^{m} x_p) * (x_{1+m} * ... * x_{k+m})bigr) * x_{k+1+m}$ by associativity of the law of composition of $G$.
$= bigl((Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{k} x_{q+m})bigr) * x_{k+1+m}$ by definition of product
$= (Pi_{p=1}^{m+k} x_p) * x_{k+1+m}$ by our inductive hypothesis
$= Pi_{p=1}^{k+1+m} x_p$ by our definition of product.
By the principle of mathematical induction, that proves the point for all $n$.
answered Dec 14 '18 at 6:04
AddisonAddison
416315
$endgroup$
add a comment |
$begingroup$
I am just getting into this book, and came up with the following proof, which I thought was simple enough. I am curious what you all would think....
Let $G$ be a monoid. Fix an integer $m ge 1$. Suppose $n=1$.
Then we have
$$(Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{n} x_{q+m})$$
$$ = (x_1 * ... * x_m) * x_{m+1}$$
$$ = Pi_{p=1}^{m+1} x_p.$$
Suppose $n=k$ for some positive integer $k$. Assume this inductive hypothesis:
$$(Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{k} x_{q+m}) = (Pi_{p=1}^{m+k} x_p)$$
Then we have
$$(Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{k+1} x_{q+m})$$
$= (Pi_{p=1}^{m} x_p) * bigl((x_{1+m} * ... * x_{k+m}) * x_{k+1+m}bigr)$ by definition of product
$= bigl((Pi_{p=1}^{m} x_p) * (x_{1+m} * ... * x_{k+m})bigr) * x_{k+1+m}$ by associativity of the law of composition of $G$.
$= bigl((Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{k} x_{q+m})bigr) * x_{k+1+m}$ by definition of product
$= (Pi_{p=1}^{m+k} x_p) * x_{k+1+m}$ by our inductive hypothesis
$= Pi_{p=1}^{k+1+m} x_p$ by our definition of product.
By the principle of mathematical induction, that proves the point for all $n$.
answered Dec 14 '18 at 6:04
AddisonAddison
416315
$endgroup$
add a comment |
$begingroup$
I am just getting into this book, and came up with the following proof, which I thought was simple enough. I am curious what you all would think....
Let $G$ be a monoid. Fix an integer $m ge 1$. Suppose $n=1$.
Then we have
$$(Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{n} x_{q+m})$$
$$ = (x_1 * ... * x_m) * x_{m+1}$$
$$ = Pi_{p=1}^{m+1} x_p.$$
Suppose $n=k$ for some positive integer $k$. Assume this inductive hypothesis:
$$(Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{k} x_{q+m}) = (Pi_{p=1}^{m+k} x_p)$$
Then we have
$$(Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{k+1} x_{q+m})$$
$= (Pi_{p=1}^{m} x_p) * bigl((x_{1+m} * ... * x_{k+m}) * x_{k+1+m}bigr)$ by definition of product
$= bigl((Pi_{p=1}^{m} x_p) * (x_{1+m} * ... * x_{k+m})bigr) * x_{k+1+m}$ by associativity of the law of composition of $G$.
$= bigl((Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{k} x_{q+m})bigr) * x_{k+1+m}$ by definition of product
$= (Pi_{p=1}^{m+k} x_p) * x_{k+1+m}$ by our inductive hypothesis
$= Pi_{p=1}^{k+1+m} x_p$ by our definition of product.
By the principle of mathematical induction, that proves the point for all $n$.
answered Dec 14 '18 at 6:04
AddisonAddison
416315
$endgroup$
I am just getting into this book, and came up with the following proof, which I thought was simple enough. I am curious what you all would think....
Let $G$ be a monoid. Fix an integer $m ge 1$. Suppose $n=1$.
Then we have
$$(Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{n} x_{q+m})$$
$$ = (x_1 * ... * x_m) * x_{m+1}$$
$$ = Pi_{p=1}^{m+1} x_p.$$
Suppose $n=k$ for some positive integer $k$. Assume this inductive hypothesis:
$$(Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{k} x_{q+m}) = (Pi_{p=1}^{m+k} x_p)$$
Then we have
$$(Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{k+1} x_{q+m})$$
$= (Pi_{p=1}^{m} x_p) * bigl((x_{1+m} * ... * x_{k+m}) * x_{k+1+m}bigr)$ by definition of product
$= bigl((Pi_{p=1}^{m} x_p) * (x_{1+m} * ... * x_{k+m})bigr) * x_{k+1+m}$ by associativity of the law of composition of $G$.
$= bigl((Pi_{p=1}^{m} x_p) * (Pi_{q=1}^{k} x_{q+m})bigr) * x_{k+1+m}$ by definition of product
$= (Pi_{p=1}^{m+k} x_p) * x_{k+1+m}$ by our inductive hypothesis
$= Pi_{p=1}^{k+1+m} x_p$ by our definition of product.
By the principle of mathematical induction, that proves the point for all $n$.
answered Dec 14 '18 at 6:04
AddisonAddison
416315
answered Dec 14 '18 at 6:04
AddisonAddison
416315
answered Dec 14 '18 at 6:04
AddisonAddison
416315
answered Dec 14 '18 at 6:04
AddisonAddison
416315
416315
add a comment |
add a comment |
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$begingroup$
He goes on to say "The proof is easy by induction, and we shall leave it as an exercise". It follows indeed, but I think this rule may be seen as an intermediate step in proving the fact the parentheses can be placed in any way.
$endgroup$
– Marcin Łoś
Feb 26 '14 at 9:47
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@MarcinŁoś- The proof that I have in mind does not involve the second step at all!
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– fierydemon
Feb 26 '14 at 9:54
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The associativity rule is for precisely 3 elements. You must prove the version for any finite number of elements using induction. Anything else is hand-waving and amounts to assuming what you intend to prove.
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– MPW
Feb 26 '14 at 9:59
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@MPW- Say we have $a_1a_2a_3a_4$. The first statement defines it as $(((a_1a_2)a_3)a_4)$. Due to the associativity of the elements of a monoid, you can arrange the parentheses however you like (for more than 3 elements too).
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– fierydemon
Feb 26 '14 at 10:03
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@AyushKhaitan That is true, but it requires proof, which is Lang's point. Though one may have hardwired intuitive beliefs why this is true, they need to be translated into a formal proof to be rigorous.
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– Bill Dubuque
Feb 26 '14 at 17:25