Is there a finite list of identites in the language of $(mathbb{N},0,1,+,times,mathrm{gcd},mathrm{lcm})$ that...
10
$begingroup$
Let $Phi$ denote the set of all identities satisfied by $(mathbb{N},0,1,+,times,mathrm{gcd},mathrm{lcm}).$
Question. Is $Phi$ finitely axiomatizable? If so, I'd like to see a list of identities.
Noteworthy elements of $Phi$:
- $(mathbb{N},0,1,+,times)$ is a commutative semiring
- $(mathbb{N},1,0,mathrm{gcd},mathrm{lcm})$ is a bounded distributive lattice (with bottom $1$ and top $0$)
- $mathrm{gcd}(a,b+a) = mathrm{gcd}(a,b)$
- $gcd(a+b,operatorname{lcm}(a,b))=gcd(a,b)$
- $mathrm{gcd}(a,b)mathrm{lcm}(a,b) = ab$
The second last identity above actually follows from the preceeding one's; see Bill's answer here.
elementary-number-theory logic universal-algebra
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 10 '16 at 18:47
goblingoblin
36.8k1159193
$endgroup$
|
show 9 more comments
10
$begingroup$
Let $Phi$ denote the set of all identities satisfied by $(mathbb{N},0,1,+,times,mathrm{gcd},mathrm{lcm}).$
Question. Is $Phi$ finitely axiomatizable? If so, I'd like to see a list of identities.
Noteworthy elements of $Phi$:
- $(mathbb{N},0,1,+,times)$ is a commutative semiring
- $(mathbb{N},1,0,mathrm{gcd},mathrm{lcm})$ is a bounded distributive lattice (with bottom $1$ and top $0$)
- $mathrm{gcd}(a,b+a) = mathrm{gcd}(a,b)$
- $gcd(a+b,operatorname{lcm}(a,b))=gcd(a,b)$
- $mathrm{gcd}(a,b)mathrm{lcm}(a,b) = ab$
The second last identity above actually follows from the preceeding one's; see Bill's answer here.
elementary-number-theory logic universal-algebra
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 10 '16 at 18:47
goblingoblin
36.8k1159193
$endgroup$
$begingroup$
"$(mathbb{N}, 1, 0, gcd, $lcm$)$ is a bounded distributive lattice (with least element 1)". And, by the way, greatest element is $0$.
$endgroup$
– amrsa
Jul 11 '16 at 11:46
1
$begingroup$
@amrsa, of course it is. I just couldn't fit that onto one line :)
$endgroup$
– goblin
Jul 11 '16 at 11:53
$begingroup$
You can fit it in if you say "with top $0$ and bottom $1$".
$endgroup$
– user21820
Jul 11 '16 at 13:25
2
$begingroup$
If this is possible it will be just barely. If you add merely a $min$ operator to the language, then the set of identities is definitely not recursively enumerable, due to Matiyasevich's theorem.
$endgroup$
– Henning Makholm
Jul 11 '16 at 13:44
$begingroup$
... combined with $$min(1+gcd(a+1,b+1),operatorname{lcm}(a+1,b+1))=1+gcd(a+1,b+1) iff ane b$$
$endgroup$
– Henning Makholm
Jul 11 '16 at 14:09
|
show 9 more comments
10
10
10
2
$begingroup$
Let $Phi$ denote the set of all identities satisfied by $(mathbb{N},0,1,+,times,mathrm{gcd},mathrm{lcm}).$
Question. Is $Phi$ finitely axiomatizable? If so, I'd like to see a list of identities.
Noteworthy elements of $Phi$:
- $(mathbb{N},0,1,+,times)$ is a commutative semiring
- $(mathbb{N},1,0,mathrm{gcd},mathrm{lcm})$ is a bounded distributive lattice (with bottom $1$ and top $0$)
- $mathrm{gcd}(a,b+a) = mathrm{gcd}(a,b)$
- $gcd(a+b,operatorname{lcm}(a,b))=gcd(a,b)$
- $mathrm{gcd}(a,b)mathrm{lcm}(a,b) = ab$
The second last identity above actually follows from the preceeding one's; see Bill's answer here.
elementary-number-theory logic universal-algebra
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 10 '16 at 18:47
goblingoblin
36.8k1159193
$endgroup$
Let $Phi$ denote the set of all identities satisfied by $(mathbb{N},0,1,+,times,mathrm{gcd},mathrm{lcm}).$
Question. Is $Phi$ finitely axiomatizable? If so, I'd like to see a list of identities.
Noteworthy elements of $Phi$:
- $(mathbb{N},0,1,+,times)$ is a commutative semiring
- $(mathbb{N},1,0,mathrm{gcd},mathrm{lcm})$ is a bounded distributive lattice (with bottom $1$ and top $0$)
- $mathrm{gcd}(a,b+a) = mathrm{gcd}(a,b)$
- $gcd(a+b,operatorname{lcm}(a,b))=gcd(a,b)$
- $mathrm{gcd}(a,b)mathrm{lcm}(a,b) = ab$
The second last identity above actually follows from the preceeding one's; see Bill's answer here.
elementary-number-theory logic universal-algebra
elementary-number-theory logic universal-algebra
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 10 '16 at 18:47
goblingoblin
36.8k1159193
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 10 '16 at 18:47
goblingoblin
36.8k1159193
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
asked Jul 10 '16 at 18:47
goblingoblin
36.8k1159193
asked Jul 10 '16 at 18:47
goblingoblin
36.8k1159193
asked Jul 10 '16 at 18:47
goblingoblin
36.8k1159193
36.8k1159193
$begingroup$
"$(mathbb{N}, 1, 0, gcd, $lcm$)$ is a bounded distributive lattice (with least element 1)". And, by the way, greatest element is $0$.
$endgroup$
– amrsa
Jul 11 '16 at 11:46
1
$begingroup$
@amrsa, of course it is. I just couldn't fit that onto one line :)
$endgroup$
– goblin
Jul 11 '16 at 11:53
$begingroup$
You can fit it in if you say "with top $0$ and bottom $1$".
$endgroup$
– user21820
Jul 11 '16 at 13:25
2
$begingroup$
If this is possible it will be just barely. If you add merely a $min$ operator to the language, then the set of identities is definitely not recursively enumerable, due to Matiyasevich's theorem.
$endgroup$
– Henning Makholm
Jul 11 '16 at 13:44
$begingroup$
... combined with $$min(1+gcd(a+1,b+1),operatorname{lcm}(a+1,b+1))=1+gcd(a+1,b+1) iff ane b$$
$endgroup$
– Henning Makholm
Jul 11 '16 at 14:09
|
show 9 more comments
$begingroup$
"$(mathbb{N}, 1, 0, gcd, $lcm$)$ is a bounded distributive lattice (with least element 1)". And, by the way, greatest element is $0$.
$endgroup$
– amrsa
Jul 11 '16 at 11:46
1
$begingroup$
@amrsa, of course it is. I just couldn't fit that onto one line :)
$endgroup$
– goblin
Jul 11 '16 at 11:53
$begingroup$
You can fit it in if you say "with top $0$ and bottom $1$".
$endgroup$
– user21820
Jul 11 '16 at 13:25
2
$begingroup$
If this is possible it will be just barely. If you add merely a $min$ operator to the language, then the set of identities is definitely not recursively enumerable, due to Matiyasevich's theorem.
$endgroup$
– Henning Makholm
Jul 11 '16 at 13:44
$begingroup$
... combined with $$min(1+gcd(a+1,b+1),operatorname{lcm}(a+1,b+1))=1+gcd(a+1,b+1) iff ane b$$
$endgroup$
– Henning Makholm
Jul 11 '16 at 14:09
$begingroup$
"$(mathbb{N}, 1, 0, gcd, $lcm$)$ is a bounded distributive lattice (with least element 1)". And, by the way, greatest element is $0$.
$endgroup$
– amrsa
Jul 11 '16 at 11:46
$begingroup$
"$(mathbb{N}, 1, 0, gcd, $lcm$)$ is a bounded distributive lattice (with least element 1)". And, by the way, greatest element is $0$.
$endgroup$
– amrsa
Jul 11 '16 at 11:46
1
1
$begingroup$
@amrsa, of course it is. I just couldn't fit that onto one line :)
$endgroup$
– goblin
Jul 11 '16 at 11:53
$begingroup$
@amrsa, of course it is. I just couldn't fit that onto one line :)
$endgroup$
– goblin
Jul 11 '16 at 11:53
$begingroup$
You can fit it in if you say "with top $0$ and bottom $1$".
$endgroup$
– user21820
Jul 11 '16 at 13:25
$begingroup$
You can fit it in if you say "with top $0$ and bottom $1$".
$endgroup$
– user21820
Jul 11 '16 at 13:25
2
2
$begingroup$
If this is possible it will be just barely. If you add merely a $min$ operator to the language, then the set of identities is definitely not recursively enumerable, due to Matiyasevich's theorem.
$endgroup$
– Henning Makholm
Jul 11 '16 at 13:44
$begingroup$
If this is possible it will be just barely. If you add merely a $min$ operator to the language, then the set of identities is definitely not recursively enumerable, due to Matiyasevich's theorem.
$endgroup$
– Henning Makholm
Jul 11 '16 at 13:44
$begingroup$
... combined with $$min(1+gcd(a+1,b+1),operatorname{lcm}(a+1,b+1))=1+gcd(a+1,b+1) iff ane b$$
$endgroup$
– Henning Makholm
Jul 11 '16 at 14:09
$begingroup$
... combined with $$min(1+gcd(a+1,b+1),operatorname{lcm}(a+1,b+1))=1+gcd(a+1,b+1) iff ane b$$
$endgroup$
– Henning Makholm
Jul 11 '16 at 14:09
|
show 9 more comments
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$begingroup$
"$(mathbb{N}, 1, 0, gcd, $lcm$)$ is a bounded distributive lattice (with least element 1)". And, by the way, greatest element is $0$.
$endgroup$
– amrsa
Jul 11 '16 at 11:46
1
$begingroup$
@amrsa, of course it is. I just couldn't fit that onto one line :)
$endgroup$
– goblin
Jul 11 '16 at 11:53
$begingroup$
You can fit it in if you say "with top $0$ and bottom $1$".
$endgroup$
– user21820
Jul 11 '16 at 13:25
2
$begingroup$
If this is possible it will be just barely. If you add merely a $min$ operator to the language, then the set of identities is definitely not recursively enumerable, due to Matiyasevich's theorem.
$endgroup$
– Henning Makholm
Jul 11 '16 at 13:44
$begingroup$
... combined with $$min(1+gcd(a+1,b+1),operatorname{lcm}(a+1,b+1))=1+gcd(a+1,b+1) iff ane b$$
$endgroup$
– Henning Makholm
Jul 11 '16 at 14:09