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Relation between primeness and co-primeness of integers

Relation between residues and primitive roots modulo $p$Interesting pairs of properties and relations in ring theoryTriangle Requirements based of triangle InequalityIs there a logic of sufficiency? Or goals?Using rules of inference (Leibniz) to prove theorems.Complicated FOL Formula {∃a,c(a≠c) ∧ ∀a,c[(a≠c)⇒(h(a,c)⟺ ¬h(c,a))] ∧ ∀a,c[h(a,c) ⇒ ∃b(h(a,b)∧h(b,c)∧b≠c)]} ⇒ ¬{∃a∀b[b≠a⇒ h(a,b)]}What is the method to solve these kind of questions?True Sentence in Calculus of Classes: Individual Domain and CardinalityShow $alpha leftrightarrow psi$ is a tautology if and only if $alpha approx psi$How to prove axiom of pairs from power set axiom and replacement?Prove or give counter example: $forall x exists y(Q(y)wedge P(x))vDash exists yforall x(Q(y)wedge P(x))$Interesting pairs of properties and relations in ring theory

7

4

$begingroup$

I wonder what this stunning formal analogy between the definitions of being co-prime (for two integers) and being prime (for one integer) might reveal – and how:

$alpha, beta$ are co-prime iff

$$(forall x) alpha|x wedge beta|x leftrightarrow alphabeta|x$$

$alpha$ is prime iff

$$(forall xy) alpha|x vee alpha|y leftrightarrow alpha|xy$$

Note that and how the two definitions are equivalent modulo swapping

• $wedge$ and $vee$ on the left side




• constants and variables on the right side together with




• the direction of divisibility $|$ with respect to the product _{[thanks to user Wojowu]}




• the direction of inference $rightarrow$ _{[thanks to user Roll up and smoke Adjoint]}

number-theory logic prime-numbers integers coprime

share|cite|improve this question

edited Nov 3 '18 at 9:18

Hans Stricker

asked Nov 2 '18 at 19:10

Hans StrickerHans Stricker

6,47943994

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note you also swap the direction of divisibility. This is an interesting observation, but I don't think there is any "stunning" meaning behind it.
  $endgroup$
  – Wojowu
  Nov 2 '18 at 19:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that you can also replace $rightarrow$ with $leftrightarrow$ since each of those propositions on either side are in isomorphism with one another.
  $endgroup$
  – BananaCats
  Nov 3 '18 at 3:29
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is most likely a nice interpretation in the language of lattices/filters.
  $endgroup$
  – Slade
  Nov 3 '18 at 9:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the poset category for integers dividing one another $text{Hom}(X,Y)$ consists of a single point or is empty. Let $(X mid Y)$ denote the nonempty hom-set. Notice that multiplication by an element $alpha in Bbb{Z}$ is a functor from the poset category to itself since $(X mid Y) implies (alpha X mid alpha Y)$ and so on proves the functoriality of multiplication.
  $endgroup$
  – BananaCats
  Nov 4 '18 at 6:34
  
  
  

  
  
  
  

add a comment | 

7

4

$begingroup$

I wonder what this stunning formal analogy between the definitions of being co-prime (for two integers) and being prime (for one integer) might reveal – and how:

$alpha, beta$ are co-prime iff

$$(forall x) alpha|x wedge beta|x leftrightarrow alphabeta|x$$

$alpha$ is prime iff

$$(forall xy) alpha|x vee alpha|y leftrightarrow alpha|xy$$

Note that and how the two definitions are equivalent modulo swapping

• $wedge$ and $vee$ on the left side




• constants and variables on the right side together with




• the direction of divisibility $|$ with respect to the product _{[thanks to user Wojowu]}




• the direction of inference $rightarrow$ _{[thanks to user Roll up and smoke Adjoint]}

number-theory logic prime-numbers integers coprime

share|cite|improve this question

edited Nov 3 '18 at 9:18

Hans Stricker

asked Nov 2 '18 at 19:10

Hans StrickerHans Stricker

6,47943994

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note you also swap the direction of divisibility. This is an interesting observation, but I don't think there is any "stunning" meaning behind it.
  $endgroup$
  – Wojowu
  Nov 2 '18 at 19:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that you can also replace $rightarrow$ with $leftrightarrow$ since each of those propositions on either side are in isomorphism with one another.
  $endgroup$
  – BananaCats
  Nov 3 '18 at 3:29
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is most likely a nice interpretation in the language of lattices/filters.
  $endgroup$
  – Slade
  Nov 3 '18 at 9:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the poset category for integers dividing one another $text{Hom}(X,Y)$ consists of a single point or is empty. Let $(X mid Y)$ denote the nonempty hom-set. Notice that multiplication by an element $alpha in Bbb{Z}$ is a functor from the poset category to itself since $(X mid Y) implies (alpha X mid alpha Y)$ and so on proves the functoriality of multiplication.
  $endgroup$
  – BananaCats
  Nov 4 '18 at 6:34
  
  
  

  
  
  
  

add a comment | 

$begingroup$

I wonder what this stunning formal analogy between the definitions of being co-prime (for two integers) and being prime (for one integer) might reveal – and how:

$alpha, beta$ are co-prime iff

$$(forall x) alpha|x wedge beta|x leftrightarrow alphabeta|x$$

$alpha$ is prime iff

$$(forall xy) alpha|x vee alpha|y leftrightarrow alpha|xy$$

Note that and how the two definitions are equivalent modulo swapping

• $wedge$ and $vee$ on the left side




• constants and variables on the right side together with




• the direction of divisibility $|$ with respect to the product _{[thanks to user Wojowu]}




• the direction of inference $rightarrow$ _{[thanks to user Roll up and smoke Adjoint]}

number-theory logic prime-numbers integers coprime

share|cite|improve this question

edited Nov 3 '18 at 9:18

Hans Stricker

asked Nov 2 '18 at 19:10

Hans StrickerHans Stricker

6,47943994

$endgroup$

share|cite|improve this question

edited Nov 3 '18 at 9:18

Hans Stricker

asked Nov 2 '18 at 19:10

Hans StrickerHans Stricker

6,47943994

share|cite|improve this question

edited Nov 3 '18 at 9:18

Hans Stricker

asked Nov 2 '18 at 19:10

Hans StrickerHans Stricker

6,47943994

asked Nov 2 '18 at 19:10

Hans StrickerHans Stricker

6,47943994

asked Nov 2 '18 at 19:10

Hans StrickerHans Stricker

6,47943994

2 Answers
2

active

oldest

votes

1

$begingroup$

In the poset category of integers dividing one another we write sometimes write $alpha to x$ instead of $alpha mid x$.

The "product" $alpha beta$ of two coprime objects $alpha, beta$ is such that $alpha to alpha beta leftarrow beta$ and for any object $x$ such that $alpha to x leftarrow beta$ then there is a unique arrow $alpha beta to x$. That is precisely the definition of coproduct in a general category.

Thus when $alpha, beta$ are coprime, then the category definitely has a coproduct for them.

Thus $wedge$ is encoded in the fact that $alpha to x leftarrow beta$, ie. both morphisms exist simultaneously. I believe product is $gcd(alpha, beta)$.

---

Actually, it turns out that the coproduct exists for any two integers and it's $text{lcm}(alpha, beta)$.

---

Primality is difficult because of the $vee$. So create a new definition. An arrow into a coproduct is prime when there exists a morphism into at least one of the coproduct's components such tha the relevant triangle commutes.

Notice we used coproduct here which is $text{lcm}(alpha, beta)$ since the definition of prime is equivalent to $p mid text{lcm}(alpha, beta) implies p mid alpha vee p mid beta$.

---

So take the contrapositive of that. $p nmid alpha wedge p nmid beta implies p nmid text{lcm}(alpha, beta)$.

Form the "negated" poset category of integers not dividing one another. It's formed by mapping each hom-set to $varnothing$ when it's nonempty and vise-versa, so it's not a functor from the original category.

share|cite|improve this answer

edited Nov 4 '18 at 7:48

answered Nov 4 '18 at 6:52

BananaCatsBananaCats

9,26452659

$endgroup$

add a comment | 

0

$begingroup$

The second can become:$$(forall xy)alphamid xytoalphamid xloralphamid y$$ The first is: $$(forall x)alphabetanmid x to alphanmid x lor betanmid x lor alphabeta gt xlor(alpha,beta)^2nmid x$$ Not sure it reveals much though.

share|cite|improve this answer

answered Mar 10 at 14:00

Roddy MacPheeRoddy MacPhee

30716

$endgroup$

add a comment | 

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1

$begingroup$

In the poset category of integers dividing one another we write sometimes write $alpha to x$ instead of $alpha mid x$.

The "product" $alpha beta$ of two coprime objects $alpha, beta$ is such that $alpha to alpha beta leftarrow beta$ and for any object $x$ such that $alpha to x leftarrow beta$ then there is a unique arrow $alpha beta to x$. That is precisely the definition of coproduct in a general category.

Thus when $alpha, beta$ are coprime, then the category definitely has a coproduct for them.

Thus $wedge$ is encoded in the fact that $alpha to x leftarrow beta$, ie. both morphisms exist simultaneously. I believe product is $gcd(alpha, beta)$.

---

Actually, it turns out that the coproduct exists for any two integers and it's $text{lcm}(alpha, beta)$.

---

Primality is difficult because of the $vee$. So create a new definition. An arrow into a coproduct is prime when there exists a morphism into at least one of the coproduct's components such tha the relevant triangle commutes.

Notice we used coproduct here which is $text{lcm}(alpha, beta)$ since the definition of prime is equivalent to $p mid text{lcm}(alpha, beta) implies p mid alpha vee p mid beta$.

---

So take the contrapositive of that. $p nmid alpha wedge p nmid beta implies p nmid text{lcm}(alpha, beta)$.

Form the "negated" poset category of integers not dividing one another. It's formed by mapping each hom-set to $varnothing$ when it's nonempty and vise-versa, so it's not a functor from the original category.

share|cite|improve this answer

edited Nov 4 '18 at 7:48

answered Nov 4 '18 at 6:52

BananaCatsBananaCats

9,26452659

$endgroup$

add a comment | 

1

$begingroup$

In the poset category of integers dividing one another we write sometimes write $alpha to x$ instead of $alpha mid x$.

The "product" $alpha beta$ of two coprime objects $alpha, beta$ is such that $alpha to alpha beta leftarrow beta$ and for any object $x$ such that $alpha to x leftarrow beta$ then there is a unique arrow $alpha beta to x$. That is precisely the definition of coproduct in a general category.

Thus when $alpha, beta$ are coprime, then the category definitely has a coproduct for them.

Thus $wedge$ is encoded in the fact that $alpha to x leftarrow beta$, ie. both morphisms exist simultaneously. I believe product is $gcd(alpha, beta)$.

---

Actually, it turns out that the coproduct exists for any two integers and it's $text{lcm}(alpha, beta)$.

---

Primality is difficult because of the $vee$. So create a new definition. An arrow into a coproduct is prime when there exists a morphism into at least one of the coproduct's components such tha the relevant triangle commutes.

Notice we used coproduct here which is $text{lcm}(alpha, beta)$ since the definition of prime is equivalent to $p mid text{lcm}(alpha, beta) implies p mid alpha vee p mid beta$.

---

So take the contrapositive of that. $p nmid alpha wedge p nmid beta implies p nmid text{lcm}(alpha, beta)$.

Form the "negated" poset category of integers not dividing one another. It's formed by mapping each hom-set to $varnothing$ when it's nonempty and vise-versa, so it's not a functor from the original category.

share|cite|improve this answer

edited Nov 4 '18 at 7:48

answered Nov 4 '18 at 6:52

BananaCatsBananaCats

9,26452659

$endgroup$

add a comment | 

$begingroup$

In the poset category of integers dividing one another we write sometimes write $alpha to x$ instead of $alpha mid x$.

The "product" $alpha beta$ of two coprime objects $alpha, beta$ is such that $alpha to alpha beta leftarrow beta$ and for any object $x$ such that $alpha to x leftarrow beta$ then there is a unique arrow $alpha beta to x$. That is precisely the definition of coproduct in a general category.

Thus when $alpha, beta$ are coprime, then the category definitely has a coproduct for them.

Thus $wedge$ is encoded in the fact that $alpha to x leftarrow beta$, ie. both morphisms exist simultaneously. I believe product is $gcd(alpha, beta)$.

---

Actually, it turns out that the coproduct exists for any two integers and it's $text{lcm}(alpha, beta)$.

---

Primality is difficult because of the $vee$. So create a new definition. An arrow into a coproduct is prime when there exists a morphism into at least one of the coproduct's components such tha the relevant triangle commutes.

Notice we used coproduct here which is $text{lcm}(alpha, beta)$ since the definition of prime is equivalent to $p mid text{lcm}(alpha, beta) implies p mid alpha vee p mid beta$.

---

So take the contrapositive of that. $p nmid alpha wedge p nmid beta implies p nmid text{lcm}(alpha, beta)$.

Form the "negated" poset category of integers not dividing one another. It's formed by mapping each hom-set to $varnothing$ when it's nonempty and vise-versa, so it's not a functor from the original category.

share|cite|improve this answer

edited Nov 4 '18 at 7:48

answered Nov 4 '18 at 6:52

BananaCatsBananaCats

9,26452659

$endgroup$

share|cite|improve this answer

edited Nov 4 '18 at 7:48

answered Nov 4 '18 at 6:52

BananaCatsBananaCats

9,26452659

answered Nov 4 '18 at 6:52

BananaCatsBananaCats

9,26452659

answered Nov 4 '18 at 6:52

BananaCatsBananaCats

9,26452659

0

$begingroup$

The second can become:$$(forall xy)alphamid xytoalphamid xloralphamid y$$ The first is: $$(forall x)alphabetanmid x to alphanmid x lor betanmid x lor alphabeta gt xlor(alpha,beta)^2nmid x$$ Not sure it reveals much though.

share|cite|improve this answer

answered Mar 10 at 14:00

Roddy MacPheeRoddy MacPhee

30716

$endgroup$

add a comment | 

0

$begingroup$

The second can become:$$(forall xy)alphamid xytoalphamid xloralphamid y$$ The first is: $$(forall x)alphabetanmid x to alphanmid x lor betanmid x lor alphabeta gt xlor(alpha,beta)^2nmid x$$ Not sure it reveals much though.

share|cite|improve this answer

answered Mar 10 at 14:00

Roddy MacPheeRoddy MacPhee

30716

$endgroup$

add a comment | 

$begingroup$

The second can become:$$(forall xy)alphamid xytoalphamid xloralphamid y$$ The first is: $$(forall x)alphabetanmid x to alphanmid x lor betanmid x lor alphabeta gt xlor(alpha,beta)^2nmid x$$ Not sure it reveals much though.

share|cite|improve this answer

answered Mar 10 at 14:00

Roddy MacPheeRoddy MacPhee

30716

$endgroup$

share|cite|improve this answer

answered Mar 10 at 14:00

Roddy MacPheeRoddy MacPhee

30716

answered Mar 10 at 14:00

Roddy MacPheeRoddy MacPhee

30716

answered Mar 10 at 14:00

Roddy MacPheeRoddy MacPhee

30716