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2

$begingroup$

We have a name for the property of integers to be $0$ or $1$ $mathrm{mod} 2$ - parity.

Is there any similar name for the remainder for any other base? Like a generalization of parity? Could I use parity in a broader sense, just to name the remainder $mathrm{mod} n$?

elementary-number-theory soft-question integers

share|cite|improve this question

asked 2 hours ago

dEmigOddEmigOd

1,5001612

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  How about congruence?
  $endgroup$
  – Umberto P.
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So should I say (variable)'s congruence class or value?
  $endgroup$
  – dEmigOd
  2 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

We have a name for the property of integers to be $0$ or $1$ $mathrm{mod} 2$ - parity.

Is there any similar name for the remainder for any other base? Like a generalization of parity? Could I use parity in a broader sense, just to name the remainder $mathrm{mod} n$?

elementary-number-theory soft-question integers

share|cite|improve this question

asked 2 hours ago

dEmigOddEmigOd

1,5001612

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  How about congruence?
  $endgroup$
  – Umberto P.
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So should I say (variable)'s congruence class or value?
  $endgroup$
  – dEmigOd
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

We have a name for the property of integers to be $0$ or $1$ $mathrm{mod} 2$ - parity.

Is there any similar name for the remainder for any other base? Like a generalization of parity? Could I use parity in a broader sense, just to name the remainder $mathrm{mod} n$?

elementary-number-theory soft-question integers

share|cite|improve this question

asked 2 hours ago

dEmigOddEmigOd

1,5001612

$endgroup$

share|cite|improve this question

asked 2 hours ago

dEmigOddEmigOd

1,5001612

share|cite|improve this question

asked 2 hours ago

dEmigOddEmigOd

1,5001612

asked 2 hours ago

dEmigOddEmigOd

1,5001612

asked 2 hours ago

dEmigOddEmigOd

1,5001612

3 Answers
3

active

oldest

votes

3

$begingroup$

Simply say "congruent to $a$ modulo $m$" to read "$equiv a pmod{m}$".

share|cite|improve this answer

answered 2 hours ago

AlessioDVAlessioDV

43413

$endgroup$

add a comment | 

2

$begingroup$

In a broader sense, when you are dealing with congruences you are dealing with an equivalence relation and its equivalence classes.

The basic idea of an equivalence relation is to collect elements that are different but behave in the same manner with respect to some property of interest.

In modular arithmetic this property is having the same rest when divided by a prescribed integer

If $a=bbmod m$ or $aequiv_m b$ you essentially say that $a$ and $b$ are in the same equivalence class whith respect to the equivalence relation $equiv_m$.

So you could say that "$a$ and $b$ are in the same equivalence class when we look at the remainder upon dividing by $m$", which definitely longer and more cumbersome to say "$a$ congruent to $b$ modulo $m$" but nonetheless another way to exrpess the same concept.

share|cite|improve this answer

answered 1 hour ago

Vinyl_coat_jawaVinyl_coat_jawa

3,0001131

$endgroup$

add a comment | 

1

$begingroup$

Either AlessioDV's good answer, or you could say: "of the form $nq+r$". as a representation of $equiv r pmod n$ for example a number $N$ with $$Nequiv 1pmod 3$$ has property $N=3q+1$.

share|cite|improve this answer

answered 1 hour ago

Rhys HughesRhys Hughes

6,9031530

$endgroup$

add a comment | 

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3

$begingroup$

Simply say "congruent to $a$ modulo $m$" to read "$equiv a pmod{m}$".

share|cite|improve this answer

answered 2 hours ago

AlessioDVAlessioDV

43413

$endgroup$

add a comment | 

3

$begingroup$

Simply say "congruent to $a$ modulo $m$" to read "$equiv a pmod{m}$".

share|cite|improve this answer

answered 2 hours ago

AlessioDVAlessioDV

43413

$endgroup$

add a comment | 

$begingroup$

Simply say "congruent to $a$ modulo $m$" to read "$equiv a pmod{m}$".

share|cite|improve this answer

answered 2 hours ago

AlessioDVAlessioDV

43413

$endgroup$

share|cite|improve this answer

answered 2 hours ago

AlessioDVAlessioDV

43413

answered 2 hours ago

AlessioDVAlessioDV

43413

answered 2 hours ago

AlessioDVAlessioDV

43413

2

$begingroup$

In a broader sense, when you are dealing with congruences you are dealing with an equivalence relation and its equivalence classes.

The basic idea of an equivalence relation is to collect elements that are different but behave in the same manner with respect to some property of interest.

In modular arithmetic this property is having the same rest when divided by a prescribed integer

If $a=bbmod m$ or $aequiv_m b$ you essentially say that $a$ and $b$ are in the same equivalence class whith respect to the equivalence relation $equiv_m$.

So you could say that "$a$ and $b$ are in the same equivalence class when we look at the remainder upon dividing by $m$", which definitely longer and more cumbersome to say "$a$ congruent to $b$ modulo $m$" but nonetheless another way to exrpess the same concept.

share|cite|improve this answer

answered 1 hour ago

Vinyl_coat_jawaVinyl_coat_jawa

3,0001131

$endgroup$

add a comment | 

2

$begingroup$

In a broader sense, when you are dealing with congruences you are dealing with an equivalence relation and its equivalence classes.

The basic idea of an equivalence relation is to collect elements that are different but behave in the same manner with respect to some property of interest.

In modular arithmetic this property is having the same rest when divided by a prescribed integer

If $a=bbmod m$ or $aequiv_m b$ you essentially say that $a$ and $b$ are in the same equivalence class whith respect to the equivalence relation $equiv_m$.

So you could say that "$a$ and $b$ are in the same equivalence class when we look at the remainder upon dividing by $m$", which definitely longer and more cumbersome to say "$a$ congruent to $b$ modulo $m$" but nonetheless another way to exrpess the same concept.

share|cite|improve this answer

answered 1 hour ago

Vinyl_coat_jawaVinyl_coat_jawa

3,0001131

$endgroup$

add a comment | 

$begingroup$

In a broader sense, when you are dealing with congruences you are dealing with an equivalence relation and its equivalence classes.

The basic idea of an equivalence relation is to collect elements that are different but behave in the same manner with respect to some property of interest.

In modular arithmetic this property is having the same rest when divided by a prescribed integer

If $a=bbmod m$ or $aequiv_m b$ you essentially say that $a$ and $b$ are in the same equivalence class whith respect to the equivalence relation $equiv_m$.

So you could say that "$a$ and $b$ are in the same equivalence class when we look at the remainder upon dividing by $m$", which definitely longer and more cumbersome to say "$a$ congruent to $b$ modulo $m$" but nonetheless another way to exrpess the same concept.

share|cite|improve this answer

answered 1 hour ago

Vinyl_coat_jawaVinyl_coat_jawa

3,0001131

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Vinyl_coat_jawaVinyl_coat_jawa

3,0001131

answered 1 hour ago

Vinyl_coat_jawaVinyl_coat_jawa

3,0001131

answered 1 hour ago

Vinyl_coat_jawaVinyl_coat_jawa

3,0001131

1

$begingroup$

Either AlessioDV's good answer, or you could say: "of the form $nq+r$". as a representation of $equiv r pmod n$ for example a number $N$ with $$Nequiv 1pmod 3$$ has property $N=3q+1$.

share|cite|improve this answer

answered 1 hour ago

Rhys HughesRhys Hughes

6,9031530

$endgroup$

add a comment | 

1

$begingroup$

Either AlessioDV's good answer, or you could say: "of the form $nq+r$". as a representation of $equiv r pmod n$ for example a number $N$ with $$Nequiv 1pmod 3$$ has property $N=3q+1$.

share|cite|improve this answer

answered 1 hour ago

Rhys HughesRhys Hughes

6,9031530

$endgroup$

add a comment | 

$begingroup$

Either AlessioDV's good answer, or you could say: "of the form $nq+r$". as a representation of $equiv r pmod n$ for example a number $N$ with $$Nequiv 1pmod 3$$ has property $N=3q+1$.

share|cite|improve this answer

answered 1 hour ago

Rhys HughesRhys Hughes

6,9031530

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Rhys HughesRhys Hughes

6,9031530

answered 1 hour ago

Rhys HughesRhys Hughes

6,9031530

answered 1 hour ago

Rhys HughesRhys Hughes

6,9031530