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Discrete math equivalence classes

Need help counting equivalence classes.Discrete Math - Equivalence ClassesFind all equivalence classesProving that two equivalence classes are disjoint?Equivalence Classes OutputFinding distinct equivalence classesTotal Number of Equivalence classes of RProving equivalence classes for a equivalence relationDiscrete Math - Equivalence Classes of a set containing all real numbersWhat are the equivalence classes of R?

0

$begingroup$

Let $A$ be a finite set of size $k$ and $R$ a relation on the power set $P(A)$ defined by $R=left{(A,B) : |A|=|B|right}$

1. Show that $A$ is an equivalence relation.


2. Let $a in A$. What is the size of the equivalence class of ${a}$?


3. Let $a, b$ be two different elements of $A$. What is the size of the equivalence class of ${a, b}$?

I’m having a lot of trouble with this problem. It says the relation is on the power set, but then I’m finding the size of the equivalence class of elements within $A$, and then of $(a,b)$? I’m honestly completely lost and don’t have any base to build off of. I think I’m a bit confused on the concept of the size of equivalence classes in general.

discrete-mathematics equivalence-relations

share|cite|improve this question

edited Mar 10 at 23:57

Brian

601114

asked Mar 10 at 22:41

Dani JoDani Jo

43

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ${a}$ and ${a,b}$ are elements of $P(A)$. The ordered pair $(a,b)$ occurs nowhere in the problem.
  $endgroup$
  – saulspatz
  Mar 10 at 22:48
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $A$ be a finite set of size $k$ and $R$ a relation on the power set $P(A)$ defined by $R=left{(A,B) : |A|=|B|right}$

1. Show that $A$ is an equivalence relation.


2. Let $a in A$. What is the size of the equivalence class of ${a}$?


3. Let $a, b$ be two different elements of $A$. What is the size of the equivalence class of ${a, b}$?

I’m having a lot of trouble with this problem. It says the relation is on the power set, but then I’m finding the size of the equivalence class of elements within $A$, and then of $(a,b)$? I’m honestly completely lost and don’t have any base to build off of. I think I’m a bit confused on the concept of the size of equivalence classes in general.

discrete-mathematics equivalence-relations

share|cite|improve this question

edited Mar 10 at 23:57

Brian

601114

asked Mar 10 at 22:41

Dani JoDani Jo

43

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ${a}$ and ${a,b}$ are elements of $P(A)$. The ordered pair $(a,b)$ occurs nowhere in the problem.
  $endgroup$
  – saulspatz
  Mar 10 at 22:48
  

  
  
  
  

add a comment | 

$begingroup$

Let $A$ be a finite set of size $k$ and $R$ a relation on the power set $P(A)$ defined by $R=left{(A,B) : |A|=|B|right}$

1. Show that $A$ is an equivalence relation.


2. Let $a in A$. What is the size of the equivalence class of ${a}$?


3. Let $a, b$ be two different elements of $A$. What is the size of the equivalence class of ${a, b}$?

I’m having a lot of trouble with this problem. It says the relation is on the power set, but then I’m finding the size of the equivalence class of elements within $A$, and then of $(a,b)$? I’m honestly completely lost and don’t have any base to build off of. I think I’m a bit confused on the concept of the size of equivalence classes in general.

discrete-mathematics equivalence-relations

share|cite|improve this question

edited Mar 10 at 23:57

Brian

601114

asked Mar 10 at 22:41

Dani JoDani Jo

43

$endgroup$

share|cite|improve this question

edited Mar 10 at 23:57

Brian

601114

asked Mar 10 at 22:41

Dani JoDani Jo

43

share|cite|improve this question

edited Mar 10 at 23:57

Brian

601114

asked Mar 10 at 22:41

Dani JoDani Jo

43

edited Mar 10 at 23:57

Brian

601114

edited Mar 10 at 23:57

Brian

601114

asked Mar 10 at 22:41

Dani JoDani Jo

43

asked Mar 10 at 22:41

Dani JoDani Jo

43

2 Answers
2

active

oldest

votes

1

$begingroup$

Informally, under this equivalence relation two subsets are equivalent when they have the same size.

Thus, the equivalence class of ${a}$ consists of all subsets of $A$ with cardinality/size equal to one. Thus the size of this equivalence class is $k=|A|$.

The equivalence class of ${a, b}$ consists of all two element subsets of $A$. Thus the size of this equivalence class is $binom{k}{2}=frac{k(k-1)}{2}$.

share|cite|improve this answer

edited Mar 10 at 22:58

answered Mar 10 at 22:52

Foobaz JohnFoobaz John

22.7k41452

$endgroup$

add a comment | 

0

$begingroup$

You mean to prove that $R$ is an equivalence relation. There are three criteria that must be followed in order to prove $R$ is an equivalence relation,

1. Reflexive


2. Symmetric


3. Transitive

To get you started, with generic sets $B,C,D in mathcal{P}(A)$, see that we always have $$|B| = |B|$$ so $R$ is reflexive. This would mean $(B,B) in R$ for some $B subseteq A$

To show it's symmetric, it requires you to show that if $|B| = |C|$, then $|C| = |B|$.

To show it's transitive, it requires you to show if $|B| = |C|$, and$ |C| = |D|$, then |B| = |D|$

share|cite|improve this answer

answered Mar 10 at 22:54

WaveXWaveX

2,7622722

$endgroup$

add a comment | 

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1

$begingroup$

Informally, under this equivalence relation two subsets are equivalent when they have the same size.

Thus, the equivalence class of ${a}$ consists of all subsets of $A$ with cardinality/size equal to one. Thus the size of this equivalence class is $k=|A|$.

The equivalence class of ${a, b}$ consists of all two element subsets of $A$. Thus the size of this equivalence class is $binom{k}{2}=frac{k(k-1)}{2}$.

share|cite|improve this answer

edited Mar 10 at 22:58

answered Mar 10 at 22:52

Foobaz JohnFoobaz John

22.7k41452

$endgroup$

add a comment | 

1

$begingroup$

Informally, under this equivalence relation two subsets are equivalent when they have the same size.

Thus, the equivalence class of ${a}$ consists of all subsets of $A$ with cardinality/size equal to one. Thus the size of this equivalence class is $k=|A|$.

The equivalence class of ${a, b}$ consists of all two element subsets of $A$. Thus the size of this equivalence class is $binom{k}{2}=frac{k(k-1)}{2}$.

share|cite|improve this answer

edited Mar 10 at 22:58

answered Mar 10 at 22:52

Foobaz JohnFoobaz John

22.7k41452

$endgroup$

add a comment | 

$begingroup$

Informally, under this equivalence relation two subsets are equivalent when they have the same size.

Thus, the equivalence class of ${a}$ consists of all subsets of $A$ with cardinality/size equal to one. Thus the size of this equivalence class is $k=|A|$.

The equivalence class of ${a, b}$ consists of all two element subsets of $A$. Thus the size of this equivalence class is $binom{k}{2}=frac{k(k-1)}{2}$.

share|cite|improve this answer

edited Mar 10 at 22:58

answered Mar 10 at 22:52

Foobaz JohnFoobaz John

22.7k41452

$endgroup$

share|cite|improve this answer

edited Mar 10 at 22:58

answered Mar 10 at 22:52

Foobaz JohnFoobaz John

22.7k41452

answered Mar 10 at 22:52

Foobaz JohnFoobaz John

22.7k41452

answered Mar 10 at 22:52

Foobaz JohnFoobaz John

22.7k41452

0

$begingroup$

You mean to prove that $R$ is an equivalence relation. There are three criteria that must be followed in order to prove $R$ is an equivalence relation,

1. Reflexive


2. Symmetric


3. Transitive

To get you started, with generic sets $B,C,D in mathcal{P}(A)$, see that we always have $$|B| = |B|$$ so $R$ is reflexive. This would mean $(B,B) in R$ for some $B subseteq A$

To show it's symmetric, it requires you to show that if $|B| = |C|$, then $|C| = |B|$.

To show it's transitive, it requires you to show if $|B| = |C|$, and$ |C| = |D|$, then |B| = |D|$

share|cite|improve this answer

answered Mar 10 at 22:54

WaveXWaveX

2,7622722

$endgroup$

add a comment | 

0

$begingroup$

You mean to prove that $R$ is an equivalence relation. There are three criteria that must be followed in order to prove $R$ is an equivalence relation,

1. Reflexive


2. Symmetric


3. Transitive

To get you started, with generic sets $B,C,D in mathcal{P}(A)$, see that we always have $$|B| = |B|$$ so $R$ is reflexive. This would mean $(B,B) in R$ for some $B subseteq A$

To show it's symmetric, it requires you to show that if $|B| = |C|$, then $|C| = |B|$.

To show it's transitive, it requires you to show if $|B| = |C|$, and$ |C| = |D|$, then |B| = |D|$

share|cite|improve this answer

answered Mar 10 at 22:54

WaveXWaveX

2,7622722

$endgroup$

add a comment | 

$begingroup$

You mean to prove that $R$ is an equivalence relation. There are three criteria that must be followed in order to prove $R$ is an equivalence relation,

1. Reflexive


2. Symmetric


3. Transitive

To get you started, with generic sets $B,C,D in mathcal{P}(A)$, see that we always have $$|B| = |B|$$ so $R$ is reflexive. This would mean $(B,B) in R$ for some $B subseteq A$

To show it's symmetric, it requires you to show that if $|B| = |C|$, then $|C| = |B|$.

To show it's transitive, it requires you to show if $|B| = |C|$, and$ |C| = |D|$, then |B| = |D|$

share|cite|improve this answer

answered Mar 10 at 22:54

WaveXWaveX

2,7622722

$endgroup$

share|cite|improve this answer

answered Mar 10 at 22:54

WaveXWaveX

2,7622722

answered Mar 10 at 22:54

WaveXWaveX

2,7622722

answered Mar 10 at 22:54

WaveXWaveX

2,7622722