Proving $binom{n}{k}=binom{n}{n-k}$ using set theory?
$begingroup$
Let $N$ be a set of $n$ items. Define a grouping to be some subset of $N$.
Let $A$ be the set of all groupings of $k$ items from $N$.
Let $B$ be the set of all groupings of $n-k$ items from $N$.
Namely, if $|A|=|B|$ then we have $binom{n}{k}=binom{n}{n-k}$.
For $|A|=|B|$, we must have there exists a bijection from $A$ to $B$. Let us define a function $f:A rightarrow B$ such that $f(a)=b$ if $a cup b = N$
We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $a_1 cup b = N = a_2 cup b$, i.e. $a_1=a_2=N-b$. So, $f$ is injective.
We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $acup b=(N-b)cup b=b$.
Since $f$ is injective and surjective, it is bijective. So $|A|=|B|$, and as such we have $binom{n}{k}=binom{n}{n-k}$ QED.
Is this proof correct?
combinatorics proof-verification elementary-set-theory
asked Jan 11 at 0:16
Hisoka MorohHisoka Moroh
420110
$endgroup$
add a comment |
$begingroup$
Let $N$ be a set of $n$ items. Define a grouping to be some subset of $N$.
Let $A$ be the set of all groupings of $k$ items from $N$.
Let $B$ be the set of all groupings of $n-k$ items from $N$.
Namely, if $|A|=|B|$ then we have $binom{n}{k}=binom{n}{n-k}$.
For $|A|=|B|$, we must have there exists a bijection from $A$ to $B$. Let us define a function $f:A rightarrow B$ such that $f(a)=b$ if $a cup b = N$
We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $a_1 cup b = N = a_2 cup b$, i.e. $a_1=a_2=N-b$. So, $f$ is injective.
We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $acup b=(N-b)cup b=b$.
Since $f$ is injective and surjective, it is bijective. So $|A|=|B|$, and as such we have $binom{n}{k}=binom{n}{n-k}$ QED.
Is this proof correct?
combinatorics proof-verification elementary-set-theory
asked Jan 11 at 0:16
Hisoka MorohHisoka Moroh
420110
$endgroup$
$begingroup$
It would be hard to tell if your proof is correct without knowing what definition of $binom nk$ you are working with, and what properties of $binom nk$ have already been established. By the way, why do you define the word "grouping" to mean "set"? What's wrong with using the word "set" to mean "set"? Usually we introduce definitions to make things shorter, not to make them longer.
$endgroup$
– bof
Jan 11 at 4:39
add a comment |
$begingroup$
Let $N$ be a set of $n$ items. Define a grouping to be some subset of $N$.
Let $A$ be the set of all groupings of $k$ items from $N$.
Let $B$ be the set of all groupings of $n-k$ items from $N$.
Namely, if $|A|=|B|$ then we have $binom{n}{k}=binom{n}{n-k}$.
For $|A|=|B|$, we must have there exists a bijection from $A$ to $B$. Let us define a function $f:A rightarrow B$ such that $f(a)=b$ if $a cup b = N$
We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $a_1 cup b = N = a_2 cup b$, i.e. $a_1=a_2=N-b$. So, $f$ is injective.
We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $acup b=(N-b)cup b=b$.
Since $f$ is injective and surjective, it is bijective. So $|A|=|B|$, and as such we have $binom{n}{k}=binom{n}{n-k}$ QED.
Is this proof correct?
combinatorics proof-verification elementary-set-theory
asked Jan 11 at 0:16
Hisoka MorohHisoka Moroh
420110
$endgroup$
Let $N$ be a set of $n$ items. Define a grouping to be some subset of $N$.
Let $A$ be the set of all groupings of $k$ items from $N$.
Let $B$ be the set of all groupings of $n-k$ items from $N$.
Namely, if $|A|=|B|$ then we have $binom{n}{k}=binom{n}{n-k}$.
For $|A|=|B|$, we must have there exists a bijection from $A$ to $B$. Let us define a function $f:A rightarrow B$ such that $f(a)=b$ if $a cup b = N$
We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $a_1 cup b = N = a_2 cup b$, i.e. $a_1=a_2=N-b$. So, $f$ is injective.
We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $acup b=(N-b)cup b=b$.
Since $f$ is injective and surjective, it is bijective. So $|A|=|B|$, and as such we have $binom{n}{k}=binom{n}{n-k}$ QED.
Is this proof correct?
combinatorics proof-verification elementary-set-theory
combinatorics proof-verification elementary-set-theory
asked Jan 11 at 0:16
Hisoka MorohHisoka Moroh
420110
asked Jan 11 at 0:16
Hisoka MorohHisoka Moroh
420110
asked Jan 11 at 0:16
Hisoka MorohHisoka Moroh
420110
asked Jan 11 at 0:16
Hisoka MorohHisoka Moroh
420110
asked Jan 11 at 0:16
Hisoka MorohHisoka Moroh
420110
420110
$begingroup$
It would be hard to tell if your proof is correct without knowing what definition of $binom nk$ you are working with, and what properties of $binom nk$ have already been established. By the way, why do you define the word "grouping" to mean "set"? What's wrong with using the word "set" to mean "set"? Usually we introduce definitions to make things shorter, not to make them longer.
$endgroup$
– bof
Jan 11 at 4:39
add a comment |
$begingroup$
It would be hard to tell if your proof is correct without knowing what definition of $binom nk$ you are working with, and what properties of $binom nk$ have already been established. By the way, why do you define the word "grouping" to mean "set"? What's wrong with using the word "set" to mean "set"? Usually we introduce definitions to make things shorter, not to make them longer.
$endgroup$
– bof
Jan 11 at 4:39
$begingroup$
It would be hard to tell if your proof is correct without knowing what definition of $binom nk$ you are working with, and what properties of $binom nk$ have already been established. By the way, why do you define the word "grouping" to mean "set"? What's wrong with using the word "set" to mean "set"? Usually we introduce definitions to make things shorter, not to make them longer.
$endgroup$
– bof
Jan 11 at 4:39
$begingroup$
It would be hard to tell if your proof is correct without knowing what definition of $binom nk$ you are working with, and what properties of $binom nk$ have already been established. By the way, why do you define the word "grouping" to mean "set"? What's wrong with using the word "set" to mean "set"? Usually we introduce definitions to make things shorter, not to make them longer.
$endgroup$
– bof
Jan 11 at 4:39
add a comment |
1 Answer
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$begingroup$
The problem in this proof is with the definition of $f(a)$:
$f(a)=b$ if $a cup b = N$
However, how do we even know that such a $bin B$ exists? The fact that this function is well-defined must be proven rigorously. Therefore, I think this alternative definition would be better:
$f(a)=b$ if $b=N-a$
Now, we will prove $b=N-a in B$: Since $ain A$, we know that $|a|=k$. Furthermore, it is given that $|N|=n$. Therefore, since $a$ and $N$ are finite sets, $|N-a|=|N|-|a|=n-k$. Thus, $|b|=|N-a|=n-k$, so $b in B$. This proves that $f(a)=N-a$ is a well-defined function from $A$ to $B$.
Using this new definition, I will now modify your proof as follows:
We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $b=N-a_1$ and $b=N-a_2$. Thus, $a_1=N-b$ and $a_2=N-b$. Therefore, $a_1=a_2=N-b$. So, $f$ is injective.
We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $a=N-b$ implies $b=N-a$.
answered Jan 11 at 0:22
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
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$begingroup$
The problem in this proof is with the definition of $f(a)$:
$f(a)=b$ if $a cup b = N$
However, how do we even know that such a $bin B$ exists? The fact that this function is well-defined must be proven rigorously. Therefore, I think this alternative definition would be better:
$f(a)=b$ if $b=N-a$
Now, we will prove $b=N-a in B$: Since $ain A$, we know that $|a|=k$. Furthermore, it is given that $|N|=n$. Therefore, since $a$ and $N$ are finite sets, $|N-a|=|N|-|a|=n-k$. Thus, $|b|=|N-a|=n-k$, so $b in B$. This proves that $f(a)=N-a$ is a well-defined function from $A$ to $B$.
Using this new definition, I will now modify your proof as follows:
We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $b=N-a_1$ and $b=N-a_2$. Thus, $a_1=N-b$ and $a_2=N-b$. Therefore, $a_1=a_2=N-b$. So, $f$ is injective.
We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $a=N-b$ implies $b=N-a$.
answered Jan 11 at 0:22
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
The problem in this proof is with the definition of $f(a)$:
$f(a)=b$ if $a cup b = N$
However, how do we even know that such a $bin B$ exists? The fact that this function is well-defined must be proven rigorously. Therefore, I think this alternative definition would be better:
$f(a)=b$ if $b=N-a$
Now, we will prove $b=N-a in B$: Since $ain A$, we know that $|a|=k$. Furthermore, it is given that $|N|=n$. Therefore, since $a$ and $N$ are finite sets, $|N-a|=|N|-|a|=n-k$. Thus, $|b|=|N-a|=n-k$, so $b in B$. This proves that $f(a)=N-a$ is a well-defined function from $A$ to $B$.
Using this new definition, I will now modify your proof as follows:
We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $b=N-a_1$ and $b=N-a_2$. Thus, $a_1=N-b$ and $a_2=N-b$. Therefore, $a_1=a_2=N-b$. So, $f$ is injective.
We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $a=N-b$ implies $b=N-a$.
answered Jan 11 at 0:22
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
The problem in this proof is with the definition of $f(a)$:
$f(a)=b$ if $a cup b = N$
However, how do we even know that such a $bin B$ exists? The fact that this function is well-defined must be proven rigorously. Therefore, I think this alternative definition would be better:
$f(a)=b$ if $b=N-a$
Now, we will prove $b=N-a in B$: Since $ain A$, we know that $|a|=k$. Furthermore, it is given that $|N|=n$. Therefore, since $a$ and $N$ are finite sets, $|N-a|=|N|-|a|=n-k$. Thus, $|b|=|N-a|=n-k$, so $b in B$. This proves that $f(a)=N-a$ is a well-defined function from $A$ to $B$.
Using this new definition, I will now modify your proof as follows:
We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $b=N-a_1$ and $b=N-a_2$. Thus, $a_1=N-b$ and $a_2=N-b$. Therefore, $a_1=a_2=N-b$. So, $f$ is injective.
We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $a=N-b$ implies $b=N-a$.
answered Jan 11 at 0:22
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
The problem in this proof is with the definition of $f(a)$:
$f(a)=b$ if $a cup b = N$
However, how do we even know that such a $bin B$ exists? The fact that this function is well-defined must be proven rigorously. Therefore, I think this alternative definition would be better:
$f(a)=b$ if $b=N-a$
Now, we will prove $b=N-a in B$: Since $ain A$, we know that $|a|=k$. Furthermore, it is given that $|N|=n$. Therefore, since $a$ and $N$ are finite sets, $|N-a|=|N|-|a|=n-k$. Thus, $|b|=|N-a|=n-k$, so $b in B$. This proves that $f(a)=N-a$ is a well-defined function from $A$ to $B$.
Using this new definition, I will now modify your proof as follows:
We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $b=N-a_1$ and $b=N-a_2$. Thus, $a_1=N-b$ and $a_2=N-b$. Therefore, $a_1=a_2=N-b$. So, $f$ is injective.
We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $a=N-b$ implies $b=N-a$.
answered Jan 11 at 0:22
Noble MushtakNoble Mushtak
15.3k1835
answered Jan 11 at 0:22
Noble MushtakNoble Mushtak
15.3k1835
answered Jan 11 at 0:22
Noble MushtakNoble Mushtak
15.3k1835
answered Jan 11 at 0:22
Noble MushtakNoble Mushtak
15.3k1835
15.3k1835
add a comment |
add a comment |
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$begingroup$
It would be hard to tell if your proof is correct without knowing what definition of $binom nk$ you are working with, and what properties of $binom nk$ have already been established. By the way, why do you define the word "grouping" to mean "set"? What's wrong with using the word "set" to mean "set"? Usually we introduce definitions to make things shorter, not to make them longer.
$endgroup$
– bof
Jan 11 at 4:39