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Is Case 2 of my proof that $sum_i=1^n a_i=sum_i=1^n a_sigma(i)$ correct?
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Let $I_n=iinmathbb Nmid 1leq ileq n$, $(a_1,cdots,a_n)$ be a finite sequence in $mathbb N$, and $sigma$ be a permutation of $I_n$. I've shown that there is a sequence $(s_1,cdots,s_n)$ such that $s_1=a_1$ and $s_i+1=s_i+a_i+1$ for all $1leq i<n$ here. As a result, we write $s_n=a_1+cdots+a_n$, or $s_n=sum_i=1^n a_i$. Prove that:
$$sum_i=1^n a_i=sum_i=1^n a_sigma(i)$$
I'm not sure whether Case 2 where $sigma(k+1)leq k$ contains error, please help me check that Case.
Please suggest any shorter or simpler approach.
My attempt:
I will prove this by induction on $n$. It's clear that the theorem is trivially true for $n=1$. Assume that the theorem is true for $n=k$, i.e. $sum_i=1^k a_i=sum_i=1^k a_sigma(i)$ where $sigma$ is any permutation of $I_k$.
Let $sigma'$ be any permutation of $I_k+1$. We have $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k a_sigma'(i)+a_sigma'(k+1)$. There are two cases in total.
- $sigma'(k+1)=k+1$
As a result, $sigma'(i)mid 1leq ileq k=I_k$, then $sigma'(i)restriction_I_k$ is obviously a permutation of $I_k$. Thus $sum_i=1^k a_sigma'(i)oversetIH=sum_i=1^k a_i$, then $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k a_sigma'(i)+a_sigma'(k+1)=sum_i=1^k a_i+a_k+1=sum_i=1^k+1 a_i$.
- $sigma'(k+1)leq k$
As a result, there exists $yleq k$ such that $sigma'(y)=k+1$. We generate sequence $(b_imid 1leq ileq k)$ such that $b_i=a_sigma'(i)$.
Thus $sum_i=1^k a_sigma'(i)=sum_i=1^k b_ioversetIH=sum_i=1,ineq y^k b_i+b_y=sum_i=1,ineq y^k a_sigma'(i)+a_sigma'(y)=sum_i=1,ineq y^k a_sigma'(i)+a_k+1$, then $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k a_sigma'(i)+a_sigma'(k+1)=(sum_i=1,ineq y^k a_sigma'(i)+a_k+1)+a_sigma'(k+1)oversetAssociativity=sum_i=1,ineq y^k a_sigma'(i)+(a_k+1+a_sigma'(k+1))oversetCommutative=sum_i=1,ineq y^k a_sigma'(i)+(a_sigma'(k+1)+a_k+1)oversetAssociativity=(sum_i=1,ineq y^k a_sigma'(i)+a_sigma'(k+1))+a_k+1.$
Notice that $sigma'(i)mid 1leq ileq ktext and ineq ycupsigma'(k+1)=I_k+1setminusk+1=I_k$. Thus $sum_i=1,ineq y^k a_sigma'(i)+a_sigma'(k+1)oversetIH=sum_i=1^k a_i$. Then $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k a_i+a_k+1=sum_i=1^k+1 a_i$.
To sum up, in either case, $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k+1 a_i$. Hence the theorem is true for $n=k+1$. This completes the proof.
proof-verification summation proof-writing recurrence-relations arithmetic
asked Jul 30 at 16:42
Le Anh Dung
740317
add a comment |Â
up vote
2
down vote
favorite
Let $I_n=iinmathbb Nmid 1leq ileq n$, $(a_1,cdots,a_n)$ be a finite sequence in $mathbb N$, and $sigma$ be a permutation of $I_n$. I've shown that there is a sequence $(s_1,cdots,s_n)$ such that $s_1=a_1$ and $s_i+1=s_i+a_i+1$ for all $1leq i<n$ here. As a result, we write $s_n=a_1+cdots+a_n$, or $s_n=sum_i=1^n a_i$. Prove that:
$$sum_i=1^n a_i=sum_i=1^n a_sigma(i)$$
I'm not sure whether Case 2 where $sigma(k+1)leq k$ contains error, please help me check that Case.
Please suggest any shorter or simpler approach.
My attempt:
I will prove this by induction on $n$. It's clear that the theorem is trivially true for $n=1$. Assume that the theorem is true for $n=k$, i.e. $sum_i=1^k a_i=sum_i=1^k a_sigma(i)$ where $sigma$ is any permutation of $I_k$.
Let $sigma'$ be any permutation of $I_k+1$. We have $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k a_sigma'(i)+a_sigma'(k+1)$. There are two cases in total.
- $sigma'(k+1)=k+1$
As a result, $sigma'(i)mid 1leq ileq k=I_k$, then $sigma'(i)restriction_I_k$ is obviously a permutation of $I_k$. Thus $sum_i=1^k a_sigma'(i)oversetIH=sum_i=1^k a_i$, then $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k a_sigma'(i)+a_sigma'(k+1)=sum_i=1^k a_i+a_k+1=sum_i=1^k+1 a_i$.
- $sigma'(k+1)leq k$
As a result, there exists $yleq k$ such that $sigma'(y)=k+1$. We generate sequence $(b_imid 1leq ileq k)$ such that $b_i=a_sigma'(i)$.
Thus $sum_i=1^k a_sigma'(i)=sum_i=1^k b_ioversetIH=sum_i=1,ineq y^k b_i+b_y=sum_i=1,ineq y^k a_sigma'(i)+a_sigma'(y)=sum_i=1,ineq y^k a_sigma'(i)+a_k+1$, then $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k a_sigma'(i)+a_sigma'(k+1)=(sum_i=1,ineq y^k a_sigma'(i)+a_k+1)+a_sigma'(k+1)oversetAssociativity=sum_i=1,ineq y^k a_sigma'(i)+(a_k+1+a_sigma'(k+1))oversetCommutative=sum_i=1,ineq y^k a_sigma'(i)+(a_sigma'(k+1)+a_k+1)oversetAssociativity=(sum_i=1,ineq y^k a_sigma'(i)+a_sigma'(k+1))+a_k+1.$
Notice that $sigma'(i)mid 1leq ileq ktext and ineq ycupsigma'(k+1)=I_k+1setminusk+1=I_k$. Thus $sum_i=1,ineq y^k a_sigma'(i)+a_sigma'(k+1)oversetIH=sum_i=1^k a_i$. Then $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k a_i+a_k+1=sum_i=1^k+1 a_i$.
To sum up, in either case, $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k+1 a_i$. Hence the theorem is true for $n=k+1$. This completes the proof.
proof-verification summation proof-writing recurrence-relations arithmetic
asked Jul 30 at 16:42
Le Anh Dung
740317
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $I_n=iinmathbb Nmid 1leq ileq n$, $(a_1,cdots,a_n)$ be a finite sequence in $mathbb N$, and $sigma$ be a permutation of $I_n$. I've shown that there is a sequence $(s_1,cdots,s_n)$ such that $s_1=a_1$ and $s_i+1=s_i+a_i+1$ for all $1leq i<n$ here. As a result, we write $s_n=a_1+cdots+a_n$, or $s_n=sum_i=1^n a_i$. Prove that:
$$sum_i=1^n a_i=sum_i=1^n a_sigma(i)$$
I'm not sure whether Case 2 where $sigma(k+1)leq k$ contains error, please help me check that Case.
Please suggest any shorter or simpler approach.
My attempt:
I will prove this by induction on $n$. It's clear that the theorem is trivially true for $n=1$. Assume that the theorem is true for $n=k$, i.e. $sum_i=1^k a_i=sum_i=1^k a_sigma(i)$ where $sigma$ is any permutation of $I_k$.
Let $sigma'$ be any permutation of $I_k+1$. We have $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k a_sigma'(i)+a_sigma'(k+1)$. There are two cases in total.
- $sigma'(k+1)=k+1$
As a result, $sigma'(i)mid 1leq ileq k=I_k$, then $sigma'(i)restriction_I_k$ is obviously a permutation of $I_k$. Thus $sum_i=1^k a_sigma'(i)oversetIH=sum_i=1^k a_i$, then $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k a_sigma'(i)+a_sigma'(k+1)=sum_i=1^k a_i+a_k+1=sum_i=1^k+1 a_i$.
- $sigma'(k+1)leq k$
As a result, there exists $yleq k$ such that $sigma'(y)=k+1$. We generate sequence $(b_imid 1leq ileq k)$ such that $b_i=a_sigma'(i)$.
Thus $sum_i=1^k a_sigma'(i)=sum_i=1^k b_ioversetIH=sum_i=1,ineq y^k b_i+b_y=sum_i=1,ineq y^k a_sigma'(i)+a_sigma'(y)=sum_i=1,ineq y^k a_sigma'(i)+a_k+1$, then $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k a_sigma'(i)+a_sigma'(k+1)=(sum_i=1,ineq y^k a_sigma'(i)+a_k+1)+a_sigma'(k+1)oversetAssociativity=sum_i=1,ineq y^k a_sigma'(i)+(a_k+1+a_sigma'(k+1))oversetCommutative=sum_i=1,ineq y^k a_sigma'(i)+(a_sigma'(k+1)+a_k+1)oversetAssociativity=(sum_i=1,ineq y^k a_sigma'(i)+a_sigma'(k+1))+a_k+1.$
Notice that $sigma'(i)mid 1leq ileq ktext and ineq ycupsigma'(k+1)=I_k+1setminusk+1=I_k$. Thus $sum_i=1,ineq y^k a_sigma'(i)+a_sigma'(k+1)oversetIH=sum_i=1^k a_i$. Then $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k a_i+a_k+1=sum_i=1^k+1 a_i$.
To sum up, in either case, $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k+1 a_i$. Hence the theorem is true for $n=k+1$. This completes the proof.
proof-verification summation proof-writing recurrence-relations arithmetic
asked Jul 30 at 16:42
Le Anh Dung
740317
Let $I_n=iinmathbb Nmid 1leq ileq n$, $(a_1,cdots,a_n)$ be a finite sequence in $mathbb N$, and $sigma$ be a permutation of $I_n$. I've shown that there is a sequence $(s_1,cdots,s_n)$ such that $s_1=a_1$ and $s_i+1=s_i+a_i+1$ for all $1leq i<n$ here. As a result, we write $s_n=a_1+cdots+a_n$, or $s_n=sum_i=1^n a_i$. Prove that:
$$sum_i=1^n a_i=sum_i=1^n a_sigma(i)$$
I'm not sure whether Case 2 where $sigma(k+1)leq k$ contains error, please help me check that Case.
Please suggest any shorter or simpler approach.
My attempt:
I will prove this by induction on $n$. It's clear that the theorem is trivially true for $n=1$. Assume that the theorem is true for $n=k$, i.e. $sum_i=1^k a_i=sum_i=1^k a_sigma(i)$ where $sigma$ is any permutation of $I_k$.
Let $sigma'$ be any permutation of $I_k+1$. We have $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k a_sigma'(i)+a_sigma'(k+1)$. There are two cases in total.
- $sigma'(k+1)=k+1$
As a result, $sigma'(i)mid 1leq ileq k=I_k$, then $sigma'(i)restriction_I_k$ is obviously a permutation of $I_k$. Thus $sum_i=1^k a_sigma'(i)oversetIH=sum_i=1^k a_i$, then $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k a_sigma'(i)+a_sigma'(k+1)=sum_i=1^k a_i+a_k+1=sum_i=1^k+1 a_i$.
- $sigma'(k+1)leq k$
As a result, there exists $yleq k$ such that $sigma'(y)=k+1$. We generate sequence $(b_imid 1leq ileq k)$ such that $b_i=a_sigma'(i)$.
Thus $sum_i=1^k a_sigma'(i)=sum_i=1^k b_ioversetIH=sum_i=1,ineq y^k b_i+b_y=sum_i=1,ineq y^k a_sigma'(i)+a_sigma'(y)=sum_i=1,ineq y^k a_sigma'(i)+a_k+1$, then $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k a_sigma'(i)+a_sigma'(k+1)=(sum_i=1,ineq y^k a_sigma'(i)+a_k+1)+a_sigma'(k+1)oversetAssociativity=sum_i=1,ineq y^k a_sigma'(i)+(a_k+1+a_sigma'(k+1))oversetCommutative=sum_i=1,ineq y^k a_sigma'(i)+(a_sigma'(k+1)+a_k+1)oversetAssociativity=(sum_i=1,ineq y^k a_sigma'(i)+a_sigma'(k+1))+a_k+1.$
Notice that $sigma'(i)mid 1leq ileq ktext and ineq ycupsigma'(k+1)=I_k+1setminusk+1=I_k$. Thus $sum_i=1,ineq y^k a_sigma'(i)+a_sigma'(k+1)oversetIH=sum_i=1^k a_i$. Then $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k a_i+a_k+1=sum_i=1^k+1 a_i$.
To sum up, in either case, $sum_i=1^k+1 a_sigma'(i)=sum_i=1^k+1 a_i$. Hence the theorem is true for $n=k+1$. This completes the proof.
proof-verification summation proof-writing recurrence-relations arithmetic
asked Jul 30 at 16:42
Le Anh Dung
740317
edited yesterday
asked Jul 30 at 16:42
Le Anh Dung
740317
asked Jul 30 at 16:42
Le Anh Dung
740317
asked Jul 30 at 16:42
Le Anh Dung
740317
740317
add a comment |Â
add a comment |Â
1 Answer
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I think the main doubt is for the $sigma$ in the case of $k+1$, is it the same $sigma$ for the case of $k$? Is the induction hypothesis valid since $sigma$ is not a permutation of $I_k$?
One possible approach is note that permutation is a products of adjacent transposition so we can focus on proving that under adjacent transposition, the sum is preserved.
Suppose $sigma$ only switches $j$ and $j+1$.
We have
beginalignsum_i=1^j+1a_sigma(i)&=left( sum_i=1^j-1a_i + a_j+1right) + a_j \
&= sum_i=1^j-1a_i + (a_j+1 + a_j), textAssociativity\
&=sum_i=1^j-1 a_i +(a_j+a_j+1), textCommutative \
&=left(sum_i=1^j-1a_i+a_jright) + a_j+1 , textAssociativity\
&=sum_i=1^j+1a_iendalign
Hence $sum_i=1^n a_sigma(i)= sum_i=1^n a_i$.
Hence the result is true by induction on the number of adjacent transposition.
answered 2 days ago
Siong Thye Goh
76.8k134794
Thank you so much for pointing out my two issues. I have fixed the $1^st$ issue by using $sigma$ for $I_k$ and $sigma'$ for $I_k+1$. For the $2^nd$ issue, I'm finding way to get over. Since this is my very first exposure to advanced mathematics, I have not studied anything such as adjacent transposition in Linear Algebra. But I understood your proof plan: there is a finite number of adjacent transpositions to go from one permutation to another, and we know that the sum is preserved between them, so as between the first permutation and the last one.
– Le Anh Dung
2 days ago
I have just figured out how to tackle the second issue ^^ I will fix it tomorrow.
– Le Anh Dung
2 days ago
that's good news =)
– Siong Thye Goh
2 days ago
I have fixed two mentioned issues. Please check my revised proof!
– Le Anh Dung
yesterday
I think it looks fine :) you might like to highlight the place where you use commutative and associative properties of addition.
– Siong Thye Goh
yesterday
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I think the main doubt is for the $sigma$ in the case of $k+1$, is it the same $sigma$ for the case of $k$? Is the induction hypothesis valid since $sigma$ is not a permutation of $I_k$?
One possible approach is note that permutation is a products of adjacent transposition so we can focus on proving that under adjacent transposition, the sum is preserved.
Suppose $sigma$ only switches $j$ and $j+1$.
We have
beginalignsum_i=1^j+1a_sigma(i)&=left( sum_i=1^j-1a_i + a_j+1right) + a_j \
&= sum_i=1^j-1a_i + (a_j+1 + a_j), textAssociativity\
&=sum_i=1^j-1 a_i +(a_j+a_j+1), textCommutative \
&=left(sum_i=1^j-1a_i+a_jright) + a_j+1 , textAssociativity\
&=sum_i=1^j+1a_iendalign
Hence $sum_i=1^n a_sigma(i)= sum_i=1^n a_i$.
Hence the result is true by induction on the number of adjacent transposition.
answered 2 days ago
Siong Thye Goh
76.8k134794
Thank you so much for pointing out my two issues. I have fixed the $1^st$ issue by using $sigma$ for $I_k$ and $sigma'$ for $I_k+1$. For the $2^nd$ issue, I'm finding way to get over. Since this is my very first exposure to advanced mathematics, I have not studied anything such as adjacent transposition in Linear Algebra. But I understood your proof plan: there is a finite number of adjacent transpositions to go from one permutation to another, and we know that the sum is preserved between them, so as between the first permutation and the last one.
– Le Anh Dung
2 days ago
I have just figured out how to tackle the second issue ^^ I will fix it tomorrow.
– Le Anh Dung
2 days ago
that's good news =)
– Siong Thye Goh
2 days ago
I have fixed two mentioned issues. Please check my revised proof!
– Le Anh Dung
yesterday
I think it looks fine :) you might like to highlight the place where you use commutative and associative properties of addition.
– Siong Thye Goh
yesterday
 |Â
show 1 more comment
up vote
1
down vote
accepted
I think the main doubt is for the $sigma$ in the case of $k+1$, is it the same $sigma$ for the case of $k$? Is the induction hypothesis valid since $sigma$ is not a permutation of $I_k$?
One possible approach is note that permutation is a products of adjacent transposition so we can focus on proving that under adjacent transposition, the sum is preserved.
Suppose $sigma$ only switches $j$ and $j+1$.
We have
beginalignsum_i=1^j+1a_sigma(i)&=left( sum_i=1^j-1a_i + a_j+1right) + a_j \
&= sum_i=1^j-1a_i + (a_j+1 + a_j), textAssociativity\
&=sum_i=1^j-1 a_i +(a_j+a_j+1), textCommutative \
&=left(sum_i=1^j-1a_i+a_jright) + a_j+1 , textAssociativity\
&=sum_i=1^j+1a_iendalign
Hence $sum_i=1^n a_sigma(i)= sum_i=1^n a_i$.
Hence the result is true by induction on the number of adjacent transposition.
answered 2 days ago
Siong Thye Goh
76.8k134794
Thank you so much for pointing out my two issues. I have fixed the $1^st$ issue by using $sigma$ for $I_k$ and $sigma'$ for $I_k+1$. For the $2^nd$ issue, I'm finding way to get over. Since this is my very first exposure to advanced mathematics, I have not studied anything such as adjacent transposition in Linear Algebra. But I understood your proof plan: there is a finite number of adjacent transpositions to go from one permutation to another, and we know that the sum is preserved between them, so as between the first permutation and the last one.
– Le Anh Dung
2 days ago
I have just figured out how to tackle the second issue ^^ I will fix it tomorrow.
– Le Anh Dung
2 days ago
that's good news =)
– Siong Thye Goh
2 days ago
I have fixed two mentioned issues. Please check my revised proof!
– Le Anh Dung
yesterday
I think it looks fine :) you might like to highlight the place where you use commutative and associative properties of addition.
– Siong Thye Goh
yesterday
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I think the main doubt is for the $sigma$ in the case of $k+1$, is it the same $sigma$ for the case of $k$? Is the induction hypothesis valid since $sigma$ is not a permutation of $I_k$?
One possible approach is note that permutation is a products of adjacent transposition so we can focus on proving that under adjacent transposition, the sum is preserved.
Suppose $sigma$ only switches $j$ and $j+1$.
We have
beginalignsum_i=1^j+1a_sigma(i)&=left( sum_i=1^j-1a_i + a_j+1right) + a_j \
&= sum_i=1^j-1a_i + (a_j+1 + a_j), textAssociativity\
&=sum_i=1^j-1 a_i +(a_j+a_j+1), textCommutative \
&=left(sum_i=1^j-1a_i+a_jright) + a_j+1 , textAssociativity\
&=sum_i=1^j+1a_iendalign
Hence $sum_i=1^n a_sigma(i)= sum_i=1^n a_i$.
Hence the result is true by induction on the number of adjacent transposition.
answered 2 days ago
Siong Thye Goh
76.8k134794
I think the main doubt is for the $sigma$ in the case of $k+1$, is it the same $sigma$ for the case of $k$? Is the induction hypothesis valid since $sigma$ is not a permutation of $I_k$?
One possible approach is note that permutation is a products of adjacent transposition so we can focus on proving that under adjacent transposition, the sum is preserved.
Suppose $sigma$ only switches $j$ and $j+1$.
We have
beginalignsum_i=1^j+1a_sigma(i)&=left( sum_i=1^j-1a_i + a_j+1right) + a_j \
&= sum_i=1^j-1a_i + (a_j+1 + a_j), textAssociativity\
&=sum_i=1^j-1 a_i +(a_j+a_j+1), textCommutative \
&=left(sum_i=1^j-1a_i+a_jright) + a_j+1 , textAssociativity\
&=sum_i=1^j+1a_iendalign
Hence $sum_i=1^n a_sigma(i)= sum_i=1^n a_i$.
Hence the result is true by induction on the number of adjacent transposition.
answered 2 days ago
Siong Thye Goh
76.8k134794
answered 2 days ago
Siong Thye Goh
76.8k134794
answered 2 days ago
Siong Thye Goh
76.8k134794
answered 2 days ago
Siong Thye Goh
76.8k134794
76.8k134794
Thank you so much for pointing out my two issues. I have fixed the $1^st$ issue by using $sigma$ for $I_k$ and $sigma'$ for $I_k+1$. For the $2^nd$ issue, I'm finding way to get over. Since this is my very first exposure to advanced mathematics, I have not studied anything such as adjacent transposition in Linear Algebra. But I understood your proof plan: there is a finite number of adjacent transpositions to go from one permutation to another, and we know that the sum is preserved between them, so as between the first permutation and the last one.
– Le Anh Dung
2 days ago
I have just figured out how to tackle the second issue ^^ I will fix it tomorrow.
– Le Anh Dung
2 days ago
that's good news =)
– Siong Thye Goh
2 days ago
I have fixed two mentioned issues. Please check my revised proof!
– Le Anh Dung
yesterday
I think it looks fine :) you might like to highlight the place where you use commutative and associative properties of addition.
– Siong Thye Goh
yesterday
 |Â
show 1 more comment
Thank you so much for pointing out my two issues. I have fixed the $1^st$ issue by using $sigma$ for $I_k$ and $sigma'$ for $I_k+1$. For the $2^nd$ issue, I'm finding way to get over. Since this is my very first exposure to advanced mathematics, I have not studied anything such as adjacent transposition in Linear Algebra. But I understood your proof plan: there is a finite number of adjacent transpositions to go from one permutation to another, and we know that the sum is preserved between them, so as between the first permutation and the last one.
– Le Anh Dung
2 days ago
I have just figured out how to tackle the second issue ^^ I will fix it tomorrow.
– Le Anh Dung
2 days ago
that's good news =)
– Siong Thye Goh
2 days ago
I have fixed two mentioned issues. Please check my revised proof!
– Le Anh Dung
yesterday
I think it looks fine :) you might like to highlight the place where you use commutative and associative properties of addition.
– Siong Thye Goh
yesterday
Thank you so much for pointing out my two issues. I have fixed the $1^st$ issue by using $sigma$ for $I_k$ and $sigma'$ for $I_k+1$. For the $2^nd$ issue, I'm finding way to get over. Since this is my very first exposure to advanced mathematics, I have not studied anything such as adjacent transposition in Linear Algebra. But I understood your proof plan: there is a finite number of adjacent transpositions to go from one permutation to another, and we know that the sum is preserved between them, so as between the first permutation and the last one.
– Le Anh Dung
2 days ago
Thank you so much for pointing out my two issues. I have fixed the $1^st$ issue by using $sigma$ for $I_k$ and $sigma'$ for $I_k+1$. For the $2^nd$ issue, I'm finding way to get over. Since this is my very first exposure to advanced mathematics, I have not studied anything such as adjacent transposition in Linear Algebra. But I understood your proof plan: there is a finite number of adjacent transpositions to go from one permutation to another, and we know that the sum is preserved between them, so as between the first permutation and the last one.
– Le Anh Dung
2 days ago
I have just figured out how to tackle the second issue ^^ I will fix it tomorrow.
– Le Anh Dung
2 days ago
I have just figured out how to tackle the second issue ^^ I will fix it tomorrow.
– Le Anh Dung
2 days ago
that's good news =)
– Siong Thye Goh
2 days ago
that's good news =)
– Siong Thye Goh
2 days ago
I have fixed two mentioned issues. Please check my revised proof!
– Le Anh Dung
yesterday
I have fixed two mentioned issues. Please check my revised proof!
– Le Anh Dung
yesterday
I think it looks fine :) you might like to highlight the place where you use commutative and associative properties of addition.
– Siong Thye Goh
yesterday
I think it looks fine :) you might like to highlight the place where you use commutative and associative properties of addition.
– Siong Thye Goh
yesterday
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