Understanding Graham's proof of theorem on Unit Fractions.
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In this paper by Ronald Graham, the theorem that every integer greater than 77 has a partition with the property that the sum of the reciprocals of the various "piles" in the partition is 1 (lovely!).
He gives a proof of this which is remarkably short but I find difficult to follow. He demonstrates that all the integers from 78 to 333 inclusive have such a representation and then demonstrates 2 transforms on a sum of reciprocals that add to 1 that retain this property, but which increase the sum of denominators from $U$ to $2U+2$ and $2U+179$.
He then essentially claims that by having a certain property (the above one) hold for all the integers from 78 to 333 and by knowing that the above transforms retain the property, it holds for all integers greater than 77. However, his method of explaining this is quite incomprehensible to me and appears to contain numerous arithmetic errors on top of this!
Can someone either give an alternative proof of the fact using what I've mentioned or simply explain Graham's argument, as it is a wonderful theorem?
number-theory proof-explanation integer-partitions
asked Jan 3 at 21:44
Isky MathewsIsky Mathews
863314
$endgroup$
add a comment |
$begingroup$
In this paper by Ronald Graham, the theorem that every integer greater than 77 has a partition with the property that the sum of the reciprocals of the various "piles" in the partition is 1 (lovely!).
He gives a proof of this which is remarkably short but I find difficult to follow. He demonstrates that all the integers from 78 to 333 inclusive have such a representation and then demonstrates 2 transforms on a sum of reciprocals that add to 1 that retain this property, but which increase the sum of denominators from $U$ to $2U+2$ and $2U+179$.
He then essentially claims that by having a certain property (the above one) hold for all the integers from 78 to 333 and by knowing that the above transforms retain the property, it holds for all integers greater than 77. However, his method of explaining this is quite incomprehensible to me and appears to contain numerous arithmetic errors on top of this!
Can someone either give an alternative proof of the fact using what I've mentioned or simply explain Graham's argument, as it is a wonderful theorem?
number-theory proof-explanation integer-partitions
asked Jan 3 at 21:44
Isky MathewsIsky Mathews
863314
$endgroup$
1
$begingroup$
The "piles" in a partition are surprisingly called "parts".
$endgroup$
– Marc van Leeuwen
Jan 16 at 14:39
add a comment |
$begingroup$
In this paper by Ronald Graham, the theorem that every integer greater than 77 has a partition with the property that the sum of the reciprocals of the various "piles" in the partition is 1 (lovely!).
He gives a proof of this which is remarkably short but I find difficult to follow. He demonstrates that all the integers from 78 to 333 inclusive have such a representation and then demonstrates 2 transforms on a sum of reciprocals that add to 1 that retain this property, but which increase the sum of denominators from $U$ to $2U+2$ and $2U+179$.
He then essentially claims that by having a certain property (the above one) hold for all the integers from 78 to 333 and by knowing that the above transforms retain the property, it holds for all integers greater than 77. However, his method of explaining this is quite incomprehensible to me and appears to contain numerous arithmetic errors on top of this!
Can someone either give an alternative proof of the fact using what I've mentioned or simply explain Graham's argument, as it is a wonderful theorem?
number-theory proof-explanation integer-partitions
asked Jan 3 at 21:44
Isky MathewsIsky Mathews
863314
$endgroup$
In this paper by Ronald Graham, the theorem that every integer greater than 77 has a partition with the property that the sum of the reciprocals of the various "piles" in the partition is 1 (lovely!).
He gives a proof of this which is remarkably short but I find difficult to follow. He demonstrates that all the integers from 78 to 333 inclusive have such a representation and then demonstrates 2 transforms on a sum of reciprocals that add to 1 that retain this property, but which increase the sum of denominators from $U$ to $2U+2$ and $2U+179$.
He then essentially claims that by having a certain property (the above one) hold for all the integers from 78 to 333 and by knowing that the above transforms retain the property, it holds for all integers greater than 77. However, his method of explaining this is quite incomprehensible to me and appears to contain numerous arithmetic errors on top of this!
Can someone either give an alternative proof of the fact using what I've mentioned or simply explain Graham's argument, as it is a wonderful theorem?
number-theory proof-explanation integer-partitions
number-theory proof-explanation integer-partitions
asked Jan 3 at 21:44
Isky MathewsIsky Mathews
863314
asked Jan 3 at 21:44
Isky MathewsIsky Mathews
863314
asked Jan 3 at 21:44
Isky MathewsIsky Mathews
863314
asked Jan 3 at 21:44
Isky MathewsIsky Mathews
863314
asked Jan 3 at 21:44
Isky MathewsIsky Mathews
863314
863314
1
$begingroup$
The "piles" in a partition are surprisingly called "parts".
$endgroup$
– Marc van Leeuwen
Jan 16 at 14:39
add a comment |
1
$begingroup$
The "piles" in a partition are surprisingly called "parts".
$endgroup$
– Marc van Leeuwen
Jan 16 at 14:39
1
1
$begingroup$
The "piles" in a partition are surprisingly called "parts".
$endgroup$
– Marc van Leeuwen
Jan 16 at 14:39
$begingroup$
The "piles" in a partition are surprisingly called "parts".
$endgroup$
– Marc van Leeuwen
Jan 16 at 14:39
add a comment |
1 Answer
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What a wonderful theorem indeed :)
You can prove this easily by strong induction. The base case of $78leq nleq333$ has been handled. Assume the theorem holds for all $78leq nleq m$, $mgeq333$. Then $m+1=2k$ or $m+1=2k+1$ for some natural number $k$.
Case $m+1=2k$: Let $U=k-1$. Then $Uleq m$ and $U=(m-1)/2geq(333-1)/2geq78$, so the theorem holds for $U$. Hence, the theorem holds for $m+1=2U+2$.
Case $m+1=2k+1$: Let $U=k-89$. Then $Uleq m$ and $U=(m-178)/2geq(333-178)/2=77frac12$. Because $U$ is an integer, $Ugeq78$, so the theorem holds for $U$. Hence, the theorem holds for $m+1=2U+179$.
answered Jan 3 at 22:10
SmileyCraftSmileyCraft
3,776519
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$begingroup$
Thank you very much! You must agree that Graham's explanation leaves something to be desired...
$endgroup$
– Isky Mathews
Jan 4 at 9:47
add a comment |
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$begingroup$
What a wonderful theorem indeed :)
You can prove this easily by strong induction. The base case of $78leq nleq333$ has been handled. Assume the theorem holds for all $78leq nleq m$, $mgeq333$. Then $m+1=2k$ or $m+1=2k+1$ for some natural number $k$.
Case $m+1=2k$: Let $U=k-1$. Then $Uleq m$ and $U=(m-1)/2geq(333-1)/2geq78$, so the theorem holds for $U$. Hence, the theorem holds for $m+1=2U+2$.
Case $m+1=2k+1$: Let $U=k-89$. Then $Uleq m$ and $U=(m-178)/2geq(333-178)/2=77frac12$. Because $U$ is an integer, $Ugeq78$, so the theorem holds for $U$. Hence, the theorem holds for $m+1=2U+179$.
answered Jan 3 at 22:10
SmileyCraftSmileyCraft
3,776519
$endgroup$
$begingroup$
Thank you very much! You must agree that Graham's explanation leaves something to be desired...
$endgroup$
– Isky Mathews
Jan 4 at 9:47
add a comment |
$begingroup$
What a wonderful theorem indeed :)
You can prove this easily by strong induction. The base case of $78leq nleq333$ has been handled. Assume the theorem holds for all $78leq nleq m$, $mgeq333$. Then $m+1=2k$ or $m+1=2k+1$ for some natural number $k$.
Case $m+1=2k$: Let $U=k-1$. Then $Uleq m$ and $U=(m-1)/2geq(333-1)/2geq78$, so the theorem holds for $U$. Hence, the theorem holds for $m+1=2U+2$.
Case $m+1=2k+1$: Let $U=k-89$. Then $Uleq m$ and $U=(m-178)/2geq(333-178)/2=77frac12$. Because $U$ is an integer, $Ugeq78$, so the theorem holds for $U$. Hence, the theorem holds for $m+1=2U+179$.
answered Jan 3 at 22:10
SmileyCraftSmileyCraft
3,776519
$endgroup$
$begingroup$
Thank you very much! You must agree that Graham's explanation leaves something to be desired...
$endgroup$
– Isky Mathews
Jan 4 at 9:47
add a comment |
$begingroup$
What a wonderful theorem indeed :)
You can prove this easily by strong induction. The base case of $78leq nleq333$ has been handled. Assume the theorem holds for all $78leq nleq m$, $mgeq333$. Then $m+1=2k$ or $m+1=2k+1$ for some natural number $k$.
Case $m+1=2k$: Let $U=k-1$. Then $Uleq m$ and $U=(m-1)/2geq(333-1)/2geq78$, so the theorem holds for $U$. Hence, the theorem holds for $m+1=2U+2$.
Case $m+1=2k+1$: Let $U=k-89$. Then $Uleq m$ and $U=(m-178)/2geq(333-178)/2=77frac12$. Because $U$ is an integer, $Ugeq78$, so the theorem holds for $U$. Hence, the theorem holds for $m+1=2U+179$.
answered Jan 3 at 22:10
SmileyCraftSmileyCraft
3,776519
$endgroup$
What a wonderful theorem indeed :)
You can prove this easily by strong induction. The base case of $78leq nleq333$ has been handled. Assume the theorem holds for all $78leq nleq m$, $mgeq333$. Then $m+1=2k$ or $m+1=2k+1$ for some natural number $k$.
Case $m+1=2k$: Let $U=k-1$. Then $Uleq m$ and $U=(m-1)/2geq(333-1)/2geq78$, so the theorem holds for $U$. Hence, the theorem holds for $m+1=2U+2$.
Case $m+1=2k+1$: Let $U=k-89$. Then $Uleq m$ and $U=(m-178)/2geq(333-178)/2=77frac12$. Because $U$ is an integer, $Ugeq78$, so the theorem holds for $U$. Hence, the theorem holds for $m+1=2U+179$.
answered Jan 3 at 22:10
SmileyCraftSmileyCraft
3,776519
answered Jan 3 at 22:10
SmileyCraftSmileyCraft
3,776519
answered Jan 3 at 22:10
SmileyCraftSmileyCraft
3,776519
answered Jan 3 at 22:10
SmileyCraftSmileyCraft
3,776519
3,776519
$begingroup$
Thank you very much! You must agree that Graham's explanation leaves something to be desired...
$endgroup$
– Isky Mathews
Jan 4 at 9:47
add a comment |
$begingroup$
Thank you very much! You must agree that Graham's explanation leaves something to be desired...
$endgroup$
– Isky Mathews
Jan 4 at 9:47
$begingroup$
Thank you very much! You must agree that Graham's explanation leaves something to be desired...
$endgroup$
– Isky Mathews
Jan 4 at 9:47
$begingroup$
Thank you very much! You must agree that Graham's explanation leaves something to be desired...
$endgroup$
– Isky Mathews
Jan 4 at 9:47
add a comment |
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1
$begingroup$
The "piles" in a partition are surprisingly called "parts".
$endgroup$
– Marc van Leeuwen
Jan 16 at 14:39