Question on a proof of no existance of monochromatic triangle in any tricoloring of edges of $K_{16}$
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Here is a question on the book "Problem -Solving Strategies" by Arthur Engel, with a provided solution:
In the solution, they partitioned the abelian group into 3 sets such that none of them is sum-free. However, in $A_3$, $(a+b+d) +(a+c+d)=b+c in A_3$.
proof-verification graph-theory pigeonhole-principle
asked Dec 12 '18 at 0:18
nafhgoodnafhgood
1,805422
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add a comment |
$begingroup$
Here is a question on the book "Problem -Solving Strategies" by Arthur Engel, with a provided solution:
In the solution, they partitioned the abelian group into 3 sets such that none of them is sum-free. However, in $A_3$, $(a+b+d) +(a+c+d)=b+c in A_3$.
proof-verification graph-theory pigeonhole-principle
asked Dec 12 '18 at 0:18
nafhgoodnafhgood
1,805422
$endgroup$
$begingroup$
yes, sorry I just wanted to know the correct version
$endgroup$
– nafhgood
Dec 12 '18 at 0:51
$begingroup$
Ok! I edited it!
$endgroup$
– nafhgood
Dec 12 '18 at 0:56
add a comment |
$begingroup$
Here is a question on the book "Problem -Solving Strategies" by Arthur Engel, with a provided solution:
In the solution, they partitioned the abelian group into 3 sets such that none of them is sum-free. However, in $A_3$, $(a+b+d) +(a+c+d)=b+c in A_3$.
proof-verification graph-theory pigeonhole-principle
asked Dec 12 '18 at 0:18
nafhgoodnafhgood
1,805422
$endgroup$
Here is a question on the book "Problem -Solving Strategies" by Arthur Engel, with a provided solution:
In the solution, they partitioned the abelian group into 3 sets such that none of them is sum-free. However, in $A_3$, $(a+b+d) +(a+c+d)=b+c in A_3$.
proof-verification graph-theory pigeonhole-principle
proof-verification graph-theory pigeonhole-principle
asked Dec 12 '18 at 0:18
nafhgoodnafhgood
1,805422
asked Dec 12 '18 at 0:18
nafhgoodnafhgood
1,805422
edited Dec 12 '18 at 0:55
nafhgood
asked Dec 12 '18 at 0:18
nafhgoodnafhgood
1,805422
asked Dec 12 '18 at 0:18
nafhgoodnafhgood
1,805422
asked Dec 12 '18 at 0:18
nafhgoodnafhgood
1,805422
1,805422
$begingroup$
yes, sorry I just wanted to know the correct version
$endgroup$
– nafhgood
Dec 12 '18 at 0:51
$begingroup$
Ok! I edited it!
$endgroup$
– nafhgood
Dec 12 '18 at 0:56
add a comment |
$begingroup$
yes, sorry I just wanted to know the correct version
$endgroup$
– nafhgood
Dec 12 '18 at 0:51
$begingroup$
Ok! I edited it!
$endgroup$
– nafhgood
Dec 12 '18 at 0:56
$begingroup$
yes, sorry I just wanted to know the correct version
$endgroup$
– nafhgood
Dec 12 '18 at 0:51
$begingroup$
yes, sorry I just wanted to know the correct version
$endgroup$
– nafhgood
Dec 12 '18 at 0:51
$begingroup$
Ok! I edited it!
$endgroup$
– nafhgood
Dec 12 '18 at 0:56
$begingroup$
Ok! I edited it!
$endgroup$
– nafhgood
Dec 12 '18 at 0:56
add a comment |
1 Answer
1
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oldest
votes
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You are right. To correct the proof, move $a+c+d$ to $A_2$ and $b+c+d$ to $A_3$. Is everything all right now?
answered Dec 12 '18 at 0:48
bofbof
51.7k558120
$endgroup$
$begingroup$
Is there an algorithm to partition a group into disjoint unions of subsets that are sum free?
$endgroup$
– nafhgood
Dec 12 '18 at 15:03
$begingroup$
I don't know.$ $
$endgroup$
– bof
Dec 12 '18 at 15:10
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are right. To correct the proof, move $a+c+d$ to $A_2$ and $b+c+d$ to $A_3$. Is everything all right now?
answered Dec 12 '18 at 0:48
bofbof
51.7k558120
$endgroup$
$begingroup$
Is there an algorithm to partition a group into disjoint unions of subsets that are sum free?
$endgroup$
– nafhgood
Dec 12 '18 at 15:03
$begingroup$
I don't know.$ $
$endgroup$
– bof
Dec 12 '18 at 15:10
add a comment |
$begingroup$
You are right. To correct the proof, move $a+c+d$ to $A_2$ and $b+c+d$ to $A_3$. Is everything all right now?
answered Dec 12 '18 at 0:48
bofbof
51.7k558120
$endgroup$
$begingroup$
Is there an algorithm to partition a group into disjoint unions of subsets that are sum free?
$endgroup$
– nafhgood
Dec 12 '18 at 15:03
$begingroup$
I don't know.$ $
$endgroup$
– bof
Dec 12 '18 at 15:10
add a comment |
$begingroup$
You are right. To correct the proof, move $a+c+d$ to $A_2$ and $b+c+d$ to $A_3$. Is everything all right now?
answered Dec 12 '18 at 0:48
bofbof
51.7k558120
$endgroup$
You are right. To correct the proof, move $a+c+d$ to $A_2$ and $b+c+d$ to $A_3$. Is everything all right now?
answered Dec 12 '18 at 0:48
bofbof
51.7k558120
answered Dec 12 '18 at 0:48
bofbof
51.7k558120
answered Dec 12 '18 at 0:48
bofbof
51.7k558120
answered Dec 12 '18 at 0:48
bofbof
51.7k558120
51.7k558120
$begingroup$
Is there an algorithm to partition a group into disjoint unions of subsets that are sum free?
$endgroup$
– nafhgood
Dec 12 '18 at 15:03
$begingroup$
I don't know.$ $
$endgroup$
– bof
Dec 12 '18 at 15:10
add a comment |
$begingroup$
Is there an algorithm to partition a group into disjoint unions of subsets that are sum free?
$endgroup$
– nafhgood
Dec 12 '18 at 15:03
$begingroup$
I don't know.$ $
$endgroup$
– bof
Dec 12 '18 at 15:10
$begingroup$
Is there an algorithm to partition a group into disjoint unions of subsets that are sum free?
$endgroup$
– nafhgood
Dec 12 '18 at 15:03
$begingroup$
Is there an algorithm to partition a group into disjoint unions of subsets that are sum free?
$endgroup$
– nafhgood
Dec 12 '18 at 15:03
$begingroup$
I don't know.$ $
$endgroup$
– bof
Dec 12 '18 at 15:10
$begingroup$
I don't know.$ $
$endgroup$
– bof
Dec 12 '18 at 15:10
add a comment |
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$begingroup$
yes, sorry I just wanted to know the correct version
$endgroup$
– nafhgood
Dec 12 '18 at 0:51
$begingroup$
Ok! I edited it!
$endgroup$
– nafhgood
Dec 12 '18 at 0:56