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What does it mean when proof by contradiction doesn't lead to a contradiction?
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I started writing a proof using the method of proof by contradiction and encountered a situation which was true. More specifically, the hypothesis that I set out to prove was:
If the first 10 positive integer is placed around a circle, in any order, there exists 3 integer in consecutive locations around the circle that have a sum greater than or equal to 17. (From Discrete Mathematics and its Applications - K. Rosen)
This is how I proceeded:
Let $a_i$ denote the $i^th$ integer on the boundary of the circle. To proceed with proof by contradiction, we assume that $forall i$
$a_i + a_i+1 + a_i+2 < 17$
Then,
$a_1 + a_2 + a_3 < 17$
$a_2 + a_3 + a_4 < 17$
$vdots$
$a_10 + a_1 + a_2 < 17$
$therefore 3 cdot (a_1 + a_2 + dots + a_10) < 17 cdot10$
$Rightarrow 3 cdot 55 < 170$
$Rightarrow 165 < 170$
which is true. What does this mean?
P.S. I am not looking for the solution to this problem. I am aware of how to prove the claim. I am just curious about what it means to arrive at a truth after assuming the negation of the hypothesis.
proof-verification proof-writing arithmetic
asked 2 hours ago
Nihal Jain
805
add a comment |Â
up vote
2
down vote
favorite
I started writing a proof using the method of proof by contradiction and encountered a situation which was true. More specifically, the hypothesis that I set out to prove was:
If the first 10 positive integer is placed around a circle, in any order, there exists 3 integer in consecutive locations around the circle that have a sum greater than or equal to 17. (From Discrete Mathematics and its Applications - K. Rosen)
This is how I proceeded:
Let $a_i$ denote the $i^th$ integer on the boundary of the circle. To proceed with proof by contradiction, we assume that $forall i$
$a_i + a_i+1 + a_i+2 < 17$
Then,
$a_1 + a_2 + a_3 < 17$
$a_2 + a_3 + a_4 < 17$
$vdots$
$a_10 + a_1 + a_2 < 17$
$therefore 3 cdot (a_1 + a_2 + dots + a_10) < 17 cdot10$
$Rightarrow 3 cdot 55 < 170$
$Rightarrow 165 < 170$
which is true. What does this mean?
P.S. I am not looking for the solution to this problem. I am aware of how to prove the claim. I am just curious about what it means to arrive at a truth after assuming the negation of the hypothesis.
proof-verification proof-writing arithmetic
asked 2 hours ago
Nihal Jain
805
It doesn't mean very much -- you've made the assumption that any three consecutive numbers will sum to no more than 16, so if after making that assumption you conclude that it's true, then you've just sort of gone in a circle
– dbx
2 hours ago
It means your proof doesn't work. Either proof by contradiction is not the right approach, or you need to find stronger statements you can make. In the present instance, you really haven't used the fact the $3$ number are consecutive in any way.
– saulspatz
2 hours ago
@Vasya $a_9+a_10+a_1$ is implicitly part of the $vdots$
– Henry
2 hours ago
@saulspatz By including $a_9+a_10+a_1$ and $a_10+a_1+a_2$ I have ensured that all sets of 3 consecutive integers are included.
– Nihal Jain
2 hours ago
Yes, that's true, and Henry's argument shows that the problem is easier than I had realized. What I meant is that the argument is the same if you have any collection of three-element subsets such that each number belongs to three of the sets. That turns out to be irrelevant, however.
– saulspatz
2 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I started writing a proof using the method of proof by contradiction and encountered a situation which was true. More specifically, the hypothesis that I set out to prove was:
If the first 10 positive integer is placed around a circle, in any order, there exists 3 integer in consecutive locations around the circle that have a sum greater than or equal to 17. (From Discrete Mathematics and its Applications - K. Rosen)
This is how I proceeded:
Let $a_i$ denote the $i^th$ integer on the boundary of the circle. To proceed with proof by contradiction, we assume that $forall i$
$a_i + a_i+1 + a_i+2 < 17$
Then,
$a_1 + a_2 + a_3 < 17$
$a_2 + a_3 + a_4 < 17$
$vdots$
$a_10 + a_1 + a_2 < 17$
$therefore 3 cdot (a_1 + a_2 + dots + a_10) < 17 cdot10$
$Rightarrow 3 cdot 55 < 170$
$Rightarrow 165 < 170$
which is true. What does this mean?
P.S. I am not looking for the solution to this problem. I am aware of how to prove the claim. I am just curious about what it means to arrive at a truth after assuming the negation of the hypothesis.
proof-verification proof-writing arithmetic
asked 2 hours ago
Nihal Jain
805
I started writing a proof using the method of proof by contradiction and encountered a situation which was true. More specifically, the hypothesis that I set out to prove was:
If the first 10 positive integer is placed around a circle, in any order, there exists 3 integer in consecutive locations around the circle that have a sum greater than or equal to 17. (From Discrete Mathematics and its Applications - K. Rosen)
This is how I proceeded:
Let $a_i$ denote the $i^th$ integer on the boundary of the circle. To proceed with proof by contradiction, we assume that $forall i$
$a_i + a_i+1 + a_i+2 < 17$
Then,
$a_1 + a_2 + a_3 < 17$
$a_2 + a_3 + a_4 < 17$
$vdots$
$a_10 + a_1 + a_2 < 17$
$therefore 3 cdot (a_1 + a_2 + dots + a_10) < 17 cdot10$
$Rightarrow 3 cdot 55 < 170$
$Rightarrow 165 < 170$
which is true. What does this mean?
P.S. I am not looking for the solution to this problem. I am aware of how to prove the claim. I am just curious about what it means to arrive at a truth after assuming the negation of the hypothesis.
proof-verification proof-writing arithmetic
proof-verification proof-writing arithmetic
asked 2 hours ago
Nihal Jain
805
asked 2 hours ago
Nihal Jain
805
asked 2 hours ago
Nihal Jain
805
asked 2 hours ago
Nihal Jain
805
asked 2 hours ago
Nihal Jain
805
805
It doesn't mean very much -- you've made the assumption that any three consecutive numbers will sum to no more than 16, so if after making that assumption you conclude that it's true, then you've just sort of gone in a circle
– dbx
2 hours ago
It means your proof doesn't work. Either proof by contradiction is not the right approach, or you need to find stronger statements you can make. In the present instance, you really haven't used the fact the $3$ number are consecutive in any way.
– saulspatz
2 hours ago
@Vasya $a_9+a_10+a_1$ is implicitly part of the $vdots$
– Henry
2 hours ago
@saulspatz By including $a_9+a_10+a_1$ and $a_10+a_1+a_2$ I have ensured that all sets of 3 consecutive integers are included.
– Nihal Jain
2 hours ago
Yes, that's true, and Henry's argument shows that the problem is easier than I had realized. What I meant is that the argument is the same if you have any collection of three-element subsets such that each number belongs to three of the sets. That turns out to be irrelevant, however.
– saulspatz
2 hours ago
add a comment |Â
It doesn't mean very much -- you've made the assumption that any three consecutive numbers will sum to no more than 16, so if after making that assumption you conclude that it's true, then you've just sort of gone in a circle
– dbx
2 hours ago
It means your proof doesn't work. Either proof by contradiction is not the right approach, or you need to find stronger statements you can make. In the present instance, you really haven't used the fact the $3$ number are consecutive in any way.
– saulspatz
2 hours ago
@Vasya $a_9+a_10+a_1$ is implicitly part of the $vdots$
– Henry
2 hours ago
@saulspatz By including $a_9+a_10+a_1$ and $a_10+a_1+a_2$ I have ensured that all sets of 3 consecutive integers are included.
– Nihal Jain
2 hours ago
Yes, that's true, and Henry's argument shows that the problem is easier than I had realized. What I meant is that the argument is the same if you have any collection of three-element subsets such that each number belongs to three of the sets. That turns out to be irrelevant, however.
– saulspatz
2 hours ago
It doesn't mean very much -- you've made the assumption that any three consecutive numbers will sum to no more than 16, so if after making that assumption you conclude that it's true, then you've just sort of gone in a circle
– dbx
2 hours ago
It doesn't mean very much -- you've made the assumption that any three consecutive numbers will sum to no more than 16, so if after making that assumption you conclude that it's true, then you've just sort of gone in a circle
– dbx
2 hours ago
It means your proof doesn't work. Either proof by contradiction is not the right approach, or you need to find stronger statements you can make. In the present instance, you really haven't used the fact the $3$ number are consecutive in any way.
– saulspatz
2 hours ago
It means your proof doesn't work. Either proof by contradiction is not the right approach, or you need to find stronger statements you can make. In the present instance, you really haven't used the fact the $3$ number are consecutive in any way.
– saulspatz
2 hours ago
@Vasya $a_9+a_10+a_1$ is implicitly part of the $vdots$
– Henry
2 hours ago
@Vasya $a_9+a_10+a_1$ is implicitly part of the $vdots$
– Henry
2 hours ago
@saulspatz By including $a_9+a_10+a_1$ and $a_10+a_1+a_2$ I have ensured that all sets of 3 consecutive integers are included.
– Nihal Jain
2 hours ago
@saulspatz By including $a_9+a_10+a_1$ and $a_10+a_1+a_2$ I have ensured that all sets of 3 consecutive integers are included.
– Nihal Jain
2 hours ago
Yes, that's true, and Henry's argument shows that the problem is easier than I had realized. What I meant is that the argument is the same if you have any collection of three-element subsets such that each number belongs to three of the sets. That turns out to be irrelevant, however.
– saulspatz
2 hours ago
Yes, that's true, and Henry's argument shows that the problem is easier than I had realized. What I meant is that the argument is the same if you have any collection of three-element subsets such that each number belongs to three of the sets. That turns out to be irrelevant, however.
– saulspatz
2 hours ago
add a comment |Â
3 Answers
3
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oldest
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up vote
2
down vote
accepted
Your goal is to show that $p$ is false.
If $p implies q$ and $q$ is true.
We can't conclude if $p$ is true or false. Hence, we get an inconclusive situation.
answered 2 hours ago
Siong Thye Goh
81.5k1453103
Could you provide some context by telling what $p$ and $q$ are in the case of this example?
– Nihal Jain
2 hours ago
your $p$ is "every $3$ in consecutive locations around the table has a sum that is less than $17$. your $q$ is $165<170$.
– Siong Thye Goh
2 hours ago
So, if $q$ evaluates to true, we cannot comment on whether $p$ is true or false. Is my understanding correct?
– Nihal Jain
1 hour ago
1
yup, if it's sunny ($p$) , i go to play ($q$). But if you observe that I go to play, you can't conclude that it's sunny.
– Siong Thye Goh
1 hour ago
add a comment |Â
up vote
4
down vote
It means that your precise approach does not work, but since this does not provide a counterexample it means the question would still be open.
Seeing $170-165$ is so small, there may be a way to save your proof. Try $$a_i + a_i+1 + a_i+2 le 16$$
leading to $$3 cdot (a_1 + a_2 + dots + a_10) le 16 cdot10$$ and $$165 le 160$$ for a contradiction
answered 2 hours ago
Henry
93.8k471149
I see it now, thanks
– Vasya
2 hours ago
Woah! I didn't know that proofs could be tweaked like this to change the conclusions. Thanks for your input! However, could you explain why arriving at a truth $(165 < 170)$ does not mean that the initial assumptions must be true?
– Nihal Jain
2 hours ago
@NihalJain If you show $p implies q$ then this excludes the possibility that $p$ is true and $q$ is false, but there remain three possibilities: (1) $p$ is true and $q$ is true; (2) $p$ is false and $q$ is true; (3) $p$ is false and $q$ is false. If you also know that $q$ is true then that excludes (3), but does not help in deciding between (1) and (2), so it is possible that $p$ is true and it is possible that $p$ is false.
– Henry
1 hour ago
@Henry yes, this point was made even below. I've understood where I went wrong. Thanks :)
– Nihal Jain
1 hour ago
add a comment |Â
up vote
0
down vote
Try this. Assuming there is a solution that always sums to less than $17$ for $3$ consecutive numbers, and building around the number $10$, the only possible number combinations adjacent to $10$ will be $4$ numbers from the subset $1,2,3,4,5$. Taking the maximum of these as $1,5$ and $2,4$ $(1,5,10,2,4)$ leaves a minimum sequence of $3,6,7,8,9$.
These $5$ numbers are impossible to arrange so all $3$ consecutive numbers in this subset are less than $17$. That is, the only $3$ numbers summing to less than $17$ are $6,3,7$ and adding an $8$ or $9$ to this sequence results in a $3$ consecutive number sum being greater than $16$. Hence the impossibility or contradiction of this assumption.
answered 1 hour ago
Phil H
2,0622311
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your goal is to show that $p$ is false.
If $p implies q$ and $q$ is true.
We can't conclude if $p$ is true or false. Hence, we get an inconclusive situation.
answered 2 hours ago
Siong Thye Goh
81.5k1453103
Could you provide some context by telling what $p$ and $q$ are in the case of this example?
– Nihal Jain
2 hours ago
your $p$ is "every $3$ in consecutive locations around the table has a sum that is less than $17$. your $q$ is $165<170$.
– Siong Thye Goh
2 hours ago
So, if $q$ evaluates to true, we cannot comment on whether $p$ is true or false. Is my understanding correct?
– Nihal Jain
1 hour ago
1
yup, if it's sunny ($p$) , i go to play ($q$). But if you observe that I go to play, you can't conclude that it's sunny.
– Siong Thye Goh
1 hour ago
add a comment |Â
up vote
2
down vote
accepted
Your goal is to show that $p$ is false.
If $p implies q$ and $q$ is true.
We can't conclude if $p$ is true or false. Hence, we get an inconclusive situation.
answered 2 hours ago
Siong Thye Goh
81.5k1453103
Could you provide some context by telling what $p$ and $q$ are in the case of this example?
– Nihal Jain
2 hours ago
your $p$ is "every $3$ in consecutive locations around the table has a sum that is less than $17$. your $q$ is $165<170$.
– Siong Thye Goh
2 hours ago
So, if $q$ evaluates to true, we cannot comment on whether $p$ is true or false. Is my understanding correct?
– Nihal Jain
1 hour ago
1
yup, if it's sunny ($p$) , i go to play ($q$). But if you observe that I go to play, you can't conclude that it's sunny.
– Siong Thye Goh
1 hour ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your goal is to show that $p$ is false.
If $p implies q$ and $q$ is true.
We can't conclude if $p$ is true or false. Hence, we get an inconclusive situation.
answered 2 hours ago
Siong Thye Goh
81.5k1453103
Your goal is to show that $p$ is false.
If $p implies q$ and $q$ is true.
We can't conclude if $p$ is true or false. Hence, we get an inconclusive situation.
answered 2 hours ago
Siong Thye Goh
81.5k1453103
answered 2 hours ago
Siong Thye Goh
81.5k1453103
answered 2 hours ago
Siong Thye Goh
81.5k1453103
answered 2 hours ago
Siong Thye Goh
81.5k1453103
81.5k1453103
Could you provide some context by telling what $p$ and $q$ are in the case of this example?
– Nihal Jain
2 hours ago
your $p$ is "every $3$ in consecutive locations around the table has a sum that is less than $17$. your $q$ is $165<170$.
– Siong Thye Goh
2 hours ago
So, if $q$ evaluates to true, we cannot comment on whether $p$ is true or false. Is my understanding correct?
– Nihal Jain
1 hour ago
1
yup, if it's sunny ($p$) , i go to play ($q$). But if you observe that I go to play, you can't conclude that it's sunny.
– Siong Thye Goh
1 hour ago
add a comment |Â
Could you provide some context by telling what $p$ and $q$ are in the case of this example?
– Nihal Jain
2 hours ago
your $p$ is "every $3$ in consecutive locations around the table has a sum that is less than $17$. your $q$ is $165<170$.
– Siong Thye Goh
2 hours ago
So, if $q$ evaluates to true, we cannot comment on whether $p$ is true or false. Is my understanding correct?
– Nihal Jain
1 hour ago
1
yup, if it's sunny ($p$) , i go to play ($q$). But if you observe that I go to play, you can't conclude that it's sunny.
– Siong Thye Goh
1 hour ago
Could you provide some context by telling what $p$ and $q$ are in the case of this example?
– Nihal Jain
2 hours ago
Could you provide some context by telling what $p$ and $q$ are in the case of this example?
– Nihal Jain
2 hours ago
your $p$ is "every $3$ in consecutive locations around the table has a sum that is less than $17$. your $q$ is $165<170$.
– Siong Thye Goh
2 hours ago
your $p$ is "every $3$ in consecutive locations around the table has a sum that is less than $17$. your $q$ is $165<170$.
– Siong Thye Goh
2 hours ago
So, if $q$ evaluates to true, we cannot comment on whether $p$ is true or false. Is my understanding correct?
– Nihal Jain
1 hour ago
So, if $q$ evaluates to true, we cannot comment on whether $p$ is true or false. Is my understanding correct?
– Nihal Jain
1 hour ago
1
1
yup, if it's sunny ($p$) , i go to play ($q$). But if you observe that I go to play, you can't conclude that it's sunny.
– Siong Thye Goh
1 hour ago
yup, if it's sunny ($p$) , i go to play ($q$). But if you observe that I go to play, you can't conclude that it's sunny.
– Siong Thye Goh
1 hour ago
add a comment |Â
up vote
4
down vote
It means that your precise approach does not work, but since this does not provide a counterexample it means the question would still be open.
Seeing $170-165$ is so small, there may be a way to save your proof. Try $$a_i + a_i+1 + a_i+2 le 16$$
leading to $$3 cdot (a_1 + a_2 + dots + a_10) le 16 cdot10$$ and $$165 le 160$$ for a contradiction
answered 2 hours ago
Henry
93.8k471149
I see it now, thanks
– Vasya
2 hours ago
Woah! I didn't know that proofs could be tweaked like this to change the conclusions. Thanks for your input! However, could you explain why arriving at a truth $(165 < 170)$ does not mean that the initial assumptions must be true?
– Nihal Jain
2 hours ago
@NihalJain If you show $p implies q$ then this excludes the possibility that $p$ is true and $q$ is false, but there remain three possibilities: (1) $p$ is true and $q$ is true; (2) $p$ is false and $q$ is true; (3) $p$ is false and $q$ is false. If you also know that $q$ is true then that excludes (3), but does not help in deciding between (1) and (2), so it is possible that $p$ is true and it is possible that $p$ is false.
– Henry
1 hour ago
@Henry yes, this point was made even below. I've understood where I went wrong. Thanks :)
– Nihal Jain
1 hour ago
add a comment |Â
up vote
4
down vote
It means that your precise approach does not work, but since this does not provide a counterexample it means the question would still be open.
Seeing $170-165$ is so small, there may be a way to save your proof. Try $$a_i + a_i+1 + a_i+2 le 16$$
leading to $$3 cdot (a_1 + a_2 + dots + a_10) le 16 cdot10$$ and $$165 le 160$$ for a contradiction
answered 2 hours ago
Henry
93.8k471149
I see it now, thanks
– Vasya
2 hours ago
Woah! I didn't know that proofs could be tweaked like this to change the conclusions. Thanks for your input! However, could you explain why arriving at a truth $(165 < 170)$ does not mean that the initial assumptions must be true?
– Nihal Jain
2 hours ago
@NihalJain If you show $p implies q$ then this excludes the possibility that $p$ is true and $q$ is false, but there remain three possibilities: (1) $p$ is true and $q$ is true; (2) $p$ is false and $q$ is true; (3) $p$ is false and $q$ is false. If you also know that $q$ is true then that excludes (3), but does not help in deciding between (1) and (2), so it is possible that $p$ is true and it is possible that $p$ is false.
– Henry
1 hour ago
@Henry yes, this point was made even below. I've understood where I went wrong. Thanks :)
– Nihal Jain
1 hour ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
It means that your precise approach does not work, but since this does not provide a counterexample it means the question would still be open.
Seeing $170-165$ is so small, there may be a way to save your proof. Try $$a_i + a_i+1 + a_i+2 le 16$$
leading to $$3 cdot (a_1 + a_2 + dots + a_10) le 16 cdot10$$ and $$165 le 160$$ for a contradiction
answered 2 hours ago
Henry
93.8k471149
It means that your precise approach does not work, but since this does not provide a counterexample it means the question would still be open.
Seeing $170-165$ is so small, there may be a way to save your proof. Try $$a_i + a_i+1 + a_i+2 le 16$$
leading to $$3 cdot (a_1 + a_2 + dots + a_10) le 16 cdot10$$ and $$165 le 160$$ for a contradiction
answered 2 hours ago
Henry
93.8k471149
answered 2 hours ago
Henry
93.8k471149
answered 2 hours ago
Henry
93.8k471149
answered 2 hours ago
Henry
93.8k471149
93.8k471149
I see it now, thanks
– Vasya
2 hours ago
Woah! I didn't know that proofs could be tweaked like this to change the conclusions. Thanks for your input! However, could you explain why arriving at a truth $(165 < 170)$ does not mean that the initial assumptions must be true?
– Nihal Jain
2 hours ago
@NihalJain If you show $p implies q$ then this excludes the possibility that $p$ is true and $q$ is false, but there remain three possibilities: (1) $p$ is true and $q$ is true; (2) $p$ is false and $q$ is true; (3) $p$ is false and $q$ is false. If you also know that $q$ is true then that excludes (3), but does not help in deciding between (1) and (2), so it is possible that $p$ is true and it is possible that $p$ is false.
– Henry
1 hour ago
@Henry yes, this point was made even below. I've understood where I went wrong. Thanks :)
– Nihal Jain
1 hour ago
add a comment |Â
I see it now, thanks
– Vasya
2 hours ago
Woah! I didn't know that proofs could be tweaked like this to change the conclusions. Thanks for your input! However, could you explain why arriving at a truth $(165 < 170)$ does not mean that the initial assumptions must be true?
– Nihal Jain
2 hours ago
@NihalJain If you show $p implies q$ then this excludes the possibility that $p$ is true and $q$ is false, but there remain three possibilities: (1) $p$ is true and $q$ is true; (2) $p$ is false and $q$ is true; (3) $p$ is false and $q$ is false. If you also know that $q$ is true then that excludes (3), but does not help in deciding between (1) and (2), so it is possible that $p$ is true and it is possible that $p$ is false.
– Henry
1 hour ago
@Henry yes, this point was made even below. I've understood where I went wrong. Thanks :)
– Nihal Jain
1 hour ago
I see it now, thanks
– Vasya
2 hours ago
I see it now, thanks
– Vasya
2 hours ago
Woah! I didn't know that proofs could be tweaked like this to change the conclusions. Thanks for your input! However, could you explain why arriving at a truth $(165 < 170)$ does not mean that the initial assumptions must be true?
– Nihal Jain
2 hours ago
Woah! I didn't know that proofs could be tweaked like this to change the conclusions. Thanks for your input! However, could you explain why arriving at a truth $(165 < 170)$ does not mean that the initial assumptions must be true?
– Nihal Jain
2 hours ago
@NihalJain If you show $p implies q$ then this excludes the possibility that $p$ is true and $q$ is false, but there remain three possibilities: (1) $p$ is true and $q$ is true; (2) $p$ is false and $q$ is true; (3) $p$ is false and $q$ is false. If you also know that $q$ is true then that excludes (3), but does not help in deciding between (1) and (2), so it is possible that $p$ is true and it is possible that $p$ is false.
– Henry
1 hour ago
@NihalJain If you show $p implies q$ then this excludes the possibility that $p$ is true and $q$ is false, but there remain three possibilities: (1) $p$ is true and $q$ is true; (2) $p$ is false and $q$ is true; (3) $p$ is false and $q$ is false. If you also know that $q$ is true then that excludes (3), but does not help in deciding between (1) and (2), so it is possible that $p$ is true and it is possible that $p$ is false.
– Henry
1 hour ago
@Henry yes, this point was made even below. I've understood where I went wrong. Thanks :)
– Nihal Jain
1 hour ago
@Henry yes, this point was made even below. I've understood where I went wrong. Thanks :)
– Nihal Jain
1 hour ago
add a comment |Â
up vote
0
down vote
Try this. Assuming there is a solution that always sums to less than $17$ for $3$ consecutive numbers, and building around the number $10$, the only possible number combinations adjacent to $10$ will be $4$ numbers from the subset $1,2,3,4,5$. Taking the maximum of these as $1,5$ and $2,4$ $(1,5,10,2,4)$ leaves a minimum sequence of $3,6,7,8,9$.
These $5$ numbers are impossible to arrange so all $3$ consecutive numbers in this subset are less than $17$. That is, the only $3$ numbers summing to less than $17$ are $6,3,7$ and adding an $8$ or $9$ to this sequence results in a $3$ consecutive number sum being greater than $16$. Hence the impossibility or contradiction of this assumption.
answered 1 hour ago
Phil H
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Try this. Assuming there is a solution that always sums to less than $17$ for $3$ consecutive numbers, and building around the number $10$, the only possible number combinations adjacent to $10$ will be $4$ numbers from the subset $1,2,3,4,5$. Taking the maximum of these as $1,5$ and $2,4$ $(1,5,10,2,4)$ leaves a minimum sequence of $3,6,7,8,9$.
These $5$ numbers are impossible to arrange so all $3$ consecutive numbers in this subset are less than $17$. That is, the only $3$ numbers summing to less than $17$ are $6,3,7$ and adding an $8$ or $9$ to this sequence results in a $3$ consecutive number sum being greater than $16$. Hence the impossibility or contradiction of this assumption.
answered 1 hour ago
Phil H
2,0622311
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Try this. Assuming there is a solution that always sums to less than $17$ for $3$ consecutive numbers, and building around the number $10$, the only possible number combinations adjacent to $10$ will be $4$ numbers from the subset $1,2,3,4,5$. Taking the maximum of these as $1,5$ and $2,4$ $(1,5,10,2,4)$ leaves a minimum sequence of $3,6,7,8,9$.
These $5$ numbers are impossible to arrange so all $3$ consecutive numbers in this subset are less than $17$. That is, the only $3$ numbers summing to less than $17$ are $6,3,7$ and adding an $8$ or $9$ to this sequence results in a $3$ consecutive number sum being greater than $16$. Hence the impossibility or contradiction of this assumption.
answered 1 hour ago
Phil H
2,0622311
Try this. Assuming there is a solution that always sums to less than $17$ for $3$ consecutive numbers, and building around the number $10$, the only possible number combinations adjacent to $10$ will be $4$ numbers from the subset $1,2,3,4,5$. Taking the maximum of these as $1,5$ and $2,4$ $(1,5,10,2,4)$ leaves a minimum sequence of $3,6,7,8,9$.
These $5$ numbers are impossible to arrange so all $3$ consecutive numbers in this subset are less than $17$. That is, the only $3$ numbers summing to less than $17$ are $6,3,7$ and adding an $8$ or $9$ to this sequence results in a $3$ consecutive number sum being greater than $16$. Hence the impossibility or contradiction of this assumption.
answered 1 hour ago
Phil H
2,0622311
edited 1 hour ago
answered 1 hour ago
Phil H
2,0622311
answered 1 hour ago
Phil H
2,0622311
answered 1 hour ago
Phil H
2,0622311
2,0622311
add a comment |Â
add a comment |Â
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It doesn't mean very much -- you've made the assumption that any three consecutive numbers will sum to no more than 16, so if after making that assumption you conclude that it's true, then you've just sort of gone in a circle
– dbx
2 hours ago
It means your proof doesn't work. Either proof by contradiction is not the right approach, or you need to find stronger statements you can make. In the present instance, you really haven't used the fact the $3$ number are consecutive in any way.
– saulspatz
2 hours ago
@Vasya $a_9+a_10+a_1$ is implicitly part of the $vdots$
– Henry
2 hours ago
@saulspatz By including $a_9+a_10+a_1$ and $a_10+a_1+a_2$ I have ensured that all sets of 3 consecutive integers are included.
– Nihal Jain
2 hours ago
Yes, that's true, and Henry's argument shows that the problem is easier than I had realized. What I meant is that the argument is the same if you have any collection of three-element subsets such that each number belongs to three of the sets. That turns out to be irrelevant, however.
– saulspatz
2 hours ago