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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.

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.

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

A proof by contradiction functions by saying "if A is false, B must be true. I can prove that B is false, so A cannot be false."

You proved that B is true. This means that A could be false, but is not necessarily so because we have made no statements that relate the truth of A to the truth of B.

You can (more or less) avoid the complications of a proof by contradiction by reasoning as follows:

The total of the $10$ sums of consecutive triplets $a_i+a_i+1+a_i+2$ is $3 times 55 = 165$. Therefore the average of the sums of consecutive triplets is $165/10 = 16.5$. But each sum is an integer, and if a set of integers has an average of $16.5$, at least one of them must be $ge 17$.

(It could perhaps be argued that the final claim here requires a proof, and that would be a proof by contradiction.)

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.