What does it mean when proof by contradiction doesn't lead to a contradiction?

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.

Solution 1:

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.

Solution 2:

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

Solution 3:

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.

Solution 4:

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.)