Can Trivial Conditions Validate an Implication in Mathematical Proofs?

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The discussion revolves around proving the implication that if |n-1| + |n+1| ≤ 1, then |n²-1| ≤ 4 for n in natural numbers. A participant attempts to disprove the statement by showing that the condition |n-1| + |n+1| ≤ 1 leads to an impossible conclusion for n in N. They conclude that since there are no natural numbers satisfying the hypothesis, the implication is vacuously true. The conversation highlights the distinction between proving a statement and providing a counterexample, emphasizing that the absence of valid n confirms the original implication. Ultimately, the assertion that the statement is true is validated by the logical structure of implications.
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Homework Statement


Let ##n\in N ##. Prove that if ##|n-1|+|n+1|\leq 1,## then ## |n^{2}-1|\leq 4##


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The Attempt at a Solution


I am trying to show by a counter example that this statement is not true.
Consider this statement:
##|n-1|+|n+1| \leq 1##
Assume :
|n-1|=n-1 >0 and |n+1|=n+1>0
Then, by our former statement, ## n-1+n+1 \leq 1,##
which gives
## 2n \leq 1 , \text{where n} \in N##
Now, divide by 2, and ##n \leq \dfrac{1}{2}##, which is not possible since ##n \in N##

Call ##|n-1|+|n+1| \leq 1## , P.
In logic, P→Q. If P is false, then Q does not matter, the implication is always true.

Is this non-elegant proof correct?
 
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You said you were "trying to prove by a counter example that this statement is not true." First, this is not a counter example. A "counter example" would be a specific n, satisfying the hypothesis, such that the conclusion is not true.

Further, you said yourself that "In logic, P→Q. If P is false, then Q does not matter, the implication is always true." So, by showing that there is NO n in N such that |n- 1|+ |n+1|\le 1, you have shown that the hypothesis is always false and so have shown that the statement itself is true.
 
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