# Trivial and Vacuous Proofs

1. Feb 18, 2014

### knowLittle

1. The problem statement, all variables and given/known data
Let $n\in N$. Prove that if $|n-1|+|n+1|\leq 1,$ then $|n^{2}-1|\leq 4$

2. Relevant equations

3. 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?

2. Feb 19, 2014

### HallsofIvy

Staff Emeritus
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.