Let $n$ be a natural number. if $|n - 1| + |n + 1| ≤ 1$, then $|n^2 - 1| ≤ 4$

Question

My professor is telling me that I can prove this with a proof by cases approach, but I can only see possible options via a contradiction in the hypothesis and via induction. Is what my professor says true? Also, is my written proof rigorous and correct?

Here is what I have:

Proof:

Let $n\in\mathbb{N}$. Let $|n - 1| + |n + 1|\leq 1$

Observe that:

$$|n-1|+|n+1| = |(n-1) +(n+1)|$$

Since $n\in\mathbb{N}$, $n\geq 1$ by definition of natural numbers.

So, $n-1\geq 0$. Therefore, $n-1$ must be non negative, by definition of $0$.

Since $n\geq 1$, $n+1\geq 2$.

Therefore, $(n-1)+( n+1)\geq 2$ and we observe that this cannot be negative.

Observe that:

$$|n-1|+|n+1| = |(n-1) +(n+1)|$$

for all $n\in\mathbb{N}$.

This leads us to the observation that the absolute values in the statement $|(n-1)+(n+1)|$ are irrelevant, so we can drop them without loss of generality. So, $|n-1|+|n+1|\cong (n-1)+( n+1)$ for all $n\in\mathbb{N}$.

Therefore,

$|n-1|+|n+1|\leq 1 $

$\cong (n-1)+( n+1)\leq 1$

$= 2n\leq 1$

$= n \leq 1/2 < 1$

and thus $n < 1$.

However, as we established above, by definition of natural numbers, $n\geq 1$.
This is a contradiction. Therefore the hypothesis is false for all $n\in\mathbb{N}$ and the implication is vacuously true.

Answer 1

Actulally $|n-1| \leq 1$ and $|n+1| \leq 1$ so $|n^{2}-1|=|n-1||n+1| \leq 1$.

We don't even need the fact that $n$ is an integer!

EDIT. Actually the question is meaningless since there is no natural number satisfying the hypothesis.

Answer 2

If $|n-1|+|n+1| \le 1$ then $|n-1| \le 1$ and $|n+1| \le 1$.

If $|n-1| \le 1$ then $n \in [0,2]$ and $|n+1| \le 1$ implies $n \in [-2,0]$ hence $n = 0$.

Just to completely clear: It is easy to verify that $n=0$ does not satisfy the inequality hence there is no real number that satisfies the inequality.

In particular, if you require $n \ge 1$ then anything on the right hand side is true :-).

Answer 3

By (the one-dimensional application of) the triangle inequality, $|a-b| + |b-c| \ge |a-c|$ for all real numbers, $a,b,$ and $c$. In particular

$$|n-1| + |n+1| = |1-n| + |n - (-1)| \ge |1 - (-1)| = 2$$

That is

$$|n-1| + |n+1| \ge 2$$

for all real numbers n.

Hence $|n - 1| + |n + 1| ≤ 1$ is never true.