Proof by induction $n^2-2n-1>0$ for $n \ge 3$
I want to use induction to prove that $n^2-2n-1>0$ for $n \ge 3$
Base case:
$3^2-2(3)-1>0$ $ \space \checkmark$
Inductive step:
$(n+1)^2-2(n+1)-1>0$ $\iff n^2+2n+1-2n-2-1>0$ $\iff n^2-2>0 \iff n^2>2$
Is it now trivial to just say that this inequality is true for all $n\ge 3$ or do I have more work to do? I am pretty new to this so I am not sure at what point I am done or at what point I have conclusively shown that the statement is true.
$\endgroup$ 42 Answers
$\begingroup$The inequality can easily be proved without induction, as $$ n^2 - 2n -1 = (n-1)^2 -2 \ge (3-1)^2 - 2 = 2 \text{ for } n \ge 3 \, . $$ so your inductive step is formally right, but does not use the inductive hypothesis.
If you write $$ (n+1)^2 - 2(n+1)- 1 = (n^2 - 2n - 1) + 2(n-1) $$ then you can use the inductive hypothesis in the inductive step.
$\endgroup$ 0 $\begingroup$Write $f(n)$ = $n^2 - 2n - 1$ = $(n - 1)^2 - 2$. This makes it easier. Though induction isn't needed:
For $n = 3$: $f(3) = 2 > 0$, so the property holds.
Assume that up to some $k > 3$, $f(k) > 0$. Let's show $f(k+1) > 0$. We have:
$f(k+1) = (k+1 - 1)^2 - 2 = (k-1 + 1)^2 - 2 = (k-1)^2 + 1^2 +2(k-1) - 2 = (k-1)^2 - 2 + 1 + 2k - 2 = f(k) + 2k - 1> 0 + 2k - 1 > 2(3) - 1 = 5 > 0$.
Therefore etc.
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