EDIT: This is the explicit text of the practice problem:
Find out, by induction, whether for $ n \geqslant 1$, it is true that $n^2 - 3n - 1 \leqslant 0$.
I am learning to use induction. I am asked to prove by induction whether the property $n^2 - 3n - 1 < 0$ holds for all $n\geqslant 1$.
This is obviously false, but I am finding it hard to prove by induction. I am not sure what I am allowed to do.
Base case: P(1) = -3 < 0.
Inductive step: Assuming the property is true for $n$, prove that it is also true for $n+1$.
$(n+1)^2 - 3(n+1) - 1 < 0$.
$n^2+ 2n + 1 -3n -3 - 1 < 0$
$n^2 -3n - 1 < -2n + 2$
Since, by inductive hypothesis, $n^2 -3n - 1 < 0$, then $0 \leqslant -2n+2$. (Is it ok to state it like this?)
Can I just plug a 2 in the inequality and end the proof there? Or do I have to disprove $-2n+2 \geqslant 0$?
I did not know what to do, so I started a new proof by induction. I tried to disprove that $-2n + 2 \geqslant 0$ for all $n \geqslant 1$.
Base case: $P(1) = 0 \geqslant 0$
Inductive step: Assuming the property is true for some $n$, prove that is is also true for $n+1$.
$-2(n+1) + 2 \geqslant 0 $
$-2n -2 + 2 = -2n < 0 $ (for all positive integers)
Is there a more straightforward way to disprove the first property? Even if there is, would it be correct to do it the way I did?