I have this statement:
If $\frac{1}{n} < -1$, so is always $\frac{1}{n^2}$ greather than $n$ ?
The development of my teacher was:
If $\frac{1}{n}$ is negative, $n$ must be negative. And if $n$ is negative, $n$ to pow of $2$ will be positive, so $\frac{1}{n^2} > 0$
And, $1 < -n$ that is equal to ($-1 > n$), that is $(-\infty,-1)$, so according to my teacher is correct.
My development was:
Since $\frac{1}{n} < -1$, to pow of $2$:
$\frac{1^2}{n^2} < 1$, that is $(-\infty, 1)$
And according to $-1 > n$, that is $(\infty,-1)$, so my deduction is that $\frac{1^2}{n^2}$ will be greather only in the $(-1, 1)$, because all other values would be the same.
So, where is my mistake on my development ?
Notice that if $$x>y$$ it follows that $$-x<-y$$ The inequality sign flipping is essential, and here it is why your development has a mistake.
As noted, $n$ must be negative. Therefore taking:$$ (\frac 1n)<-1$$ and squaring both sides, we must flip the inequality as we are multiplying by negatives. Therefore $\frac{1}{n^2}>1$, which leads to $|n|<1$.