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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 ?

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  • $\begingroup$ $-5 < -3$ but $(-5)^2=25 > (-3)^2=9$ $\endgroup$ – Vasya Jun 18 '18 at 20:20
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    $\begingroup$ When you multiply both sides of an inequality by a negative, you must reverse the inequality. $\endgroup$ – Adrian Keister Jun 18 '18 at 20:21
  • $\begingroup$ where i multiply by a negative? $\endgroup$ – Eduardo Sebastian Jun 18 '18 at 20:30
  • $\begingroup$ when you raise $x$ to the second power , you are multiplying $x$ by $x$. If $x$ is negative, you are multiplying by a negative number. $\endgroup$ – karmalu Jun 18 '18 at 20:46
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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 $|n|<1$.

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