# Where is the mistake in my development?

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 ?

• $-5 < -3$ but $(-5)^2=25 > (-3)^2=9$ – Vasya Jun 18 '18 at 20:20
• When you multiply both sides of an inequality by a negative, you must reverse the inequality. – Adrian Keister Jun 18 '18 at 20:21
• where i multiply by a negative? – Eduardo Sebastian Jun 18 '18 at 20:30
• 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. – karmalu Jun 18 '18 at 20:46

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 $|n|<1$.