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 ?

  • $\begingroup$ $-5 < -3$ but $(-5)^2=25 > (-3)^2=9$ $\endgroup$ – Vasya Jun 18 '18 at 20:20
  • 4
    $\begingroup$ 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

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 $|n|<1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.