How to solve this problem on the number line?

Question

I have been having a lot of trouble trying to solve this problem, I have tried to work with some numbers, but I find many ways to do it, I do not arrive at anything concrete.

According to the information given in the attached number line, which of the following statements is (are) true? $$I) n^4 > \frac{n}{4}$$ $$II) 2n<n$$ $$III)n^{(-3)} > n^2$$ $$IV) (n - \frac{1}{n})^{(-2)} <n^{(-3)}$$

Original at https://i.stack.imgur.com/9LMxT.png

Answer 1

Okay.... we have $n^2 < n < n^{-1}$.

Now $n^2 \ge 0$ so we have $0 \le n^2 < n < n^{-1}$ so $n > 0$.

Since $n > 0$ we can multiply or divide by $n$ and maintain the inequality.

If we multiply all by $n$ we get $n^2\cdot n < n\cdot n < n^{-1}\cdot n$ or $n^3 < n^2 < 1$ and if we divide all by $n$ we get $\frac {n^2}n < \frac nn < \frac {n^{-1}}n$ or $n < 1 < n^{-2}$.

So we have $0 < n < 1$.

I'll leave it to you to do the rest....

Well, no... I'll do II) $n> 0$ so $n+n > n+0$ so $2n > n$. So II) which say $2n < n$ is always false.

We could also have done that since $n> 0$ then $2n < n$ is true only if $\frac {2n}n < \frac nn$ or if $2 < 1$. Since $2 > 1$ it can not be true.

Answer 2

I'll keep the letter $n$ although using it to denote a real number is utterly unsatisfying. The image tells you that $$n^2<n<\frac{1}{n}.$$ The first inequality is attained iff $0<n<1$ (¿why?). So the second inequality tells you nothing.

At first glance, (II) is wrong. Now, you can plug some specific values of $n$ (e. g. $n=1/2$) in the remaining inequalities to (possibly) get rid of the wrong ones. Spoiler: all of them are wrong.

Answer 3

The information from the numberline gives two conditions

$$\begin{array}{rcl} n-n^2 &>&0 \\ 1/n-n&>&0 \end{array}$$

These conditions are satisfied when $0<n<1$

Then you need to check the four statements

$$\begin{array}{rrcl} \text{I)}& n^4-n/4 &>&0 \\ \text{II)}& n-2n &>&0 \\ \text{III)}& n^{-3}-n^2 &>&0 \\ \text{IV)}& n^{-3}-(n-1/n)^{-2} &>&0 \\ \end{array}$$

The image below is a spoiler which shows that statement III is true, statement II is false, and statements I and IV can not be determined with the information. You can proof it algebraically for all those four cases. For example case II is simple $n-2n = -n>0$ or $n<0$ which is false when $0<n<1$.