If $x^3 - x > 0$, so $x > - 1$. How do I prove that by contrapositive? My attempt:
$Q \implies P$, by contrapositive we have: $\neg Q \implies \neg P$, so, if $x < - 1$, $x^3 - x < 0$
I don't know how to finish. Can someone help me?
 A: First, if not $a > b$, then $a \leq b$.  There are three ordering states: less than, equal, and greater than.  If we assert that one is not the case, then the other two remain.  So, from
$$  x^3 - x > 0 \implies x > -1  $$
the contrapositive is (notice the exchange of the two clauses as we negate them)
$$  \neg (x > -1) \implies \neg (x^3 - x > 0)  $$
and using the properties of inequalities, this simplifies to
$$  x \leq -1 \implies x^3 - x \leq 0  \text{.}  $$
Factoring, we obtain
$$  x \leq -1 \implies x(x^2 - 1) \leq 0  \text{.}  $$
Notice that our assumption forces $x$ to be negative.  This means the first factor, $x$, is negative.  For the second factor,

*

*if $x^2 > 1$, the second factor is positive,

*if $x^2 = 1$, the second factor is zero, and

*if $x^2 < 1$, the second factor is negative.

The assumed inequality, $x \leq -1$ gives us two cases: $x < -1$ or $x = -1$.

*

*($x = -1$ case)  If $x = -1$, then $x^2 = 1$ and the second factor is $0$, so $x(x^2-1) = (-1)(0) = 0$, which is compatible with $x(x^2-1) \leq 0$.

*($x < -1$ case)  If $x < -1$, then $|x| > 1$, so $x^2 > 1$, so $x(x^2 - 1)$ is the product of a negative and a positive number, so is negative, which is compatible with $x(x^2 - 1) \leq 0$.

Combining the two cases, we have shown the validity of $x \leq -1 \implies x^3 - x \leq 0$.
A: Take the simpler route:
Suppose that, $x^3-x>0$, but $x≤-1$. Then we have:

*

*If $x=-1$, then $x^3-x=0$.

A contradiction.

*

*If $x<-1$, then $x<0$ and $x^2>1$. This implies, $x^2-1>0$.
This means, $x(x^2-1)=x^3-x<0$.

Again a contradiction.
Therefore, if $x^3-x>0$, then $x>-1$.
