Proof by contrapositive: $x^4 + 2x^2 - 2x \lt 0 \Rightarrow 0 \lt x \lt 1$

Question

I managed to prove this using a direct proof but my prof suggested I try proving it using the contrapositive. Here's what I have so far:

Contrapositive: $(x \le 0) \lor (x \ge 1) \Rightarrow x^4 + 2x^2 - 2x \ge 0$

Splitting this into two, ($P_1 \Rightarrow Q)\land(P_2 \Rightarrow Q)$:

$$x \ge 1 \Rightarrow x^4 + 2x^2 - 2x \ge 0$$

1- $x \ge 1 \Rightarrow [x^4 \ge 1, 2x^2 \ge 2, -2x \ge -2]$

2- $x^4 + 2x^2 - 2x \ge 1 + 2 -2$

3- $x^4 + 2x^2 - 2x \ge 1$

Ok, that was easy enough... but here's where my brain gets stuck.

$$x \le 0 \Rightarrow x^4 + 2x^2 - 2x \ge 0$$

Where should I begin here? In my direct proof I started with the equation and worked towards the x value. I am not sure how to go about proving it the other way. Any help is appreciated.

Answer 1

Remember that when you multiply both sides of an inequality by a negative number, the direction of the inequality is reversed.

The second part of your proof is actually the easier part. First, $x^2 \geq 0$ for every real number $x,$ and likewise $x^4 \geq 0.$ So $x \leq 0 \implies x^4 + 2x^2 \geq 0$ (the right-hand side of the implication is true, therefore the implication is true). Next, $x \leq 0 \implies -2x \geq 0$ (multiplying by a negative number reverses the implication). Putting them together, $x \leq 0 \implies x^4 + 2x^2 - 2x\geq 0.$

Your proof for the first half goes wrong due to the same fact about multiplication by a negative number: $x \geq 1 \implies -2x \leq -2.$ But if you plot $x^2 - x$ you may notice that it's never negative when $x \geq 1.$ If you prove that fact, can you see how to build on that to get the rest of the proof?

Answer 2

As David K commented, you need to rework the first part.

For the second part, it helps to factor. We assume $x \le 0$ and want to show that $x^4 + 2x^2 - 2x \ge 0$; i.e. that $x(x^3 + 2x - 2) \ge 0$. It is easy to explain why each factor is negative.