Prove that for real numbers $x$, if $x^2 - 5x + 4 \ge 0$, then either $x \le 1$ or $x \ge 4$.

Question

Its another homework question that I'm having trouble understanding.

The full question is write a detailed structured proof that uses a proof by cases to prove that for real numbers $x$, if $x^2 - 5x + 4 \ge 0$, then either $x \le 1$ or $x \ge 4$.

You may use without proof the fact that:

$\forall a \in \mathbb R, \forall b \in \mathbb R, [(ab \ge 0) \longleftrightarrow (((a \ge 0) \wedge (b \ge 0)) \vee ((a \le 0) \wedge (b \le 0)))]$.

I actually don't understand what is going on here $\forall a \in \mathbb R, \forall b \in \mathbb R, [(ab \ge 0) \longleftrightarrow (((a \ge 0) \wedge (b \ge 0)) \vee ((a \le 0) \wedge (b \le 0)))]$.

I know that $x^2 - 5x + 4 \ge 0$, then either $x \le 1$ or $x \ge 4$ can be factored with the quadratic equation to give me $x^2 - 5x +4 = (x-1)(x-4)$. I also know that I can use $(x-1)(x-4)$ in the place of $ab \ge 0$, where $a$ would most likely be $(x-1)$, and $b$ being $(x-4)$.

What's throwing me off is it says $((a \ge 0) \wedge (b \ge 0)) \vee ((a \le 0) \wedge (b \le 0))$. My math is rusty admittedly and I don't understand why both $a$ and $b$ both need to be $\ge \vee \le$. I'm unable to actually start the homework question because I don't understand what it's trying to tell me. Can anyone help me understand/clarify it for me.

Answer 1

$x^2-5x+4=(x-1)(x-4)\ge0$. So in order for the expression to be positive it has to be the product of two positive numbers or two negative numbers (which the hint states). If it is the product of two negative numbers what is the inequality for x? If it is the product of two positive numbers what is the inequality for x?

Answer 2

The symbolic statement that you are allowed to use without proof is simply saying nothing more than "if the product of two terms is positive, then either they are both positive, or they are both negative".

So with that in mind, You want to take this thing which you have clearly factored, $$(x-1)(x-4) \geq 0,$$

and consider cases. One way might be to find critical values where

$$(x-1)(x-4)=0.$$ Clearly these are $x=1$, or $x=4$. Now test a case for each region in $x<1$, $1<x<4$, and $x>4$. You are of course looking for the positive regions. Your solution will follow from this effort.

Answer 3

So the answer was that the notation says both $a$ and $b$ need to be positive OR negative, no positive and negative, but both at the same time. And after reading this highschool calculus should be flooding back into your head.