# Thread: Understanding logic of proof

1. ## Understanding logic of proof

Prove: If $x^2 + x - 6 \geq 0$, then $x \leq -3$ or $x \geq 2.$

Im not understanding the logic behind this questions answer. My professor went ahead and:

Suppose that $x^2 + x - 6 \geq 0$ and $x > -3$. It follows that $(x-2)(x+3) \geq 0$ and, since $x + 3 > 0$, it must be that $x -2 \geq 0$. That is, $x \geq 2$.

im not understanding why they suppose x > -3, and how the bolded makes sense. I think im just not getting something easy here, please help.

Thanks!

2. Hello,
Originally Posted by p00ndawg
im not understanding why they suppose x > -3,
For any real number $x$ we have either $x\leq -3$ or $x>-3$, in particular this is also true if $x$ satisfies $x^2 + x - 6 \geq 0$. Thus instead of showing "If $x^2 + x - 6 \geq 0$, then $x \leq -3$ or $x \geq 2$." one can show these two statements:
• If $x^2 + x - 6 \geq 0$ and $x\leq -3$, then $x \leq -3$ or $x \geq 2$.
• If $x^2 + x - 6 \geq 0$ and $x>-3$, then $x \leq -3$ or $x \geq 2$.

As the first one is obviously true they do not mention it and only prove the last one.
Originally Posted by p00ndawg
It follows that $(x-2)(x+3) \geq 0$
This is true because $(x-2)(x+3) = x^2-2x+3x-6 = x^2+x-6\geq0$.
Originally Posted by p00ndawg
and, since $x + 3 > 0$
This is a consequence of this inequality: $x>-3$.
Originally Posted by p00ndawg
it must be that $x -2 \geq 0$.
We have $(x-2)(x+3)\geq 0$ hence $x-2$ and $x+3$ must have the same sign. Since $x+3$ is non-negative, $x-2$ has to be non-negative too.

3. Originally Posted by p00ndawg
Prove: If $x^2 + x - 6 \geq 0$, then $x \leq -3$ or $x \geq 2.$

Im not understanding the logic behind this questions answer. My professor went ahead and:

Suppose that $x^2 + x - 6 \geq 0$ and $x > -3$. It follows that $(x-2)(x+3) \geq 0$ and, since $x + 3 > 0$, it must be that $x -2 \geq 0$. That is, $x \geq 2$.

im not understanding why they suppose x > -3, and how the bolded makes sense. I think im just not getting something easy here, please help.

Thanks!

Put...( $x^2 + x -6\geq 0$) = p, ( $x\leq -3$)= q ,( $x\geq 2$) =r.

and we asked to prove :

if p ,then q or r which in symbols is : $p\Longrightarrow (q\vee r)$.................................................. ......................................1

But q or r is logically equivalent to $\neg q\Longrightarrow r$ then (1) becomes :

if p then ( if not q then r) which in symbols is: $p\Longrightarrow (\neg q\Longrightarrow r)$.

But.

$p\Longrightarrow (\neg q\Longrightarrow)$ is logically equivalent to $(p\wedge\neg q)\Longrightarrow r$.

That in logic means that whether we prove the 1st or the 2nd formula it is the same thing.

And in using the 2nd formula we must assume p and not q ( $p\wedge \neg q$) and try to prove r.

OR .

assume : $x^2 + x - 6\geq 0$ and not $x\leq -3$....... $\neg (x\leq -3)$.... and try to prove $x\geq 2$.

But since $\neg(x\leq -3)\Longrightarrow x>- 3$ ,because $x\leq -3$ or x>-3,instead assuming :

$x^2 + x -6\geq 0$ and $\neg(x\leq -3)$ we can assume:

$x^2 + x -6\geq 0$ and x>-3 or x+3>0.

Now since $x\geq 2$ or x<2 or x-2<0,if x-2<0 and since x+3>0 we have (x-2)(x+3) >0 or $x^2 + x -6< 0$ which contradicts the fact that $x^2 + x -6\geq 0$.

THUS $x\geq 2$