# Understanding logic of proof

• Sep 9th 2009, 05:42 PM
p00ndawg
Understanding logic of proof
Prove: If $\displaystyle x^2 + x - 6 \geq 0$, then $\displaystyle x \leq -3$ or $\displaystyle x \geq 2.$

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

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

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!
• Sep 10th 2009, 01:20 AM
flyingsquirrel
Hello,
Quote:

Originally Posted by p00ndawg
im not understanding why they suppose x > -3,

For any real number $\displaystyle x$ we have either $\displaystyle x\leq -3$ or $\displaystyle x>-3$, in particular this is also true if $\displaystyle x$ satisfies $\displaystyle x^2 + x - 6 \geq 0$. Thus instead of showing "If $\displaystyle x^2 + x - 6 \geq 0$, then $\displaystyle x \leq -3$ or $\displaystyle x \geq 2$." one can show these two statements:
• If $\displaystyle x^2 + x - 6 \geq 0$ and $\displaystyle x\leq -3$, then $\displaystyle x \leq -3$ or $\displaystyle x \geq 2$.
• If $\displaystyle x^2 + x - 6 \geq 0$ and $\displaystyle x>-3$, then $\displaystyle x \leq -3$ or $\displaystyle x \geq 2$.

As the first one is obviously true they do not mention it and only prove the last one.
Quote:

Originally Posted by p00ndawg
It follows that $\displaystyle (x-2)(x+3) \geq 0$

This is true because $\displaystyle (x-2)(x+3) = x^2-2x+3x-6 = x^2+x-6\geq0$.
Quote:

Originally Posted by p00ndawg
and, since $\displaystyle x + 3 > 0$

This is a consequence of this inequality: $\displaystyle x>-3$.
Quote:

Originally Posted by p00ndawg
it must be that $\displaystyle x -2 \geq 0$.

We have $\displaystyle (x-2)(x+3)\geq 0$ hence $\displaystyle x-2$ and $\displaystyle x+3$ must have the same sign. Since $\displaystyle x+3$ is non-negative, $\displaystyle x-2$ has to be non-negative too.
• Sep 10th 2009, 09:33 AM
xalk
Quote:

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

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

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

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

and we asked to prove :

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

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

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

But.

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

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 ( $\displaystyle p\wedge \neg q$) and try to prove r.

OR .

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

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

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

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

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

THUS $\displaystyle x\geq 2$