Understanding logic of proof

**p00ndawg** · September 9th 2009, 05:42 PM

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!

**flyingsquirrel** · September 10th 2009, 01:20 AM

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.

**xalk** · September 10th 2009, 09:33 AM

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 $x\geq 2$

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