# absolute value proof

• Sep 20th 2007, 07:50 PM
annalif
absolute value proof
Prove that if c is a positive real number and x is any real number, then -c </= x </= c if and only if |x|</= C.

</= greater than or equal to

Thank you!
• Sep 20th 2007, 07:51 PM
Krizalid
That's the absolute value definition.
• Sep 20th 2007, 07:58 PM
annalif
yeah, but i cant figure out how to formalize it into a proof :(

i get x </= |c| , so

-|c| </= x </= |c|

and then my head hurts and i want to give up on life
• Sep 20th 2007, 08:04 PM
Jhevon
Quote:

Originally Posted by annalif
yeah, but i cant figure out how to formalize it into a proof :(

i get x </= |c| , so

-|c| </= x </= |c|

and then my head hurts and i want to give up on life

yeah. we often take this fact for granted around here. it's hard to come up with a rigorous proof. but here is a possible one for the reverse implication.

Recall that $|x| = \left \{ \begin {array}{cc} x & \mbox { if } x \ge 0 \\ -x & \mbox { if } x < 0 \end {array} \right.$

Thus, by the definition of $|x|$:

$|x| \le C$

$\implies x \le C$ or $-x \le C$

$\implies x \le C$ or $x \ge -C$

combining these two inequalities, we obtain:

$-C \le x \le C$

now try to go the other way. i would probably try to reverse this exact proof. so split the inequality into two, and work on each case by case (if x > 0 or x < 0)