Absolute value question

geekay

New member
If 0<a<2b applies, then

|a-2b| = 2b-a

Could someone please explain me why the above is true?

Sorry if this is posted in the wrong section.

Much appreciated.

Cubist

Full Member
Is the following true? And if so, then can you simplify the RHS?

|a-2b| = |-(a-2b)|

geekay

New member

if I open the parentheses on the RHS, it then becomes |-1*(a-2b)|=|-a+2b|, would it not?

Guess this is wrong way of thinking. So should one instead use |-1*(a-2b)|=|-1|*|a-2b|=|a-2b| ?

Perhaps my problem is this: don’t know how to get rid of the absolute value signs.

geekay

New member
Hey, guess I got it now! Thank you so much!

Dr.Peterson

Elite Member
If 0<a<2b applies, then

|a-2b| = 2b-a

Could someone please explain me why the above is true?

Sorry if this is posted in the wrong section.

Much appreciated.
Recall the piecewise definition of absolute value: if x>=0, then |x| = x; if x<0, then |x| = -x.

Which case applies here?

JeffM

Elite Member
$$\displaystyle \text {Definition: } c < 0 \implies |\ c \ | = - c;\ c \ge 0 \implies |\ c \ | = c.$$

Good so far?

$$\displaystyle \text {Let } d \text { be any real number.}$$

$$\displaystyle \therefore d < 0, d = 0, \text { or else } d > 0.$$

$$\displaystyle \text {CASE I: } d < 0 \implies \\ |\ d \ | = - d > 0 \implies |\ - d \ | = - d \implies |\ d \ | = |\ - d \ |.$$

$$\displaystyle \text {CASE II: } d = 0 \implies \\ |\ d \ | = 0 = - 0 \implies |\ - d \ | = |\ - 0 \ | = |\ 0 \ | = 0 \implies |\ d \ | = |\ - d \ |.$$

$$\displaystyle \text {CASE III: } d > 0 \implies \\ |\ d \ | = d > 0 \implies - d < 0 \implies |\ - d \ | = - (-d) = d \implies |\ d \ | = |\ - d \ |.$$
$$\displaystyle \therefore \text {in all cases, } |\ d \ | = |\ - d \ |.$$

Any questions?

Now apply that theorem to your problem.

geekay

New member
Thanks again for the help, guys. Great community!