# Math Help - [SOLVED] Abosolute Values, precalc

1. ## [SOLVED] Abosolute Values, precalc

Hello; I currently working on absolute values and was doing good until I now.

Can anyone please explain to me, why; the absolute value of the square route of 2 -5 is 5- the square route of 2?

Thanks a bunch

Carolyn

2. Originally Posted by Carolyng66
Hello; I currently working on absolute values and was doing good until I now.

Can anyone please explain to me, why; the absolute value of the square route of 2 -5 is 5- the square route of 2?

Thanks a bunch

Carolyn
$|a-b| = |b| - |a|$

3. Originally Posted by Mush
$|a-b| = |b| - |a|$
THAT IS FALSE!
Prehaps, Mush means $\left| {a - b} \right| = \left| {b - a} \right|$?
That is true for all a & b

4. Originally Posted by Plato
THAT IS FALSE!
Prehaps, Mush means $\left| {a - b} \right| = \left| {b - a} \right|$?
That is true for all a & b
Now that I've gone over my notes gain, I understand this now. I believe this is right.

Thank You both for your help.

5. Originally Posted by Carolyng66
Hello; I currently working on absolute values and was doing good until I now. Can anyone please explain to me, why; the absolute value of the square route of 2 -5 is 5- the square route of 2?
By definition $\left| a \right| = \left\{ {\begin{array}{*{20}c}
{a,} & {0 \leqslant a} \\ { - a,} & {a < 0} \\ \end{array} } \right.$
.
Now $\sqrt 2 - 5 < 0\;\text{ so }\;\left| {\sqrt 2 - 5} \right| = - \left( {\sqrt 2 - 5} \right) = 5 - \sqrt 2 .
$

6. ## Math symbols

Originally Posted by Plato
By definition $\left| a \right| = \left\{ {\begin{array}{*{20}c}
{a,} & {0 \leqslant a} \\ { - a,} & {a < 0} \\ \end{array} } \right.$
.
Now $\sqrt 2 - 5 < 0\;\text{ so }\;\left| {\sqrt 2 - 5} \right| = - \left( {\sqrt 2 - 5} \right) = 5 - \sqrt 2 .
$
I'm finding all of this quite helpful. One more question about this sight (this is my first time using it). Can someone please explain to me how to make math symbols so I don't have to spell out words such as; "square root and absolute value". It would make future postings easier to read. I tried unsuccessfully locating a "cheat sheet" if you will.

Regards
Carolyn

7. Originally Posted by Plato
By definition $\left| a \right| = \left\{ {\begin{array}{*{20}c}
{a,} & {0 \leqslant a} \\ { - a,} & {a < 0} \\ \end{array} } \right.$
.
Now $\sqrt 2 - 5 < 0\;\text{ so }\;\left| {\sqrt 2 - 5} \right| = - \left( {\sqrt 2 - 5} \right) = 5 - \sqrt 2 .
$
So would be the same as saying the absolute value of 1-3 is 3-1?

8. Originally Posted by Plato
THAT IS FALSE!
Prehaps, Mush means $\left| {a - b} \right| = \left| {b - a} \right|$?
That is true for all a & b
Indeed, apologies.