1. ## Without Absolute Values

Write the following expressions without using absolute values.

(1) |b - 3| if ≥ 3

(2) |u - v|/|v - u| if u does not = v, u does not = 0 and v does not = 0.

(3) |a - 5| if a < 5

2. Originally Posted by sologuitar
Write the following expressions without using absolute values.

(1) |b - 3| if ≥ 3

(2) |u - v|/|v - u| if u does not = v, u does not = 0 and v does not = 0.

(3) |a - 5| if a < 5
I like to think about the absolute value of the difference of two numbers as the same thing as the bigger number minus to smaller one. |b-a|=|a-b| but depending on whether a>b or b>a the difference without the absolute value could be positive or negative.

For #1 you are just considering when b is greater than or equal to 3. What do you notice about numbers that are bigger than 3? Try plugging in b=4,5,10 and so on. Do they all have the same sign? If so, does the absolute value change anything?

3. ## but...

Originally Posted by Jameson
I like to think about the absolute value of the difference of two numbers as the same thing as the bigger number minus to smaller one. |b-a|=|a-b| but depending on whether a>b or b>a the difference without the absolute value could be positive or negative.

For #1 you are just considering when b is greater than or equal to 3. What do you notice about numbers that are bigger than 3? Try plugging in b=4,5,10 and so on. Do they all have the same sign? If so, does the absolute value change anything?
Thanks but I'm looking for a more algebraic reply. Can you solve at least one them algebraically?

4. 1) from the definition of absolute value, we get

$\displaystyle \left\vert b-3\right\vert =\left\{ \begin{array} [c]{cc}% (b-3), & for\text{ }(b-3)\geq0\text{ or }b\geq3\text{ }\\ (3-b), & for\text{ }(b-3)<0\text{ or }b<3\text{ }% \end{array} \right.$

hence

$\displaystyle \left\vert b-3\right\vert =(b-3)$ for $\displaystyle b\geq3\$

5. ## ok...

Originally Posted by dedust
1) from the definition of absolute value, we get

$\displaystyle \left\vert b-3\right\vert =\left\{ \begin{array} [c]{cc}% (b-3), & for\text{ }(b-3)\geq0\text{ or }b\geq3\text{ }\\ (3-b), & for\text{ }(b-3)<0\text{ or }b<3\text{ }% \end{array} \right.$

hence

$\displaystyle \left\vert b-3\right\vert =(b-3)$ for $\displaystyle b\geq3\$
Good. That's more like it.

6. Originally Posted by sologuitar
Thanks but I'm looking for a more algebraic reply. Can you solve at least one them algebraically?
I spent too much time just now looking up Latex syntax and reviewing logic rules so I'm posting this even though it probably isn't exactly what you want.

$\displaystyle \left \{ \forall b| b\in \mathbb{R} \text{ and } b \ge 3 \right \} \rightarrow \Big( \Big [ (b-3) \ge 0 \Big] \Leftrightarrow |b-3|=b-3 \Big)$

What do you mean by algebraically? You have to refer to the definition of absolute value at some point.

7. ## Great stuff!

Originally Posted by Jameson
I spent too much time just now looking up Latex syntax and reviewing logic rules so I'm posting this even though it probably isn't exactly what you want.

$\displaystyle \left \{ \forall b| b\in \mathbb{R} \text{ and } b \ge 3 \right \} \rightarrow \Big( \Big [ (b-3) \ge 0 \Big] \Leftrightarrow |b-3|=b-3 \Big)$

What do you mean by algebraically? You have to refer to the definition of absolute value at some point.
This is an extra question in the study booklet for the state math exam. I thank you for your time and effort.