How to simplify quotient of absolute values?

[DEmaden]

I am wondering how to simplify a quotient of absolute values, because this comes up when working with logarithms. For example, is |x-3| / |x-1| the same as |(x-3)/(x-1)| ?

 

[tkhunny]

Consider a number line:
<------------(1)--------------(3)-------------->
Think more about parity, rather than value.

Maybe it comes up when working with logarithms. What is the actual problem?

 

[JeffM]

There are four cases.

[MATH]\text {CASE I: }x < 0 \text { and } y < 0 \implies |\ x \ | = -\ x \text { and } |\ y \ | = -\ y.[/MATH]
[MATH]x < 0 \text { and } y < 0 \implies \dfrac{x}{y} > 0 \implies \left | \ \dfrac{x}{y} \ \right | = \dfrac{x}{y}.[/MATH]
[MATH]\dfrac{x}{y} = \dfrac{x}{y} * 1 = \dfrac{x}{y} * \dfrac{-\ 1}{-\ 1} = \dfrac{-\ x}{-\ y} = \dfrac{|\ x \ |}{|\ y \ |}.[/MATH]
[MATH]\therefore \left | \ \dfrac{x}{y} \ \right | = \dfrac{|\ x \ |}{|\ y \ |}.[/MATH]
I'll let you figure out the other three cases.

 

[DEmaden]

There are four cases.

[MATH]\text {CASE I: }x < 0 \text { and } y < 0 \implies |\ x \ | = -\ x \text { and } |\ y \ | = -\ y.[/MATH]
[MATH]x < 0 \text { and } y < 0 \implies \dfrac{x}{y} > 0 \implies \left | \ \dfrac{x}{y} \ \right | = \dfrac{x}{y}.[/MATH]
[MATH]\dfrac{x}{y} = \dfrac{x}{y} * 1 = \dfrac{x}{y} * \dfrac{-\ 1}{-\ 1} = \dfrac{-\ x}{-\ y} = \dfrac{|\ x \ |}{|\ y \ |}.[/MATH]
[MATH]\therefore \left | \ \dfrac{x}{y} \ \right | = \dfrac{|\ x \ |}{|\ y \ |}.[/MATH]
I'll let you figure out the other four cases.

Thanks, that helped a lot.

 

[Steven G]

For some value of x (not x=1), |x-3| will either be (x-3) or -(x-3). Same for |x-1|, either it is (x-1) or -(x-1). In the end |x-3|/|x-1| will NOT be negative.

Let's consider |(x-3)/(x-1)|. Either this will be (x-3)/(x-1) or -(x-3)/(x-1). In the end |(x-3)/(x-1)| will NOT be negative.

If you think about it the most that the |x-3|/|x-1| and |(x-3)/(x-1)| can possible differ by will be a negative sign. However if the they are to both be non-negative then they can't even differ by a sign. Hence |x-3|/|x-1| = |(x-3)/(x-1)|

 

[DEmaden]

Problem solved, you may close the thread.