So I've been given a set $V = \{x \in \mathbb{R}|x\geq0\}$ with the following addition and scalar multiplication operations for the set (scalar in this case referring to $c \in \mathbb{R})$:
$x \oplus y = |x-y|$
$c \odot x = |c|x$
I need to determine if the set's addition is associative and if the set's scalar multiplication distributes over the addition.
I have already determined that the defined addition is not associative since if it was associative then: $x \oplus (y \oplus z) = |x – |y-z|| = ||x-y|-z| = (x \oplus y) \oplus z$ but for $x=1$, $y=2$ and $z=3$ then $x,y,z \in \mathbb{R}$:
$|1 – |2-3| = |1-|-1||=|1-1|=|0|=0$
$||1-2|-3|=||-1|-3|=|1-3|=|-2|=2$
$0 \neq 2$
However, I am getting stuck on the distributivity part. I have tried many combinations of real number and positive real number only to get the same answer from both sides of the equation
$c \odot (x \oplus y) = |c||x-y|=||c|x-|c|y|=(c\odot x) \oplus (c\odot y)$
Moreover, I tried graphing both $|c||x-y|$ and $||c|x-|c|y|$ and they appear to be the same graph by setting c and y and they appear to be the same graph. However, I'm pretty sure that I can't just distribute $|c|$ over $|x-y|$ and call it a day due to the absolute value functions. Any help would be appreciated!
Best Answer
The property holds.
We have $|ab|=|a||b|$.
Hence,
$$|c(x-y)|=|c||x-y|$$