Absolute value question

Question

Is it true that:$$\left|\,a-b+c-c\,\right|=\left|\,(c-a)+(c-b)\,\right|,$$ or, alternatively, $$\left|\,a-b+c-c\,\right| = \left|\,(a-c)+(c-b)\, \right|?$$

Why is this the case?

Answer 1

It's false. Example: $a = b = 1; c = 2$> left hand side is |a-b+c-c| = 0; right hand side is $|(2-1) + (2-1)| = |2| = 2$.

The second claim is correct, because $a - b + c - c = (a-c) + (b-c)$ by commutative and distributive properties, and since they are equal, their absolute values are equal as well.

Answer 2

Since we have

a - b + c - c = a - b

We get a counterexample for the first equation with

a = b , $a\ne c$

The second equation is trivially right.

Answer 3

$|a-b+c-c| \neq |(c-a)+(c-b)|$. As a counterexample, let $a=0, b=10, c=5$.

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$|a-b+c-c| = |a+(-b)+c+(-c)| = |(a+(-c))+(c+(-b))| = |(a-c)+(c-b)|$