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My question is from Apostol's Vol. 1 One-variable calculus with introduction to linear algebra textbook.

Page 43. Problem 1 Prove each of the following properties of absolute values.

(c) $|x-y|=|y-x|$.

The attempt at a solution: I solved similar problem, which was this: $|x|-|y|\le|x-y|$, by manipulating triangle inequality, I guess this one might be similar but I don't see it. Please help.

So far I have proven following properties:

$|x|=0$ if and only if $x=0$.

$|-x|=|x|$.

Also, absolute value is defined in such way: If $x$ is a real number, the absolute value of $x$ is a nonnegative real number denoted by $|x|$ and defined as follows: $|x|=\begin{cases} x, & \text{if $x\ge0$,} \\ -x, & \text{if $x\le0$.} \end{cases}$.

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    $\begingroup$ If you've shown that the absolute value is multiplicative, then you could say $|x-y|=|(-1)(y-x)|=|-1||y-x|=|y-x$. $\endgroup$ – Hayden Oct 26 '14 at 13:31
  • $\begingroup$ No, I have not done that yet, that's (f) part of problem 1. $\endgroup$ – George Apriashvili Oct 26 '14 at 13:33
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    $\begingroup$ it might be helpful to include the properties you have proven, including, for example, how you're defining the absolute value (i.e. as either the square root of the square, or as a piecewise function, although these are clearly equivalent) $\endgroup$ – Hayden Oct 26 '14 at 13:34
  • $\begingroup$ @Hayden Yes, Sorry for not being clear, I listed all that now. $\endgroup$ – George Apriashvili Oct 26 '14 at 13:42
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$$|x-y|=|x-y|$$ $$|x-y|=|1|\cdot|x-y|$$ $$|x-y|=|-1|\cdot|x-y|$$ $$|x-y|=|-1\cdot(x-y)|$$ $$|x-y|=|y-x|$$


Without $|x||y|=|xy|$

If $x>y$
Since $y-x<0$ that means $|y-x|=-(y-x)=x-y$ $$|y-x|=x-y$$ Since $x-y>0$ that means $|x-y|=x-y$ $$|x-y|=x-y$$ Equality is transitive $$|x-y|=|y-x|$$

If $y>x$
Since $x-y<0$ that means $|x-y|=-(x-y)=y-x$ $$|x-y|=y-x$$ Since $y-x>0$ that means $|y-x|=y-x$ $$|y-x|=y-x$$ Equality is transitive $$|x-y|=|y-x|$$

The case of $x=y$ is left as an exercise for the reader.

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  • $\begingroup$ I have not proven property $|xy|=|x||y|$ yet, so is there any other way to achieve the result? $\endgroup$ – George Apriashvili Oct 26 '14 at 13:36
  • $\begingroup$ @GeorgeDirac Look at my edit $\endgroup$ – Alice Ryhl Oct 26 '14 at 13:45
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You say that you have proven that $|x|=|-x|$, then it immediately follows that

$$|x-y| = |-(x-y)| =|-x+y| = |y-x|. $$

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  • $\begingroup$ Yeah, thats true. Well I feel stupid now, that I didn't think of that, thanks for simple explanation. $\endgroup$ – George Apriashvili Oct 26 '14 at 13:54
  • $\begingroup$ @GeorgeDirac You're welcome :-) $\endgroup$ – Eff Oct 26 '14 at 13:56

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