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Let $f(x)$ and $g(x)$ be two functions.

Does the following property hold true? $$|f(x) - g(x)| = |g(x) - f(x)|$$

On the surface, it would seem like it, but then some other properties of absolute values don't seem to hold.

$$|f(x) - g(x)| \le a$$ $$-a \le f(x) - g(x) \le a$$

But this doesn't seem to be equivalent to the following $$|g(x) - f(x) | \le a$$ $$-a \le g(x) - f(x) \le a$$

Why doesn't the property $|f(x) - g(x)| = |g(x) - f(x)|$ hold true?

Christopher Crutchfield
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    Why do you think $-a\leq f(x)-g(x)\leq a$ isn't equivalent to $-a\leq g(x)-f(x)\leq a$? "Seem" isn't good enough ;) – symplectomorphic May 19 '20 at 05:02
  • I guess that's a good point. I just tried to find some values of $f(x)$, $g(x)$, and $a$ that would provide a counter-example to the above property being true and haven't been able to find any. I know that's far from proof that none exists. Does there exist a proof that shows that the above holds true? – Christopher Crutchfield May 19 '20 at 05:08
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    Yes, there's a proof... can you think of anything you might do to one inequality to turn it into the other? – symplectomorphic May 19 '20 at 05:09
  • @symplectomorphic Thank you, that question helps me a lot. – Christopher Crutchfield May 19 '20 at 05:19

1 Answers1

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Does the following property hold true?

Yes. $$|a-b|=|-(b-a)|=|-1||b-a|=|b-a|,$$ for any $a,b\in V$, where $V$ is normed vector space, and $|\cdot|$ is norm (which is absolute value if $V=\mathbb{R}$) on $V$.

Jingeon An-Lacroix
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