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This question already has an answer here:

I'm trying to prove that $||a| - |b|| \leq |a - b|$. So far, by using the triangle inequality, I've got: $$|a| = |\left(a - b\right) + b| \leq |a - b| + |b|$$ Subtracting $|b|$ from both sides yields, $$|a| - |b| \leq |a - b|$$ The book I'm working from claims you can achieve this proof by considering just two cases: $|a| - |b| \geq 0$ and $|a| - |b| < 0$. The first case is pretty straightforward: $$|a| - |b| \geq 0 \implies ||a| - |b|| = |a| - |b| \leq |a - b|$$ But I'm stuck on the case where $|a| - |b| < 0$

Cool, I think I got it (thanks for the hints!). So, $$|b| - |a| \leq |b - a| = |a - b|$$ And when $|a| - |b| < 0$, $$||a| - |b|| = -\left(|a| - |b|\right) = |b| - |a| \leq |a - b|$$

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marked as duplicate by Namaste, Jim, P.., Alexander Gruber, Stefan Hansen Feb 23 '13 at 8:51

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You had it

$$|a| - |b| \leq |a - b|$$

Similarly

$$|b| - |a| \leq |b - a|$$

Using the fact $|b-a| = |a-b|$ and multiplying by $-1$,

$$-|a - b| \leq |a| - |b|$$

Combining these, we get

$$-|a-b| \leq |a| - |b| \leq |a - b|$$

and so finally we see

$$ ||a|-|b|| \leq |a-b| $$

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  • $\begingroup$ That $|a|=|-a|$ is really key, but then, I guess that's whole point of absolute value, eh? Thanks for the help! $\endgroup$ – ivan Feb 23 '13 at 5:58
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Hint: If $|a|-|b|<0$, rename $a$ to $b'$ and $b$ to $a'$.

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