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I saw this used in a proof in Rudin's Principles of Mathematical Analysis, but I can't figure out why it is true. Can someone explain? $$\lvert a - b\rvert < \frac{1}{2}\lvert b \rvert \implies \lvert a \rvert > \frac{1}{2}\lvert b \rvert$$

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  • $\begingroup$ What have you tried? Do you know what $|\cdot|$ means? How about writing out the different cases? $\endgroup$ – TMM Oct 2 '16 at 22:35
  • $\begingroup$ Do you mean the cases for $a$ and $b$? As in positive and negative? $\endgroup$ – b_pcakes Oct 2 '16 at 22:36
  • $\begingroup$ What does $|a - b|$ mean? This should give you two distinct cases to consider. Similarly $|b|$ can mean two things, depending on a certain value. That's at most four simple cases to consider to see if it always holds. $\endgroup$ – TMM Oct 2 '16 at 22:38
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Perhaps a picture might help:

enter image description here

$a$ must be strictly less than $\dfrac{|b|}{2}$ from $b$

and so must be more than $\dfrac{|b|}{2}$ from $0$

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Hint. It follows by the identity $$ |b| - |a| ≤ |a-b| $$ which is true for any $a,b∈\Bbb R$.

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  • $\begingroup$ and that follows from the triangle inequality $|x+y| \le |x| +|y|$ letting $a=x$ and $b=x+y$ $\endgroup$ – Henry Oct 2 '16 at 22:40
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Here's a geometric visualization:

Draw the x-axis. Place some $b$ somewhere. Now place some $a$ such that the distance from $a$ to $b$ is less than half $|b|$. You see immediately that $\lvert a \rvert > \frac{1}{2}\lvert b \rvert$.

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