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Given $ a, b, \varepsilon \in \mathbb{R} $ prove that

$$|a-b|<\varepsilon \implies |b| - \varepsilon < |a| < |b|+\varepsilon. $$

Hi, I need help for proof this expression, which could be used arguments or results. I would appreciate any suggestions.

user76838
  • 127

2 Answers2

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What you want to show is $-\varepsilon<|a|-|b|<\varepsilon$, which is equivalent to $||a|-|b||<\varepsilon$. From triangle inequality you know that $|a|\leq |b|+|a-b|$, so $|a|-|b|\leq |a-b|$. Similarly $|b|-|a|\leq |a-b|$. This means $||a|-|b||\leq |a-b|<\varepsilon$, as you want.

pi66
  • 7,164
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By reverse triangle inequality, we have $||a|-|b||\leq |a-b|<\varepsilon$. Hence, $|a|-|b|<\varepsilon$ (i.e. $|a|<|b|+\varepsilon$) and $|b|-|a|<\varepsilon$ (i.e. $|b|-\varepsilon < |a|$).

paf
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