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Math Help - Inequality involving modulus

  1. #1
    Ant
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    Inequality involving modulus

    This is a simple step in a proof but I can't see why it works (sorry, I'm no good at dealing with modulus sign...). Here's the step, if anyone could explain it to me that would be great!

    |\frac{a}{b}-c | \leq \epsilon \Rightarrow | \frac{a}{b} | \leq \epsilon + |c|

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  2. #2
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    Re: Inequality involving modulus

    Quote Originally Posted by Ant View Post
    This is a simple step in a proof but I can't see why it works (sorry, I'm no good at dealing with modulus sign...). Here's the step, if anyone could explain it to me that would be great!

    |\frac{a}{b}-c | \leq \epsilon \Rightarrow | \frac{a}{b} | \leq \epsilon + |c|
    Just take note of the fact |x|-|y|\le |x-y| for all x~\&~y in \mathbb{R}.
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  3. #3
    Ant
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    Re: Inequality involving modulus

    Quote Originally Posted by Plato View Post
    Just take note of the fact |x|-|y|\le |x-y| for all x~\&~y in \mathbb{R}.
    So we have:

    |\frac{a}{b}| - |c| \leq |\frac{a}{b}-c| \leq \epsilon \Rightarrow |\frac{a}{b}| - |c| \leq \epsilon \Rightarrow  | \frac{a}{b} | \leq \epsilon + |c|
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    Re: Inequality involving modulus

    Quote Originally Posted by Ant View Post
    So we have:

    |\frac{a}{b}| - |c| \leq |\frac{a}{b}-c| \leq \epsilon \Rightarrow |\frac{a}{b}| - |c| \leq \epsilon \Rightarrow  | \frac{a}{b} | \leq \epsilon + |c|
    You got it! Way to go.
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