Inequality involving modulus

**Ant** · April 4th 2012, 03:31 PM

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!

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

Thanks

**Plato** · April 4th 2012, 03:57 PM

Originally Posted by Ant

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 $\mathbb{R}$ .

**Ant** · April 4th 2012, 04:04 PM

Originally Posted by Plato

Just take note of the fact $|x|-|y|\le |x-y|$ for all $x~\&~y$ in $\mathbb{R}$ .

So we have:

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

**Plato** · April 4th 2012, 04:32 PM

Originally Posted by Ant

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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