# Inequality involving modulus

• April 4th 2012, 02:31 PM
Ant
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!

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

Thanks
• April 4th 2012, 02:57 PM
Plato
Re: Inequality involving modulus
Quote:

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 $\mathbb{R}$.
• April 4th 2012, 03:04 PM
Ant
Re: Inequality involving modulus
Quote:

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:

$|\frac{a}{b}| - |c| \leq |\frac{a}{b}-c| \leq \epsilon \Rightarrow |\frac{a}{b}| - |c| \leq \epsilon \Rightarrow | \frac{a}{b} | \leq \epsilon + |c|$
• April 4th 2012, 03:32 PM
Plato
Re: Inequality involving modulus
Quote:

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.