Thread: 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!

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

Thanks

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!

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

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

Originally Posted by Plato

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

So we have:

$\displaystyle |\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| $

Originally Posted by Ant

So we have:

$\displaystyle |\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.