# Continuity Question

• Feb 2nd 2010, 08:43 PM
vuze88
Continuity Question
Discuss the continuity of the function defined by:

$\displaystyle f(x)=\begin{cases} & x\text{ if } x\,\in\,Q \\ & -x\text{ if } x\,\notin \,Q \end{cases}$
• Feb 2nd 2010, 09:08 PM
Prove It
Quote:

Originally Posted by vuze88
Discuss the continuity of the function defined by:

$\displaystyle f(x)=\begin{cases} & x\text{ if } x\,\in\,Q \\ & -x\text{ if } x\,\notin \,Q \end{cases}$

Every rational number lies between two irrational numbers, and vice versa.

So the function is discontinuous everywhere...
• Feb 2nd 2010, 09:16 PM
vuze88
Can you explain why every rational number is between two irrational numbers? Also, according to the textbook, it is discontinuous everywhere but is continuous at 0.
• Feb 3rd 2010, 04:58 AM
HallsofIvy
Yes, the function is continuous at 0. Given any $\displaystyle \epsilon$> 0, [tex]|f(x)- f(0)| is |x- 0|= |x| if x is rational, and |-x- 0|= |x| if x is irrational. Take $\displaystyle \delta$ to be any number less than $\displaystyle \epsilon$.

If a is any rational number other than 0, f(x)= a so |f(x)- f(a)= |x-a| if x is rational and |-x-a|= |x+a| if x is irrational. No matter how small $\displaystyle \delta$ is, there will exist both rational and irrational numbers between $\displaystyle a-\delta$ and $\displaystyle a+ \delta$ so with $\displaystyle \epsilon$ less than half the difference between them, there will exist x such that |f(x)- f(a)|> [tex]\epsilon[/b]. The same is true if a is irrational.