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}$

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- Feb 2nd 2010, 08:43 PMvuze88Continuity 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 PMProve It
- Feb 2nd 2010, 09:16 PMvuze88
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 AMHallsofIvy
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.