# Thread: Determine points of continuity

1. ## Determine points of continuity

Define $\displaystyle f:\mathbb{R}\to\mathbb{R}$ as follows:
$\displaystyle f(x)=\begin{cases}x-\left\lfloor x\right\rfloor, & \text{if} \ \left\lfloor x\right\rfloor \text{is even}\\x-\left\lfloor x+1\right\rfloor, & \text{if} \ \left\lfloor x+1\right\rfloor \ \text{is odd}\end{cases}$
Determine the points where f(x) is continuous.

I graphed the function and obtain $\displaystyle y=x, \ [0,1) \ \text{and} \ y=x-1, \ [0,1)$.
Nothing in the interval (1,2) and the same lines shifted to the right 2 units and this repeats in both directions.
Therefore, the function is discontinuous everywhere. How can I prove this?

2. ## Re: Determine points of continuity

Originally Posted by dwsmith
Define $\displaystyle f:\mathbb{R}\to\mathbb{R}$ as follows:
$\displaystyle f(x)=\begin{cases}x-\left\lfloor x\right\rfloor, & \text{if} \ \left\lfloor x\right\rfloor \text{is even}\\x-\left\lfloor x+1\right\rfloor, & \text{if} \ \left\lfloor x+1\right\rfloor \ \text{is odd}\end{cases}$
Determine the points where f(x) is continuous.
Note that $\displaystyle \left\lfloor 2 \right\rfloor\text{ is even and }\left\lfloor 2+1 \right\rfloor\text{ is odd}$
Thus $\displaystyle f(2)$ has two values.

So this function is ill-defined.

Note that $\displaystyle \left\lfloor 2 \right\rfloor\text{ is even and }\left\lfloor 2+1 \right\rfloor\text{ is odd}$
Thus $\displaystyle f(2)$ has two values.