The function: $f(x)=\begin{cases} 1 & ext{x is irrational} \ -1 & ext{x is rational} \end{cases}$ is

Q: The function: $f(x)=\begin{cases} 1 & \text{x is irrational} \\ -1 & \text{x is rational} \end{cases}$ is

The function: $f(x)=\begin{cases} 1 & \text{x is irrational} \\ -1 & \text{x is rational} \end{cases}$ is

Question

The function: $f(x)=\begin{cases} 1 & 	ext{x is irrational} \ -1 & 	ext{x is rational} \end{cases}$ is

Accepted Answer

The function $ f(x) = \begin{cases} 1 & \text{x is irrational} \ -1 & \text{x is rational} \end{cases} $ is a classic example of a function that is discontinuous everywhere on the real number line. Let's explore why Option 4 is the correct answer and why the other options are incorrect.

### Explanation for Option 4: Discontinuous for every real $ x $

1. **Definition of Continuity**: A function $ f(x) $ is continuous at a point $ x = c $ if the following three conditions are satisfied:
   - $ f(c) $ is defined.
   - The limit of $ f(x) $ as $ x $ approaches $ c $ exists.
   - The limit of $ f(x) $ as $ x $ approaches $ c $ is equal to $ f(c) $.

2. **Behavior of $ f(x) $**:
   - For any real number $ x $, if you consider a sequence of rational numbers converging to $ x $, $ f(x) $ will take the value $-1$ for all terms of this sequence.
   - Similarly, if you consider a sequence of irrational numbers converging to $ x $, $ f(x) $ will take the value $ 1 $ for all terms of this sequence.

3. **Discontinuity at Every Point**:
   - At any real number $ x $, you can find both rational and irrational numbers arbitrarily close to $ x $.
   - Therefore, the limit of $ f(x) $ as $ x $ approaches any real number $ c $ does not exist because the values of $ f(x) $ oscillate between $-1$ and $ 1$ depending on whether the approaching sequence is rational or irrational.
   - Since the limit does not exist, the function cannot be continuous at any point on the real number line.

### Why Other Options are Incorrect:

1. **Option 1: Discontinuous at $ x=0 $ only**:
   - This option suggests that the function is continuous everywhere except at $ x=0 $. However, as explained, the function is discontinuous at every real number, not just at $ x=0 $.

2. **Option 2: Discontinuous at rationals only**:
   - This option implies that the function is continuous at irrational numbers. However, due to the oscillating nature of the function values between rational and irrational sequences, the function is discontinuous at both rational and irrational numbers.

3. **Option 3: Discontinuous at irrationals only**:
   - This option suggests that the function is continuous at rational numbers. Similar to the reasoning for Option 2, the function is discontinuous at both rational and irrational numbers due to the same oscillating behavior.

In conclusion, the function $ f(x) $ is discontinuous for every real $ x $ because the limit does not exist at any point due to the oscillation between $-1$ and $ 1$ for sequences of rational and irrational numbers. Thus, Option 4 is the correct answer.