Is the function `f(x)={{:("{x}",xge0),("{-x}",xlt0):}` (where {.} denotes the fractional part of x is even?

Answer

To determine if the function \( f(x) \) is even, we need to analyze the function defined as:

\[ f(x) = \begin{cases} \{x\} & \text{if } x \geq 0 \\ \{-x\} & \text{if } x < 0 \end{cases} \]

where \( \{x\} \) denotes the fractional part of \( x \).

### Step 1: Understand the Definition of Even Function
A function \( f(x) \) is considered even if it satisfies the condition:
\[ f(x) = f(-x) \quad \text{for all } x \]
This means the function is symmetric about the y-axis.

### Step 2: Evaluate \( f(x) \) for Positive \( x \)
For \( x \geq 0 \):
\[ f(x) = \{x\} \]
The fractional part \( \{x\} \) is simply \( x - \lfloor x \rfloor \), which is non-negative and less than 1.

### Step 3: Evaluate \( f(-x) \) for Positive \( x \)
Now, consider \( -x \) where \( x \geq 0 \) (thus \( -x < 0 \)):
\[ f(-x) = \{-(-x)\} = \{x\} \]
This means:
\[ f(-x) = \{x\} \]

### Step 4: Compare \( f(x) \) and \( f(-x) \)
From the evaluations:
\[ f(x) = \{x\} \quad \text{and} \quad f(-x) = \{x\} \]
Thus, for \( x \geq 0 \):
\[ f(x) = f(-x) \]

### Step 5: Evaluate \( f(x) \) for Negative \( x \)
Now consider \( x < 0 \):
\[ f(x) = \{-x\} \]
For \( x < 0 \), \( -x \) is positive, so:
\[ f(-x) = \{(-x)\} = \{-x\} \]

### Step 6: Compare \( f(x) \) and \( f(-x) \) for Negative \( x \)
From the evaluations:
\[ f(x) = \{-x\} \quad \text{and} \quad f(-x) = \{-x\} \]
Thus, for \( x < 0 \):
\[ f(x) = f(-x) \]

### Conclusion
Since \( f(x) = f(-x) \) holds true for both cases (when \( x \geq 0 \) and \( x < 0 \)), we conclude that the function \( f(x) \) is even.

### Final Answer
Yes, the function \( f(x) \) is even.

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