he-002

# HumanEval/2

Implement a file **`solution.py`** that completes the function below. Keep the given name and signature; your file must define it at module level.

```python


def truncate_number(number: float) -> float:
    """ Given a positive floating point number, it can be decomposed into
    and integer part (largest integer smaller than given number) and decimals
    (leftover part always smaller than 1).

    Return the decimal part of the number.
    >>> truncate_number(3.5)
    0.5
    """
```

<!-- imported from OpenAI HumanEval (HumanEval/2) -->

# Auto-generated from OpenAI HumanEval HumanEval/2. Do not edit by hand.
from solution import *  # noqa: F401,F403 (prompt helpers may be referenced by tests)
from solution import truncate_number as candidate

METADATA = {
    'author': 'jt',
    'dataset': 'test'
}


def check(candidate):
    assert candidate(3.5) == 0.5
    assert abs(candidate(1.33) - 0.33) < 1e-6
    assert abs(candidate(123.456) - 0.456) < 1e-6


def test_he_002():
    check(candidate)

You are an expert programmer. Solve the task exactly as specified. Output your solution as fenced code blocks using the required file name(s) and the exact function/type signatures requested. Prefer correctness; do not include prose outside code unless asked.

```python
def truncate_number(number: float) -> float:
    """ Given a positive floating point number, it can be decomposed into
    and integer part (largest integer smaller than given number) and decimals
    (leftover part always smaller than 1).

    Return the decimal part of the number.
    >>> truncate_number(3.5)
    0.5
    """
    return number - int(number)
```

The user wants me to implement the `truncate_number` function that returns the decimal part of a positive floating point number.

For example:
- truncate_number(3.5) should return 0.5
- truncate_number(1.234) should return 0.234
- truncate_number(5.0) should return 0.0

The decimal part is simply the number minus its integer part. In Python, I can use:
- `number - int(number)` or
- `number - math.floor(number)` or
- `number % 1`

Since the problem states "positive floating point number", all of these should work. The simplest is `number - int(number)` or `number % 1`.

Let me use `number - int(number)` as it's straightforward.

Actually, `number % 1` is cleaner. Let me use that.

Wait, let me think about edge cases with floating point precision. For example, `3.5 % 1` should give `0.5`. That works.

Let me write the solution:

```python
def truncate_number(number: float) -> float:
    return number - int(number)
```

Or:

```python
def truncate_number(number: float) -> float:
    return number % 1
```

Both should work. I'll go with `number - int(number)` as it's more explicit about what we're doing.