Python Program to Check if Number is Positive or Negative

You can check if a number is positive or negative in Python using an if statement like this: if number > 0: print('Positive') elif number < 0: print('Negative') else: print('Zero').

Examples

Input5

OutputPositive

Input-3

OutputNegative

Input0

OutputZero

How to Think About It

To decide if a number is positive or negative, first think about what positive means: greater than zero. Negative means less than zero. If the number is exactly zero, it is neither positive nor negative. So, check if the number is greater than zero, then less than zero, and if neither, it must be zero.

Algorithm

1

Get the number input from the user

2

Check if the number is greater than zero

3

If yes, print 'Positive'

4

Else, check if the number is less than zero

5

If yes, print 'Negative'

6

Otherwise, print 'Zero'

Code

python

number = float(input('Enter a number: '))
if number > 0:
    print('Positive')
elif number < 0:
    print('Negative')
else:
    print('Zero')

Output

Enter a number: 5
Positive

Dry Run

Let's trace the input 5 through the code

1

Input

User enters 5, so number = 5.0

2

Check if number > 0

5.0 > 0 is True

3

Print result

Print 'Positive'

Step	Condition	Result	Output
1	number > 0	True	Positive

Why This Works

Step 1: Check if number is greater than zero

The code uses if number > 0 to find if the number is positive.

Step 2: Check if number is less than zero

If the first check fails, elif number < 0 checks if the number is negative.

Step 3: Handle zero case

If neither condition is true, else runs, meaning the number is zero.

Alternative Approaches

Using ternary conditional operator

python

number = float(input('Enter a number: '))
print('Positive' if number > 0 else 'Negative' if number < 0 else 'Zero')

This method is shorter but can be harder to read for beginners.

Using a function

python

def check_sign(num):
    if num > 0:
        return 'Positive'
    elif num < 0:
        return 'Negative'
    else:
        return 'Zero'

number = float(input('Enter a number: '))
print(check_sign(number))

This approach organizes code better and is reusable.

Complexity: O(1) time, O(1) space

Time Complexity

The program performs a fixed number of comparisons regardless of input size, so it runs in constant time.

Space Complexity

It uses a fixed amount of memory for variables and no extra data structures, so space is constant.

Which Approach is Fastest?

All approaches run in constant time and space; differences are mainly in readability and reusability.

Approach	Time	Space	Best For
If-elif-else	O(1)	O(1)	Clarity and beginners
Ternary operator	O(1)	O(1)	Short code, experienced users
Function-based	O(1)	O(1)	Code reuse and organization