Check Prime Number in Python

What You’ll Learn

A prime number is an integer greater than 1 whose only positive divisors are 1 and itself. Examples: 2, 3, 5, 7, 11. Non-examples: 1, 4, 9, 15. This tutorial covers trial division up to sqrt(n), a live checker, worked Python examples, edge cases, and complexity.

📝 Problem & Approach

Given an integer n, return whether it is prime.

# 17 -> prime   (no divisor up to sqrt(17))
# 15 -> not     (3 divides 15)
# 2  -> prime   (smallest / only even)
# 1  -> not     (primes start at 2)

Inputs & Outputs

Minimal workflow

function isPrime(n):
    if n < 2:
        return false
    for i from 2 while i * i <= n:
        if n mod i = 0:
            return false
    return true

Method comparison

🔮 Live Preview

Follows the same algorithm as the Python function: handle small values, then test divisors while i * i <= n.

Press “Run check” to classify n.

Examples Gallery

Three complete Python programs — check 17, list primes from 1 to 20, and a faster odd-divisor variant. Click View Output to reveal sample console results.

def is_prime(n: int) -> bool:
    if n <= 1:
        return False
    i = 2
    while i * i <= n:
        if n % i == 0:
            return False
        i += 1
    return True


test_number = 17
if is_prime(test_number):
    print(f"{test_number} is a prime number.")
else:
    print(f"{test_number} is not a prime number.")

17 is a prime number.

def is_prime(num: int) -> bool:
    if num <= 1:
        return False
    i = 2
    while i * i <= num:
        if num % i == 0:
            return False
        i += 1
    return True


start, end = 1, 20
print(f"Prime numbers in the range {start} to {end}:")
for value in range(start, end + 1):
    if is_prime(value):
        print(value, end=" ")
print()

Prime numbers in the range 1 to 20:
2 3 5 7 11 13 17 19

def is_prime(n: int) -> bool:
    if n <= 1:
        return False
    if n == 2:
        return True
    if n % 2 == 0:
        return False
    i = 3
    while i * i <= n:
        if n % i == 0:
            return False
        i += 2
    return True


for value in (2, 17, 18, 1, 19):
    label = "prime" if is_prime(value) else "not prime"
    print(f"{value}: {label}")

2: prime
17: prime
18: not prime
1: not prime
19: prime

🔎 Worked Walkthrough — 17 vs 15

Why the sqrt bound catches composites early and lets primes through.

If n = a * b with a, b > 1, at least one factor is <= sqrt(n) — so the loop is enough.

⚖️ Facts Worth Knowing

Handy follow-ups interviewers sometimes ask.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

For interview demos, quote O(sqrt(n)) for a single check and mention sieves for bulk listing.

💡 Best Practices