Maximal element on first ten prime numbers on divisibility order

The first ten prime numbers are:

\[ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 \]

When ordered by divisibility, one number \( a \) is said to divide another \( b \) if \( a \mid b \), meaning that \( b \) is divisible by \( a \). In the case of prime numbers, the only divisors of a prime number are 1 and itself. Thus, no prime number divides another, except for the number itself.

Therefore, in the divisibility order, **each prime number is incomparable** with the others. Since there is no proper relation where one prime divides another (other than itself), each prime is a maximal element in this set. 

Hence, **all of the first ten primes**—2, 3, 5, 7, 11, 13, 17, 19, 23, and 29—are maximal elements in the divisibility order.

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Would you like more details or have any questions? Here are five related questions:

1. What are minimal elements in divisibility order for non-prime numbers?
2. How do prime factorizations relate to divisibility order?
3. Can composite numbers have maximal elements in divisibility order?
4. What is a Hasse diagram and how can it represent divisibility?
5. How does divisibility relate to the concept of greatest common divisors?

**Tip:** When working with prime numbers, remember they have no divisors other than 1 and themselves, making them crucial in divisibility hierarchies.

Discover how the first ten prime numbers—2, 3, 5, 7, 11, 13, 17, 19, 23, and 29—are all maximal elements in divisibility order. This exploration in number theory shows that no prime number divides another except itself, making each prime number incomparable in the divisibility hierarchy. Suitable for Grades 9-12.

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