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Let \( F \subseteq \mathbb{N} \) be the set of all factors of 210. In the poset \((F, \mid)\), find the meet of 30 and 21, denoted as \( 30 \wedge 21 \).

Answer :

Final answer:

In the poset of the factors of 210 using division, the meet of 30 and 21 is the greatest common divisor of the two numbers, which is 3.

Explanation:

The meet of two elements in a poset is the greatest lower bound of the elements. For the poset (F, |), where F is the set of all factors of 210, the meet operation in poset is calculated using the Greatest Common Divisor (GCD) because the order is based on division. In this case, you're asked to calculate the meet of 30 and 21. Hence, you're computing the greatest common divisor of 30 and 21 and you find it using the Euclidean algorithm.

Step 1: Divide the larger number (30) by the smaller number (21) to get a quotient and a remainder.
30 ÷ 21 gives a quotient of 1 and a remainder of 9.

Step 2: Now divide the previous divisor (21) by the remainder from step 1 (9).
21 ÷ 9 gives a quotient of 2 and a remainder of 3.

Step 3: Repeat this process until you get a remainder of 0.
9 ÷ 3 gives a quotient of 3 and a remainder of 0.

When we get 0 as remainder, the divisor at this stage (3) will be the GCD. Therefore, the meet of 30 and 21 under the mentioned poset is 3.

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