Leta,b∈N.  Consider the setS={x∈Nsuch thatadividesxandbdividesx}1.  Show that the setShas a minimal elementm.2.  Show thata/gcd(a,b) andb/gcd(a,b) are relatively prime integers.3.  Using the previous part, show thatm=abgcd(a,b 

prove only question 3 but rely on 1 and 2

### Question 3: Prove that the minimal element \( m = \frac{ab}{\gcd(a,b)} \)

**Given:**  
- \( a, b \in \mathbb{N} \).
- \( S = \{ x \in \mathbb{N} : a \text{ divides } x \text{ and } b \text{ divides } x \} \).
- \( m \) is the minimal element in the set \( S \), as shown in part 1.
- \( a/\gcd(a,b) \) and \( b/\gcd(a,b) \) are relatively prime integers, as shown in part 2.

We aim to prove that the minimal element \( m \) in the set \( S \) is given by:
\[
m = \frac{ab}{\gcd(a,b)}.
\]

### Step-by-step Proof:

#### Step 1: General form of elements in \( S \)
By definition, the set \( S \) contains all natural numbers divisible by both \( a \) and \( b \). Therefore, any element \( x \in S \) must satisfy:
\[
a \mid x \quad \text{and} \quad b \mid x.
\]
This means \( x \) must be a common multiple of \( a \) and \( b \). Specifically, the smallest element of \( S \) must be the **least common multiple (LCM)** of \( a \) and \( b \), since the least common multiple is the smallest number divisible by both.

#### Step 2: Relationship between LCM and GCD
We know from number theory the relationship between the least common multiple (LCM) and the greatest common divisor (GCD) of two numbers \( a \) and \( b \):
\[
\text{lcm}(a,b) = \frac{ab}{\gcd(a,b)}.
\]
This formula tells us that the least common multiple of \( a \) and \( b \) is precisely \( \frac{ab}{\gcd(a,b)} \).

#### Step 3: Minimal element \( m \)
From part 1, we know that the set \( S \) has a minimal element, \( m \). This minimal element must be the least common multiple of \( a \) and \( b \), as it is the smallest number divisible by both. Hence, we conclude that:
\[
m = \text{lcm}(a,b).
\]

#### Step 4: Conclusion
Using the result from Step 2, the least common multiple of \( a \) and \( b \) is \( \frac{ab}{\gcd(a,b)} \). Therefore, the minimal element \( m \) is given by:
\[
m = \frac{ab}{\gcd(a,b)}.
\]
This completes the proof.

### Summary:
We proved that the minimal element \( m \) in the set \( S \), which contains all numbers divisible by both \( a \) and \( b \), is equal to \( \frac{ab}{\gcd(a,b)} \). This result follows from the relationship between the LCM and GCD of two numbers.

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Would you like more details or have any questions? 

Here are five related questions to expand on this:

1. How can the GCD and LCM of two numbers be computed efficiently?
2. Why are \( \frac{a}{\gcd(a,b)} \) and \( \frac{b}{\gcd(a,b)} \) relatively prime?
3. How does this proof relate to the Euclidean algorithm for finding GCD?
4. Can this formula be generalized to more than two numbers?
5. What is the significance of the relationship between LCM and GCD in number theory?

**Tip:** Whenever solving problems involving divisibility, make sure to explore the connections between LCM and GCD—they often provide key insights!

Leta, b ∈ N. Consider the set S = {x ∈ N such that a divides x and b divides x}. 1. Show that the set S has a minimal element m. 2. Show that a/gcd(a, b) and b/gcd(a, b) are relatively prime integers. 3. Using the previous part, show that m = ab/gcd(a, b). Prove only question 3 but rely on 1 and 2.

Explore a proof that the minimal element m = ab/gcd(a,b) in a set of natural numbers divisible by both a and b. This detailed solution highlights key concepts in number theory, including the relationship between GCD and LCM. Suitable for students in Grades 10-12.

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