Using division with remainder, show that any common multiple of a, b is divisible by lcm(a, b). Just use the division with remainder theorem, no gcd

We are tasked with proving that any common multiple of two integers \( a \) and \( b \) is divisible by their least common multiple \( \text{lcm}(a, b) \), using only the division with remainder theorem, and without relying on the greatest common divisor (gcd). Here's a step-by-step proof:

### Step 1: Division with Remainder Theorem
The division with remainder theorem states that for any two integers \( m \) and \( n \), there exist unique integers \( q \) (the quotient) and \( r \) (the remainder) such that:
\[
m = qn + r
\]
where \( 0 \leq r < n \).

### Step 2: Define the Common Multiple \( m \)
Let \( m \) be any common multiple of \( a \) and \( b \). By definition, this means that \( m \) can be written as:
\[
m = k_1 a = k_2 b
\]
for some integers \( k_1 \) and \( k_2 \). Thus, \( m \) is divisible by both \( a \) and \( b \).

### Step 3: Division of \( m \) by \( \text{lcm}(a, b) \)
We now apply the division with remainder theorem to \( m \) and \( \text{lcm}(a, b) \). By the theorem, there exist integers \( q \) (quotient) and \( r \) (remainder) such that:
\[
m = q \cdot \text{lcm}(a, b) + r
\]
where \( 0 \leq r < \text{lcm}(a, b) \).

### Step 4: Show the Remainder is Zero
Since \( m \) is a common multiple of \( a \) and \( b \), both \( a \) and \( b \) divide \( m \). Therefore, both \( a \) and \( b \) must also divide the remainder \( r \). This is because if \( r \) were not zero, then both \( a \) and \( b \) would have to divide a number smaller than \( \text{lcm}(a, b) \), which contradicts the definition of the least common multiple.

Since \( r \) is a non-negative integer smaller than \( \text{lcm}(a, b) \), and both \( a \) and \( b \) divide \( r \), the only possible value for \( r \) is 0. Therefore:
\[
r = 0
\]

### Step 5: Conclusion
Since \( r = 0 \), we have:
\[
m = q \cdot \text{lcm}(a, b)
\]
which shows that \( m \) is divisible by \( \text{lcm}(a, b) \).

Thus, any common multiple \( m \) of \( a \) and \( b \) is divisible by \( \text{lcm}(a, b) \), completing the proof.

---

Would you like further details or explanations? Here are 5 questions that could expand on this topic:

1. How do you compute the least common multiple of two numbers using prime factorizations?
2. Can you prove the relationship between gcd and lcm in terms of divisibility properties?
3. How does this proof relate to the concept of the greatest common divisor?
4. What role does the remainder play in modular arithmetic, and how does it relate to divisibility?
5. How does the Euclidean algorithm relate to the division with remainder theorem?

**Tip**: The least common multiple of two numbers can also be expressed using their product and gcd: \( \text{lcm}(a, b) = \frac{|a \cdot b|}{\text{gcd}(a, b)} \).

Using division with remainder, show that any common multiple of a, b is divisible by lcm(a, b). Just use the division with remainder theorem, no gcd.

Discover how to prove that any common multiple of two integers a and b is divisible by their least common multiple (LCM) using the division with remainder theorem. This step-by-step proof demonstrates key concepts from number theory and divisibility. Suitable for students in Grades 9-12.

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