Proof by direct proof.
Proposition: if a, b,c is an element of natural numbers, then lcm (ca, cb) = c × lcm (a,b)
Let m= lcm(ca, cb) and n= c×. Lcm(a, b)



### Proposition:
If \( a \), \( b \), and \( c \) are elements of natural numbers, then:
\[
\text{lcm}(ca, cb) = c \times \text{lcm}(a, b)
\]
Where \( \text{lcm}(x, y) \) denotes the least common multiple of \( x \) and \( y \).

We want to prove this using direct proof. Let:
- \( m = \text{lcm}(ca, cb) \)
- \( n = c \times \text{lcm}(a, b) \)

### Proof:

#### Step 1: Express \( m \) and \( n \)
- \( m = \text{lcm}(ca, cb) \), by definition of the least common multiple, means \( m \) is the smallest positive integer such that both \( ca \) and \( cb \) divide \( m \).
- \( n = c \times \text{lcm}(a, b) \), where \( \text{lcm}(a, b) \) is the least common multiple of \( a \) and \( b \), and multiplying it by \( c \) scales it accordingly.

We need to show that \( m = n \), i.e., \( \text{lcm}(ca, cb) = c \times \text{lcm}(a, b) \).

#### Step 2: Factorization of \( a \), \( b \), and \( c \)
To better understand the relationship between \( ca \), \( cb \), and their least common multiple, we can express \( a \), \( b \), and \( c \) in terms of their prime factorizations:
- Let \( a = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k} \)
- Let \( b = p_1^{f_1} p_2^{f_2} \dots p_k^{f_k} \)
- Let \( c = p_1^{g_1} p_2^{g_2} \dots p_k^{g_k} \), where the prime factors are shared or different.

The least common multiple of two numbers takes the highest power of each prime factor from either number. Thus, the least common multiple of \( a \) and \( b \) is:
\[
\text{lcm}(a, b) = p_1^{\max(e_1, f_1)} p_2^{\max(e_2, f_2)} \dots p_k^{\max(e_k, f_k)}
\]

Multiplying both \( a \) and \( b \) by \( c \) gives:
- \( ca = c \times a = p_1^{g_1 + e_1} p_2^{g_2 + e_2} \dots p_k^{g_k + e_k} \)
- \( cb = c \times b = p_1^{g_1 + f_1} p_2^{g_2 + f_2} \dots p_k^{g_k + f_k} \)

#### Step 3: LCM of \( ca \) and \( cb \)
The least common multiple of \( ca \) and \( cb \) is:
\[
\text{lcm}(ca, cb) = p_1^{\max(g_1 + e_1, g_1 + f_1)} p_2^{\max(g_2 + e_2, g_2 + f_2)} \dots p_k^{\max(g_k + e_k, g_k + f_k)}
\]
This simplifies to:
\[
\text{lcm}(ca, cb) = p_1^{g_1 + \max(e_1, f_1)} p_2^{g_2 + \max(e_2, f_2)} \dots p_k^{g_k + \max(e_k, f_k)}
\]
This shows that the least common multiple of \( ca \) and \( cb \) is the same as multiplying \( c \) by the least common multiple of \( a \) and \( b \), which is:
\[
c \times \text{lcm}(a, b) = c \times p_1^{\max(e_1, f_1)} p_2^{\max(e_2, f_2)} \dots p_k^{\max(e_k, f_k)}
\]
Thus, we have:
\[
\text{lcm}(ca, cb) = c \times \text{lcm}(a, b)
\]

### Conclusion:
We have shown that the least common multiple of \( ca \) and \( cb \) is equal to \( c \) times the least common multiple of \( a \) and \( b \). Hence, the proposition is proven:
\[
\boxed{\text{lcm}(ca, cb) = c \times \text{lcm}(a, b)}
\]

Would you like more details or clarification on any step?

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**Related Questions:**
1. What is the definition of the least common multiple (LCM)?
2. How do you compute the LCM of two numbers using prime factorizations?
3. What properties of the LCM are used in direct proofs?
4. Can you extend this proof to three numbers, i.e., \( \text{lcm}(ca, cb, cc) \)?
5. How does the greatest common divisor (GCD) relate to the LCM?

**Tip:** When dealing with LCM, always consider the prime factorizations of the numbers involved, as LCM is based on the highest powers of common prime factors.

Proof by direct proof. Proposition: if a, b, c is an element of natural numbers, then lcm (ca, cb) = c × lcm (a, b). Let m = lcm(ca, cb) and n = c × lcm(a, b).

Explore the direct proof of the proposition that lcm(ca, cb) = c × lcm(a, b) for natural numbers a, b, and c. This proof uses prime factorizations and key number theory concepts. Suitable for Grades 10-12 students studying LCM and prime factorizations.

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