Math Problem Statement

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)

Solution

Proposition:

If aa, bb, and cc are elements of natural numbers, then: lcm(ca,cb)=c×lcm(a,b)\text{lcm}(ca, cb) = c \times \text{lcm}(a, b) Where lcm(x,y)\text{lcm}(x, y) denotes the least common multiple of xx and yy.

We want to prove this using direct proof. Let:

  • m=lcm(ca,cb)m = \text{lcm}(ca, cb)
  • n=c×lcm(a,b)n = c \times \text{lcm}(a, b)

Proof:

Step 1: Express mm and nn

  • m=lcm(ca,cb)m = \text{lcm}(ca, cb), by definition of the least common multiple, means mm is the smallest positive integer such that both caca and cbcb divide mm.
  • n=c×lcm(a,b)n = c \times \text{lcm}(a, b), where lcm(a,b)\text{lcm}(a, b) is the least common multiple of aa and bb, and multiplying it by cc scales it accordingly.

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

Step 2: Factorization of aa, bb, and cc

To better understand the relationship between caca, cbcb, and their least common multiple, we can express aa, bb, and cc in terms of their prime factorizations:

  • Let a=p1e1p2e2pkeka = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}
  • Let b=p1f1p2f2pkfkb = p_1^{f_1} p_2^{f_2} \dots p_k^{f_k}
  • Let c=p1g1p2g2pkgkc = 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 aa and bb is: lcm(a,b)=p1max(e1,f1)p2max(e2,f2)pkmax(ek,fk)\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 aa and bb by cc gives:

  • ca=c×a=p1g1+e1p2g2+e2pkgk+ekca = c \times a = p_1^{g_1 + e_1} p_2^{g_2 + e_2} \dots p_k^{g_k + e_k}
  • cb=c×b=p1g1+f1p2g2+f2pkgk+fkcb = 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 caca and cbcb

The least common multiple of caca and cbcb is: lcm(ca,cb)=p1max(g1+e1,g1+f1)p2max(g2+e2,g2+f2)pkmax(gk+ek,gk+fk)\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: lcm(ca,cb)=p1g1+max(e1,f1)p2g2+max(e2,f2)pkgk+max(ek,fk)\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 caca and cbcb is the same as multiplying cc by the least common multiple of aa and bb, which is: c×lcm(a,b)=c×p1max(e1,f1)p2max(e2,f2)pkmax(ek,fk)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: lcm(ca,cb)=c×lcm(a,b)\text{lcm}(ca, cb) = c \times \text{lcm}(a, b)

Conclusion:

We have shown that the least common multiple of caca and cbcb is equal to cc times the least common multiple of aa and bb. Hence, the proposition is proven: lcm(ca,cb)=c×lcm(a,b)\boxed{\text{lcm}(ca, cb) = c \times \text{lcm}(a, b)}

Would you like more details or clarification on any step?


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., lcm(ca,cb,cc)\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.

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Math Problem Analysis

Mathematical Concepts

Number Theory
Least Common Multiple (LCM)
Prime Factorization

Formulas

lcm(ca, cb) = c × lcm(a, b)
lcm(a, b) = p_1^max(e_1, f_1) × p_2^max(e_2, f_2) × ... × p_k^max(e_k, f_k)

Theorems

Least Common Multiple (LCM) Theorem
Prime Factorization Theorem

Suitable Grade Level

Grades 10-12