Math Problem Statement

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

Solution

We are tasked with proving that any common multiple of two integers aa and bb is divisible by their least common multiple lcm(a,b)\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 mm and nn, there exist unique integers qq (the quotient) and rr (the remainder) such that: m=qn+rm = qn + r where 0r<n0 \leq r < n.

Step 2: Define the Common Multiple mm

Let mm be any common multiple of aa and bb. By definition, this means that mm can be written as: m=k1a=k2bm = k_1 a = k_2 b for some integers k1k_1 and k2k_2. Thus, mm is divisible by both aa and bb.

Step 3: Division of mm by lcm(a,b)\text{lcm}(a, b)

We now apply the division with remainder theorem to mm and lcm(a,b)\text{lcm}(a, b). By the theorem, there exist integers qq (quotient) and rr (remainder) such that: m=qlcm(a,b)+rm = q \cdot \text{lcm}(a, b) + r where 0r<lcm(a,b)0 \leq r < \text{lcm}(a, b).

Step 4: Show the Remainder is Zero

Since mm is a common multiple of aa and bb, both aa and bb divide mm. Therefore, both aa and bb must also divide the remainder rr. This is because if rr were not zero, then both aa and bb would have to divide a number smaller than lcm(a,b)\text{lcm}(a, b), which contradicts the definition of the least common multiple.

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

Step 5: Conclusion

Since r=0r = 0, we have: m=qlcm(a,b)m = q \cdot \text{lcm}(a, b) which shows that mm is divisible by lcm(a,b)\text{lcm}(a, b).

Thus, any common multiple mm of aa and bb is divisible by lcm(a,b)\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: lcm(a,b)=abgcd(a,b)\text{lcm}(a, b) = \frac{|a \cdot b|}{\text{gcd}(a, b)}.

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Math Problem Analysis

Mathematical Concepts

Number Theory
Divisibility
Common Multiples
Least Common Multiple (LCM)
Division with Remainder

Formulas

Division with Remainder: m = qn + r, where 0 <= r < n
LCM expression: m = q * lcm(a, b) + r

Theorems

Division with Remainder Theorem

Suitable Grade Level

Grades 9-12