Let a and b be natural numbers and M a common multiple of a and B. This gives us that a*b|gcd(a,b)*M. Now show that (a*b)\gcd(a,b) <= M 

To prove the inequality \(\frac{a \cdot b}{\gcd(a, b)} \leq M\), where \(M\) is a common multiple of \(a\) and \(b\), we can approach this step by step using the properties of the greatest common divisor (gcd) and least common multiple (lcm).

### Key Facts:
- The least common multiple (denoted \(\text{lcm}(a, b)\)) of two natural numbers \(a\) and \(b\) is the smallest number \(M\) that is divisible by both \(a\) and \(b\).
- There is a fundamental relationship between the gcd and lcm of two numbers given by:
  \[
  \text{lcm}(a, b) = \frac{a \cdot b}{\gcd(a, b)}
  \]
  This follows directly from the definitions of gcd and lcm.

### Proof:
Let \(M\) be a common multiple of both \(a\) and \(b\), i.e., \(M\) satisfies:
\[
a \mid M \quad \text{and} \quad b \mid M.
\]

From the relationship between gcd and lcm, we know that the least common multiple of \(a\) and \(b\) is given by:
\[
\text{lcm}(a, b) = \frac{a \cdot b}{\gcd(a, b)}.
\]

Now, since \(M\) is a common multiple of \(a\) and \(b\), and the least common multiple \(\text{lcm}(a, b)\) is the smallest such multiple, it follows that:
\[
\text{lcm}(a, b) \leq M.
\]

Substituting the expression for \(\text{lcm}(a, b)\), we get:
\[
\frac{a \cdot b}{\gcd(a, b)} \leq M.
\]

Thus, the inequality \(\frac{a \cdot b}{\gcd(a, b)} \leq M\) holds, as required.

### Conclusion:
The inequality is proven by using the fundamental relationship between the gcd and lcm of two numbers. Since \(\text{lcm}(a, b)\) is the smallest common multiple of \(a\) and \(b\), any common multiple \(M\) must be at least as large as \(\text{lcm}(a, b)\), which is \(\frac{a \cdot b}{\gcd(a, b)}\).

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Let me know if you'd like more details on any step of this proof or any questions you may have. Here are 5 related questions to deepen your understanding:

1. What is the relationship between gcd and lcm in terms of divisibility?
2. How does the inequality \(\frac{a \cdot b}{\gcd(a, b)} \leq M\) change if \(M = \text{lcm}(a, b)\)?
3. Why is the lcm the smallest common multiple, and how does it relate to divisibility properties?
4. Can you generalize this result for three or more natural numbers?
5. What is the significance of the gcd in reducing the product \(a \cdot b\) to find the lcm?

**Tip:** Understanding the connection between gcd and lcm is crucial in number theory and helps simplify complex arithmetic operations on numbers.

Let a and b be natural numbers and M a common multiple of a and b. This gives us that a*b | gcd(a, b)*M. Now show that (a*b)/gcd(a, b) <= M.

Explore the mathematical proof that shows (a * b) / gcd(a, b) ≤ M, where M is a common multiple of a and b. This solution explains the relationship between the greatest common divisor and least common multiple with detailed steps, suitable for advanced high school or early college students.

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