Suppose that n = pe1 1 · · · pek k and m = pf1 1 · · · pfk k , where p1, . . . , pk are distinct primes and ei, fi ≥ 0. Show that gcd(m, n) = pmin{e1,f1}. The answer should be a proof of why this equation works 1 · · · pmin{ek ,fk } k

To prove that for \( n = p_1^{e_1} \cdots p_k^{e_k} \) and \( m = p_1^{f_1} \cdots p_k^{f_k} \), the greatest common divisor (gcd) of \( m \) and \( n \) is given by:
\[
\gcd(m, n) = p_1^{\min(e_1, f_1)} \cdots p_k^{\min(e_k, f_k)},
\]
we proceed with the following steps:

### Step 1: Understanding the Structure of \( n \) and \( m \)

- Let \( n = p_1^{e_1} \cdots p_k^{e_k} \), where \( p_1, \ldots, p_k \) are distinct prime numbers, and \( e_i \geq 0 \) are the powers associated with each prime.
- Let \( m = p_1^{f_1} \cdots p_k^{f_k} \), where the same primes \( p_1, \ldots, p_k \) appear, with possibly different powers \( f_i \geq 0 \).

### Step 2: Definition of gcd

The gcd of two integers is defined as the largest integer that divides both \( m \) and \( n \). Given that \( n \) and \( m \) are expressed as products of powers of distinct primes, the gcd must also be a product of powers of these same primes.

### Step 3: Prime Factor Decomposition of gcd

For the prime factor decomposition of the gcd, consider any prime \( p_i \). The power of \( p_i \) in \( n \) is \( e_i \), and the power of \( p_i \) in \( m \) is \( f_i \). The gcd must have \( p_i \) raised to the minimum of these two powers because:
- \( p_i^{e_i} \) is divisible by \( p_i^k \) for any \( 0 \leq k \leq e_i \), and similarly, \( p_i^{f_i} \) is divisible by \( p_i^k \) for any \( 0 \leq k \leq f_i \).
- The largest power of \( p_i \) that divides both \( p_i^{e_i} \) and \( p_i^{f_i} \) is therefore \( p_i^{\min(e_i, f_i)} \).

### Step 4: Forming the gcd for All Primes

The gcd of \( m \) and \( n \) is formed by taking the product of all primes \( p_i \) raised to the minimum power appearing in \( n \) and \( m \). Therefore, we have:
\[
\gcd(m, n) = p_1^{\min(e_1, f_1)} \cdots p_k^{\min(e_k, f_k)}.
\]
This result is consistent with the fundamental theorem of arithmetic, which states that every integer greater than 1 can be uniquely factorized into a product of prime powers.

### Conclusion

The proof shows that the gcd of two numbers with prime factor decompositions \( n = p_1^{e_1} \cdots p_k^{e_k} \) and \( m = p_1^{f_1} \cdots p_k^{f_k} \) is indeed given by:
\[
\gcd(m, n) = p_1^{\min(e_1, f_1)} \cdots p_k^{\min(e_k, f_k)}.
\]
This formula leverages the properties of divisibility and the uniqueness of prime factorization.

Would you like more details or have any questions?

Here are five related questions to deepen your understanding:
1. How can the concept of gcd be applied to finding the least common multiple (lcm) of two numbers?
2. How does the prime factorization approach help in solving gcd problems with more than two numbers?
3. Can this method be extended to non-integer values, such as fractions? How would it work?
4. What happens to the gcd if all powers \( e_i \) and \( f_i \) are equal for all \( i \)?
5. How does the gcd relate to the Euclidean algorithm for computing gcd?

**Tip:** When finding the gcd of two numbers using prime factorization, always look for the lowest powers for each prime factor involved. This ensures the largest number that divides both values is found.

Suppose that n = p1^e1 · · · pk^ek and m = p1^f1 · · · pk^fk, where p1, . . . , pk are distinct primes and ei, fi ≥ 0. Show that gcd(m, n) = pmin{e1,f1} · · · pmin{ek,fk}.

Explore a detailed proof for finding the greatest common divisor (gcd) of two numbers based on their prime factorization. This solution explains how the gcd of two numbers n and m, with distinct prime factors, is determined by the lowest powers of shared primes. Ideal for students studying number theory at the Grades 11-12 level.

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