Prove the following statement.
For all nonzero integers a and b, 
gcd(a, b) | lcm(a, b).
Proof:
Let a and b be any integers. Then 
gcd(a, b) | a
 by 
---Select---
 , and 
a | lcm(a, b)
 by 
---Select---
 . Hence by transitivity of divisibility, 
gcd(a, b) | lcm(a, b).


Let's fill in the missing parts of the proof:

Let \( a \) and \( b \) be any nonzero integers. Then:

1. \( \gcd(a, b) \mid a \) by **definition of greatest common divisor**, since the greatest common divisor divides both \( a \) and \( b \).
2. \( a \mid \operatorname{lcm}(a, b) \) by **definition of least common multiple**, since the least common multiple is a multiple of both \( a \) and \( b \).

Hence, by the transitivity of divisibility, since \( \gcd(a, b) \mid a \) and \( a \mid \operatorname{lcm}(a, b) \), we have:

\[ \gcd(a, b) \mid \operatorname{lcm}(a, b). \]

This completes the proof.

Would you like more details on any of these concepts or any part of the proof?

### Related Questions:
1. What is the relationship between gcd and lcm for two numbers?
2. How does the Euclidean algorithm help in finding gcd?
3. Can the gcd of two numbers be larger than either of the numbers?
4. What happens to the gcd if two numbers are coprime?
5. How can you express lcm in terms of gcd?

**Tip**: The product of \( \gcd(a, b) \) and \( \operatorname{lcm}(a, b) \) is always equal to the product of \( a \) and \( b \).

Prove the following statement. For all nonzero integers a and b, gcd(a, b) | lcm(a, b). Proof: Let a and b be any integers. Then gcd(a, b) | a by ---Select--- , and a | lcm(a, b) by ---Select--- . Hence by transitivity of divisibility, gcd(a, b) | lcm(a, b).

Learn how to prove that for all nonzero integers a and b, gcd(a, b) divides lcm(a, b). This problem involves concepts from number theory, such as the greatest common divisor, least common multiple, and the transitivity of divisibility. Suitable for high school students in Grades 9-12.

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