GCD times LCM equals product
Dependencies:
Dependencies:
- gcd(a1/d, a2/d, ..., an/d) = gcd(a1, a_2, ..., an)/d
- Product of coprime divisors is divisor
- LCM divides common multiple
Let $a$ and $b$ be positive numbers. Then $\gcd(a, b)\operatorname{lcm}(a, b) = ab$.
Proof
\begin{align} & \operatorname{PFEL}(\gcd(a, b)\operatorname{lcm}(a, b)) \\ &= \operatorname{PFEL}(\gcd(a, b)) + \operatorname{PFEL}(\operatorname{lcm}(a, b)) \\ &= \min(\operatorname{PFEL}(a), \operatorname{PFEL}(b)) + \max(\operatorname{PFEL}(a), \operatorname{PFEL}(b)) \\ &= \operatorname{PFEL}(a) + \operatorname{PFEL}(b) \\ &= \operatorname{PFEL}(ab) \end{align}
Therefore, $\gcd(a, b)\operatorname{lcm}(a, b) = ab$.
Alternate Proof
Let $g = \gcd(a, b)$ and $l = \operatorname{lcm}(a, b)$.
\[ (a \mid l \wedge b \mid l) \implies \left( \frac{a}{g} \mid \frac{l}{g} \wedge \frac{b}{g} \mid \frac{l}{g} \right) \] \[ \gcd\left( \frac{a}{g}, \frac{b}{g} \right) = \frac{\gcd(a, b)}{g} = 1 \] Since product of coprime divisors is also a divisor, \[ \frac{a}{g}\frac{b}{g} \mid \frac{l}{g} \implies ab \mid lg \]
\[ \frac{ab}{g} = a\left(\frac{b}{g}\right) = \left(\frac{a}{g}\right)b \] Therefore, $\frac{ab}{g}$ is a common multiple of $a$ and $b$. Therefore, \[ l \mid \frac{ab}{g} \implies lg \mid ab \]
Since $ab \mid lg$ and $lg \mid ab$, $lg = ab$. Therefore, $\gcd(a, b)\operatorname{lcm}(a, b) = ab$.
Dependency for:
Info:
- Depth: 8
- Number of transitive dependencies: 15
Transitive dependencies:
- Every number has a prime factorization
- Integer Division Theorem
- GCD is the smallest Linear Combination
- Common divisor divides GCD
- gcd(a1/d, a2/d, ..., an/d) = gcd(a1, a_2, ..., an)/d
- Euclid's lemma
- Fundamental Theorem of Arithmetic
- Prime Factorization Exponent List (PFEL)
- PFEL of ratio is difference of PFEL
- PFEL of product is sum of PFEL
- Divisible iff PFEL is less than or equal
- PFEL of lcm is max of PFEL
- PFEL of gcd is min of PFEL
- LCM divides common multiple
- Product of coprime divisors is divisor