GCD is MIN of Exponents of Prime Factors, LCM is MAX of Exponents of Prime Factors.

Question

Let $a=p_1^{x_1}p_2^{x_2}\cdots\cdot p_q^{x_q}$ and $b=p_1^{y_1}p_2^{y_2}\cdots p_r^{x_r}$ http://www.cut-the-knot.org/blue/gcd_fta.shtml, https://math.stackexchange.com/a/349867/85100 say —
Since gcd(a,b) is the largest common divisor of a and b and is divisible by any other common divisor of the two,

GCD$(a,b)=p_1^{\min(x_1,y_1)}p_2^{\min (x_2,y_2)} \cdots p_q^{\min (x_q,y_q)}$

LCM$(a,b)= p_1^{\max(x_1,y_1)}p_2^{\max(x_2,y_2)}\cdots p_r^{\max (x_r,y_r)}$

I understand GCD(a,b) has to divide both $a,b$. Therefore the exponent of any prime factor $p_i$ in GCD has to be in both $a,b$ — therefore $p_i^{\min(x_i, y_i)}$. But I'm muddled and anxious. Why does $\min$ appear in GREATEST common divisor? Why does $\max$ appear in LOWEST common multiple?

Answer 1

If we are several persons, the GREATEST height that we can all reach is the MINIMUM of our heights (the smallest person amongst us is the problem), the lowest door that we can all pass through is the MAXIMUM of our heights (the tallest person is the problem).

Answer 2

Hint $ $ Assume $\rm\,p\nmid a,b,\,$ and write $\ \rm(x,y) := \gcd(x,y),\ \ [x,y] := {\rm lcm}(x,y)$. By FTA (existence and uniqueness of prime factorizations) we can recursively compute gcd and lcm by "peeling off" one-prime-power at at time as follows

$\rm\ (ap^j,bp^k) = (a,b)(p^j,p^k) = (a,b)\,p^{\large \min(j,k)}\,$ by $\,\rm p^i\mid p^j,p^k\!\!\iff i\le j,k \!\iff\! i\le \min(j,k)$

$\rm\ \, [ap^j,bp^k] =\, [a,b]\,[p^j,p^k]\, =\, [a,b]\,p^{\large \max(j,k)}$ by $\,\rm p^j,p^k\mid p^i\!\! \iff j,k\le i\! \iff\! \max(j,k)\le i$

Answer 3

Although the GCD is the greatest common divisor of $a$ and $b$, it still has to divide both of them! If the exponent of $p_i$ was greater than $\min(x_i,y_i)$ then the GCD wouldn't divide both $a$ and $b$.

Take for instance $40 = 2^3 \cdot 5$ and $100 = 2^2 \cdot 5^2$. Then their GCD is $20 = 2^2 \cdot 5 = 2^{\min(3,2)} \cdot 5^{\min(1,2)}$.

Answer 4

Why does min appear in GREATEST common divisor? Why does max appear in LOWEST common multiple?

Because "min" is the greatest lower bound, and "max" is the least upper bound, for linearly ordered sets ... and even for partially ordered sets, if you wish to extend the definition like so. Since the correspondence you're describing preserves ordering, when taken with respect to suitably-defined ordering relations, then one should expect "greatest" to go with "greatest" and "least" with "least".

One ordering relation is the factoring relation. Greatest Common Divisor and Least Common Multiple are its respective greatest lower bound and least upper bound.

The other ordering relation is the one for exponents - taken in tandem; which is applied to lists of exponents $a = \left(a_2,a_3,a_5,⋯\right)$ and $b = \left(b_2,b_3,b_5,⋯\right)$ in such a way that $a < b$ if and only if $a_2 < b_2$, $a_3 < b_3$, $a_5 < b_5$, and so on. Since the ordering relation is member-by-member-wise, and since the ordering relation for each member is linear, then the greatest lower bound $a ∧ b$ and least upper bound $a ∨ b$ are formed by the respective minimum and maximum as: $$ a ∧ b = \left(\min\left(a_2,b_2\right), \min\left(a_3,b_3\right), \min\left(a_5,b_5\right), ⋯\right),\quad a ∨ b = \left(\max\left(a_2,b_2\right), \max\left(a_3,b_3\right), \max\left(a_5,b_5\right), ⋯\right). $$

For numbers, this correspondence can be broadened to include positive rational numbers, and not just positive integers, because the "min" and "max" relations can be broadened to include negative integers and because the fundamental theorem of arithmetic, which guarantees the uniqueness of expansion into products of exponents of primes, can be broadened to include rational numbers. So, they also have greatest common divisors and least common multiples.