Editing Least common multiple (section)

=== Fundamental theorem of arithmetic ===
According to the [[fundamental theorem of arithmetic]], every integer greater than 1 can be represented uniquely as a product of prime numbers, [[up to]] the order of the factors:

:<math>n = 2^{n_2} 3^{n_3} 5^{n_5} 7^{n_7} \cdots = \prod_p p^{n_p},</math>

where the exponents ''n''<sub>2</sub>, ''n''<sub>3</sub>, ... are non-negative integers; for example, 84 = 2<sup>2</sup> 3<sup>1</sup> 5<sup>0</sup> 7<sup>1</sup> 11<sup>0</sup> 13<sup>0</sup> ...

Given two positive integers <math display="inline">a = \prod_p p^{a_p}</math> and <math display="inline">b = \prod_p p^{b_p}</math>, their greatest common divisor and least common multiple are given by the formulas
:<math>\gcd(a,b) = \prod_p p^{\min(a_p, b_p)}</math>

and
:<math>\operatorname{lcm}(a,b) = \prod_p p^{\max(a_p, b_p)}.</math>

Since
:<math>\min(x,y) + \max(x,y) = x + y,</math>
this gives
:<math>\gcd(a,b) \operatorname{lcm}(a,b) = ab.</math>

In fact, every rational number can be written uniquely as the product of primes, if negative exponents are allowed. When this is done, the above formulas remain valid. For example:
:<math>\begin{align}
  4 &= 2^2 3^0,                   & 6 &= 2^1 3^1,    & \gcd(4, 6) &= 2^1 3^0 = 2,    & \operatorname{lcm}(4,6) &= 2^2  3^1 = 12. \\[8pt]
  \tfrac{1}{3} &= 2^0 3^{-1} 5^0, & \tfrac{2}{5} &= 2^1 3^0 5^{-1}, & \gcd\left(\tfrac13, \tfrac{2}{5}\right) &= 2^0 3^{-1} 5^{-1} = \tfrac{1}{15}, & \operatorname{lcm}\left(\tfrac{1}{3}, \tfrac{2}{5}\right) &= 2^1 3^0 5^0 = 2, \\[8pt]
  \tfrac{1}{6} &= 2^{-1} 3^{-1},  & \tfrac{3}{4} &= 2^{-2} 3^1, & \gcd\left(\tfrac{1}{6}, \tfrac{3}{4}\right) &= 2^{-2} 3^{-1} = \tfrac{1}{12}, & \operatorname{lcm}\left(\tfrac{1}{6}, \tfrac{3}{4}\right) &= 2^{-1} 3^1 = \tfrac{3}{2}.
\end{align}</math>