បង្ហាញថា
( m , n ) [ m , n ] = m n {\displaystyle (m,n)[m,n]=mn}
តាង p {\displaystyle p} ជាចំនួនបឋមដែល
p e = m {\displaystyle p^{e}=m}
p f = n {\displaystyle p^{f}=n}
ប្រសិនបើ e ≤ f {\displaystyle e\leq f}
នោះ
( m , n ) = p e {\displaystyle (m,n)=p^{e}}
[ m , n ] = p f {\displaystyle [m,n]=p^{f}}
( m , n ) [ m , n ] = p e + f {\displaystyle (m,n)[m,n]=p^{e+f}}
តែ
m n = p e + f {\displaystyle mn=p^{e+f}}
មានន័យថា
![{\displaystyle (m,n)[m,n]=mn}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53d5837e4b79355627242cfe72e6576a1b427d30)
ជាចំនួនបឋមដែល
![{\displaystyle [m,n]=p^{f}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c9df9c4641229d5dc148ea1e4b0b45605200208)
![{\displaystyle (m,n)[m,n]=p^{e+f}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84c6dff11fd1f57d13d36c4b96c28103f39e46ee)
