prove-that-ppcm-a-b-pgcd-a-b-ab-

Question Number 162584 by SANOGO last updated on 30/Dec/21

p r o v e t h a t

p p c m (a, b) \times p g c d (a, b) =∣ a b ∣

Answered by mr W last updated on 30/Dec/21

p_{k} = p r i m e n u m b e r s

m_{k}, h_{k} ⩾ 0

a = \underset{k = 1}{\prod^{n}} p_{k}^{m_{k}}

b = \underset{k = 1}{\prod^{n}} p_{k}^{h_{k}}

g c d (a, b) = \underset{k = 1}{\prod^{n}} p_{k}^{m i n (m_{k}, h_{k})}

l c m (a, b) = \underset{k = 1}{\prod^{n}} p_{k}^{m a x (m_{k}, h_{k})}

g c d (a, b) \times l c m (a, b) = \underset{k = 1}{\prod^{n}} p_{k}^{m i n (m_{k}, h_{k}) + m a x (m_{k}, h_{k})}

g c d (a, b) \times l c m (a, b) = \underset{k = 1}{\prod^{n}} p_{k}^{m_{k} + h_{k}}

g c d (a, b) \times l c m (a, b) = (\underset{k = 1}{\prod^{n}} p_{k}^{m_{k}}) (\underset{k = 1}{\prod^{n}} p_{k}^{h_{k}})

\Rightarrow g c d (a, b) \times l c m (a, b) = a \times b

Commented by SANOGO last updated on 30/Dec/21

m e r c i b i e n

Commented by Rasheed.Sindhi last updated on 31/Dec/21

G_{Sir!}^{rea} T