prove that ppcm(a,b)×pgcd(a,b)=∣ab∣


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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

	$G_{Sir!}^{rea} T$
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