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Question Number 162584 by SANOGO last updated on 30/Dec/21
prove that  ppcm(a,b)×pgcd(a,b)=∣ab∣
$${prove}\:{that} \\ $$$${ppcm}\left({a},{b}\right)×{pgcd}\left({a},{b}\right)=\mid{ab}\mid \\ $$
Answered by mr W last updated on 30/Dec/21
p_k =prime numbers  m_k ,h_k ≥0  a=Π_(k=1) ^n p_k ^m_k    b=Π_(k=1) ^n p_k ^h_k    gcd(a,b)=Π_(k=1) ^n p_k ^(min(m_k ,h_k ))   lcm(a,b)=Π_(k=1) ^n p_k ^(max(m_k ,h_k ))   gcd(a,b)×lcm(a,b)=Π_(k=1) ^n p_k ^(min(m_k ,h_k )+max(m_k ,h_k ))   gcd(a,b)×lcm(a,b)=Π_(k=1) ^n p_k ^(m_k +h_k )   gcd(a,b)×lcm(a,b)=(Π_(k=1) ^n p_k ^m_k  )(Π_(k=1) ^n p_k ^h_k  )  ⇒gcd(a,b)×lcm(a,b)=a×b
$${p}_{{k}} ={prime}\:{numbers} \\ $$$${m}_{{k}} ,{h}_{{k}} \geqslant\mathrm{0} \\ $$$${a}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{p}_{{k}} ^{{m}_{{k}} } \\ $$$${b}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{p}_{{k}} ^{{h}_{{k}} } \\ $$$${gcd}\left({a},{b}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{p}_{{k}} ^{{min}\left({m}_{{k}} ,{h}_{{k}} \right)} \\ $$$${lcm}\left({a},{b}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{p}_{{k}} ^{{max}\left({m}_{{k}} ,{h}_{{k}} \right)} \\ $$$${gcd}\left({a},{b}\right)×{lcm}\left({a},{b}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{p}_{{k}} ^{{min}\left({m}_{{k}} ,{h}_{{k}} \right)+{max}\left({m}_{{k}} ,{h}_{{k}} \right)} \\ $$$${gcd}\left({a},{b}\right)×{lcm}\left({a},{b}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{p}_{{k}} ^{{m}_{{k}} +{h}_{{k}} } \\ $$$${gcd}\left({a},{b}\right)×{lcm}\left({a},{b}\right)=\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{p}_{{k}} ^{{m}_{{k}} } \right)\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{p}_{{k}} ^{{h}_{{k}} } \right) \\ $$$$\Rightarrow{gcd}\left({a},{b}\right)×{lcm}\left({a},{b}\right)={a}×{b} \\ $$
Commented by SANOGO last updated on 30/Dec/21
merci bien
$${merci}\:{bien} \\ $$
Commented by Rasheed.Sindhi last updated on 31/Dec/21
G_(Sir!) ^(rea) T
$$\mathbb{G}_{\mathrm{Sir}!} ^{\mathrm{rea}} \mathbb{T} \\ $$

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