Take any two coprime numbers and find their product. Also find out if the product of their HCF and LCM is equal to product of the numbers.

Saturday, June 14, 2014

Take any two coprime numbers and find their product. Also find out if the product of their HCF and LCM is equal to product of the numbers.

Hello!

Well, consider $\displaystyle{a}={12}$ and $\displaystyle{b}={25}.$ They are not prime but coprime (have no common dividers except $\displaystyle{1}$ ). $\displaystyle{a}={{2}}^{{2}}\cdot{3},$ $\displaystyle{b}={{5}}^{{2}}.$

Their product is $\displaystyle{a}{b}={12}\cdot{25}={300}.$ Their $\displaystyle{H}{C}{F}$ is $\displaystyle{1}$ as for any coprime numbers. Their $\displaystyle{L}{C}{M}$ must include all their prime factors with their degrees, so $\displaystyle{L}{C}{M}={{2}}^{{2}}\cdot{3}\cdot{{5}}^{{2}}={300}.$

And yes, $\displaystyle{a}{b}={H}{C}{F}{\left({a},{b}\right)}\cdot{L}{C}{M}{\left({a},{b}\right)}$ for these $\displaystyle{a}$ and $\displaystyle{b}.$

But wait, this identity is true for any natural $\displaystyle{a}$ and $\displaystyle{b}$ ! HCF is a factor of both $\displaystyle{a}$ and $\displaystyle{b},$ so $\displaystyle{a}={a}_{{1}}\cdot{H}{C}{F},$ $\displaystyle{b}={b}_{{1}}\cdot{H}{C}{F}.$ And because it is the highest common factor, the numbers $\displaystyle{a}_{{1}}$ and $\displaystyle{b}_{{1}}$ are coprime. Therefore LCM must include factors $\displaystyle{H}{C}{F},$ $\displaystyle{a}_{{1}}$ and $\displaystyle{b}_{{1}},$ i.e. $\displaystyle{L}{C}{M}={H}{C}{F}\cdot{a}_{{1}}\cdot{b}_{{1}}$ and

$\displaystyle{a}{b}={a}_{{1}}\cdot{b}_{{1}}\cdot{H}{C}{{F}}^{{2}}$ and $\displaystyle{L}{C}{M}\cdot{H}{C}{F}={a}_{{1}}\cdot{b}_{{1}}\cdot{H}{C}{{F}}^{{2}}$ also.

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Saturday, June 14, 2014