MATH312 December 2005
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Question 10

Prove
In this section, prove the statement given to you.
Suppose a and b are positive integers. Show that
 $\displaystyle (a,b)[a,b]=ab.$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Use the Fundamental Theorem of Arithmetic to write
$\displaystyle a=\prod _{i=1}^{n}p_{i}^{e_{i}}$
and
$\displaystyle b=\prod _{i=1}^{m}p_{i}^{f_{i}}$
where we use all the same primes above in the two expansions so we will allow the possibility of $\displaystyle e_{i}$ or $\displaystyle f_{i}$ to be 0 (but of course not both). Then
$\displaystyle \gcd(a,b)=\prod _{i=1}^{n}p_{i}^{\min(e_{i},f_{i})}$
and
$\displaystyle {\text{lcm}}(a,b)=\prod _{i=1}^{n}p_{i}^{\max(e_{i},f_{i})}$.
See how to finish the argument.

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Solution

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Use the Fundamental Theorem of Arithmetic to write
$\displaystyle a=\prod _{i=1}^{n}p_{i}^{e_{i}}$
and
$\displaystyle b=\prod _{i=1}^{m}p_{i}^{f_{i}}$
where we use all the same primes above in the two expansions so we will allow the possibility of $\displaystyle e_{i}$ or $\displaystyle f_{i}$ to be 0 (but of course not both). Then
$\displaystyle \gcd(a,b)=\prod _{i=1}^{n}p_{i}^{\min(e_{i},f_{i})}$
and
$\displaystyle {\text{lcm}}(a,b)=\prod _{i=1}^{n}p_{i}^{\max(e_{i},f_{i})}$.
Now, we have
$\displaystyle {\begin{aligned}\gcd(a,b){\text{lcm}}(a,b)&=\prod _{i=1}^{n}p_{i}^{\min(e_{i},f_{i})}\prod _{i=1}^{n}p_{i}^{\max(e_{i},f_{i})}\\&=\prod _{i=1}^{n}p_{i}^{\min(e_{i},f_{i})+\max(e_{i},f_{i})}\end{aligned}}$
As
$\displaystyle \min(e_{i},f_{i})+\max(e_{i},f_{i})=e_{i}+f_{i}$
we have that
$\displaystyle {\begin{aligned}\gcd(a,b){\text{lcm}}(a,b)&=\prod _{i=1}^{n}p_{i}^{\min(e_{i},f_{i})}\prod _{i=1}^{n}p_{i}^{\max(e_{i},f_{i})}\\&=\prod _{i=1}^{n}p_{i}^{\min(e_{i},f_{i})+\max(e_{i},f_{i})}\\&=\prod _{i=1}^{n}p_{i}^{e_{i}+f_{i}}\\&=\prod _{i=1}^{n}p_{i}^{e_{i}}\prod _{i=1}^{n}p_{i}^{f_{i}}\\&=ab\end{aligned}}$
completing the proof.

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