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With $\{a,b,c\}\subset \mathbb{Z}$ given that $\text{lcm}(a,b) = \frac{a \cdot b}{\gcd(a,b)}$. Prove that: $$\text{lcm}(a,b,c) = \frac{a \cdot b \cdot c \cdot \gcd(a,b,c)}{\gcd(a,b)\gcd(b,c)\gcd(a,c)}$$

Michael Rozenberg
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clay
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    This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. –  Aug 08 '17 at 06:51
  • This answer is somewhat related to your question. – bof Aug 08 '17 at 07:18
  • For a proof using gcd laws combine the two identities in the "alternative" proof in this answer. – Bill Dubuque Aug 08 '17 at 23:16
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    Or we can also prove it using unique prime factorizations as in said answer. Then it boils down to the following equality on exponents $\ \gamma = (\alpha+\beta+\gamma)+\alpha-(\alpha+\beta+\alpha)\ \ \ $ – Bill Dubuque Feb 21 '20 at 01:28

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Let $a=p_1^{\alpha_1}p_2^{\alpha_2}...p_n^{\alpha_n}$, $b=p_1^{\beta_1}p_2^{\beta_2}...p_n^{\beta_n}$ and $c=p_1^{\gamma_1}p_2^{\gamma_2}...p_n^{\gamma_n}$, where $\alpha_i\geq0$, $\beta_i\geq0$, $\gamma_i\geq0$ be integers and $p_i$ be different prime numbers.

Hence, we need to prove that $$p_1^{\max\{\alpha_1,\beta_1,\gamma_1\}}p_2^{\max\{\alpha_2,\beta_2,\gamma_2\}}...p_n^{\max\{\alpha_n,\beta_n,\gamma_n\}}=$$ $$=p_1^{\alpha_1+\beta_1+\gamma_1+\min\{\alpha_1,\beta_1,\gamma_1\}-\min\{\alpha_1,\beta_1\}-\min\{\alpha_1,\gamma_1\}-\min\{\beta_1,\gamma_1\}}\cdot...$$ $$\cdot p_n^{\alpha_n+\beta_n+\gamma_n+\min\{\alpha_n,\beta_n,\gamma_n\}-\min\{\alpha_n,\beta_n\}-\min\{\alpha_n,\gamma_n\}-\min\{\beta_n,\gamma_n\}}.$$

Thus, it's enough to prove that $$\max\{x,y,z\}+\min\{x,y\}+\min\{x,z\}+\min\{y,z\}=x+y+z+\min\{x,y,z\},$$ which is obvious because we can assume $x\geq y\geq z$.

Done!

Michael Rozenberg
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  • Simplest possible proof. +1 – Paramanand Singh Aug 08 '17 at 09:00
  • @clay What do you think about my solution? Thank you! – Michael Rozenberg Aug 08 '17 at 12:53
  • Thank you! The solution is perfect. I was stuck on this for hours. After the first sentence of your solution, I could see where you were going and finish the proof, but I didn't think of that route. I don't understand why this question was flagged as off topic and missing context, It's a textbook math problem from http://a.co/0yQQDOb. – clay Aug 08 '17 at 17:41
  • @clay Because you need to show you trying. Also it's better to show in your starting list from here taken this problem. – Michael Rozenberg Aug 08 '17 at 18:19
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    The cited reason for closing this was, "off topic". You are saying the real reason was lack of demonstration of effort which is a valid reason but it is completely different from what is stated. – clay Aug 08 '17 at 21:49
  • @clay They are standard words. – Michael Rozenberg Aug 09 '17 at 02:35