How might I prove that LCM$(m) \geq 2^m$?

Asked Mar 15 '18 at 13:18

Active Mar 15 '18 at 13:45

Viewed 447 times

Denoting by LCM$(m)$ the lowest common multiple of the first $m$ numbers, can anyone suggest a way in which I might prove that, for $m \geq 7$, $$ \text{LCM}(m) \geq 2^m $$

I believe that a proof of this may be found within the proof of Theorem 2 of the paper 'On Chebyshev-Type Inequalities for Primes' by M. Nair, but I am currently struggling to follow this.

edited Mar 15 '18 at 13:45

asked Mar 15 '18 at 13:18

M Smith

2,727

1

Counterexample: LCM(3)=6, which is less than 2^3. There are more (LCM(4), LCM(6), ...) – Cristian Lupascu Mar 15 '18 at 13:41
Apologies, I forgot to mention that $m \geq 7$. I have amended the question now. – M Smith Mar 15 '18 at 13:45
https://math.stackexchange.com/questions/851328/textlcm1-2-3-ldots-n-geq-2n-for-n-geq-7 – Marco Cantarini Mar 15 '18 at 14:33

How might I prove that LCM$(m) \geq 2^m$?

0 Answers0