Talk:Bertrand-Chebyshev Theorem/Lemma 1

One can also prove that $\dbinom{2n}{n}\ge\dfrac{2^{2n}}{2n}$ for all $n\ge1$. Can I add this fact to the page?

This does somewhat shorten the proof of the Bertrand-Chebyshev Theorem, by finding a much lower bound than the present one. Simcha Waldman (talk) 09:17, 12 March 2024 (UTC)

I'd prefer it if you would set up another lemma (Lemma 4?), then add a second proof of the Bertrand-Chebyshev Theorem that uses that new lemma. --prime mover (talk) 09:33, 12 March 2024 (UTC)

The proof here is not very different than yours, but its end result is much simpler. Simcha Waldman (talk) 15:36, 12 March 2024 (UTC)

okay, so feel free to implement it here, in $\mathsf{Pr} \infty \mathsf{fWiki}$ format. --prime mover (talk) 19:44, 12 March 2024 (UTC)

Thanks! If you are OK with this, It will require the slight change in Lemma 1. Simcha Waldman (talk) 06:09, 13 March 2024 (UTC)

No, as I said, I would prefer it if you create another lemma. --prime mover (talk) 07:43, 13 March 2024 (UTC)

I have rolled back your edits, as you will see. Instead, as I suggested earlier, you are requested not to change what is there, but instead implement another proof.

Please review existing pages on $\mathsf{Pr} \infty \mathsf{fWiki}$ to familiarise yourself with house presentation and source code style. --prime mover (talk) 07:55, 13 March 2024 (UTC)

I have set up the page structure for this to be accomplished. Please feel free to put your new version of the proof into Bertrand-Chebyshev Theorem/Proof 2. --prime mover (talk) 08:02, 13 March 2024 (UTC)

I have in fact implemented your version of this proof. --prime mover (talk) 09:20, 13 March 2024 (UTC)