# Is this approach valid for using least common multiple to establish Bertrand's postulate

Let $$\text{lcm}(x)$$ be the least common multiple of $$\{1,2,3,\dots,x\}$$.

Denis Hanson showed that $$\text{lcm}(x) < 3^x$$ and M. Nair showed that $$\text{lcm}(x) > 2^x$$.

Neither used Bertrand's Postulate in their result. Hanson later showed that there is always a prime between $$3n$$ and $$4n$$ without using Nair's result.

The argument for Bertrand's Postulate depends on the following idea:

$$\text{lcm}(\sqrt{2x})\frac{2x\#}{x\#}\ge \frac{\text{lcm}(2x)}{\text{lcm}(x)}$$

where $$2x\#$$ and $$x\#$$ are the primorial for $$2x$$ and for $$x$$.

Here is the argument:

• If a prime $$\sqrt{2x}, then it cancels out in $$\dfrac{\text{lcm}(2x)}{\text{lcm}(x)}$$.

• If a prime $$x < p \le 2x$$, then it divides $$\dfrac{2x\#}{x\#}$$.

• If $$a \ge 2$$ and $$x < p^a \le 2x$$, then $$p^{a-1} < \dfrac{x}{p} < x$$ and it divides out and $$p^{a+1} > 2x$$.

• The lemma follows.

Here is the argument for Bertrand's Postulate:

(1) From Hanson and Nair:

$$\frac{\text{lcm}(2x)}{\text{lcm}(x)} > \frac{2^{2x}}{3^{x}} = \left(\frac{4}{3}\right)^x$$

(2) Assume that there is no prime between $$2x$$ and $$x$$.

(3) Then we have the following:

$$3^{\sqrt{2x}} > \text{lcm}(\sqrt{2x})\frac{2x\#}{x\#} > \left(\frac{4}{3}\right)^x$$

(4) Which simplifies to:

$$\frac{\ln(4)}{\ln(3)} < \frac{x+\sqrt{2x}}{x}$$

which is invalid for $$x \ge 30$$ since:

$$\frac{\ln(4)}{\ln(3)} > 1.26 > \frac{30+\sqrt{60}}{30} \approx 1.258$$

and

$$\frac{d}{dx}\left(\frac{x+\sqrt{2x}}{x}\right) = -\frac{1}{\sqrt{2}x^{3/2}}$$

Am I wrong?

• The result of Nair is that $n\geq 7\implies lcm(n)>2^n$, as stated in your link. Nov 17, 2018 at 8:16

I think that the inequality of the idea is true, but it seems that your proof has an error. The third bullet point looks wrong since $$x\lt p^a\le 2x$$ does not imply $$p^{a-1}\lt \frac xp\lt x$$.
You've already dealt with the primes $$p$$ such that $$\sqrt{2x}\lt p\le 2x$$ in the first and the second bullet points.
Let us consider the case where $$p^a\le \sqrt{2x}\lt p^{a+1}\qquad\text{and}\qquad p^b\le x\lt p^{b+1}$$ Here, $$p$$ is a prime and $$a,b$$ are positive integers such that $$a\le b$$.
If $$a=b$$. Then, we get $$\frac{p^{2a}}{2}\le x\lt p^{b+1}=p^{a+1}\implies p^{a-1}\lt 2\implies a=b=1$$ So, we have $$p\le \sqrt{2x}\lt p^{2},\qquad p\le x\lt p^{2},\qquad p^2\le 2x\lt 2p^2\le p^3$$ So, in this case, the inequality holds.
If $$b\ge a+1$$, then we get $$p^{2a}\le 2x\lt p^{2a+2}$$ and $$p^ap^b\ge p^{2a+1}\ge p^{2a}$$ So, in this case, the inequality holds.