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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 $\sqrt{2x}<p\le x$, 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?

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Larry Freeman
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1 Answers1

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I think that you are right.

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.

mathlove
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