I am reading Kenneth Ireland and Michael Rosen's A Classical Introduction to Modern Number Theory, and I'm having trouble with a step in one of their proofs.

The proof in question is of Corollary 1 in chapter 2, section 4. The corollary is stated as follows:

There is a positive constant $c_{1}$ such that $\pi(x) < c_{1}x/$log $x$ for $x \ge 2$.

It goes without saying, but for the benefit of readers new to number theory, $\pi(x)$ is the number of primes less than or equal to $x$ and is known as the prime counting function. The proof employs a function on the natural numbers defined as $\theta(x) = \sum_{p \le x}$ log $p$, the sum being over all primes at most x. The proof begins as follows:

$\theta(x) \ge \sum^{p \le x}_{p > \sqrt{x}}$ log $p \ge $ (log$\sqrt{x})(\pi(x) - \pi(\sqrt{x})) \ge $ (log$\sqrt{x})\pi(x) - \sqrt{x}$log$\sqrt{x}$.

The first inequality follows from the definition of $\theta(x)$, the next follows from the fact that log $p > $ (log$\sqrt{x})$ and $\pi(x) - \pi(\sqrt{x})$ is the number of primes $p$ that satisfy $\sqrt{x} < p \le x$. However, the last inequality seems to imply that $\pi(\sqrt{x}) \ge \sqrt{x}$ if you multiply out the (log$\sqrt{x}$) term, and that can't be true.

I'd like to know if there's some logic I'm missing here, like maybe they aren't multiplying out the (log$\sqrt{x}$) term and are instead using some trick to get that last inequality. Perhaps they're using one of the earlier propositions about $\pi(x)$, namely Proposition 2.4.1: $\pi(x) \ge $ log(log$x$) or Proposition 2.4.2: $\pi(x) \ge $ log$x/(2$log$2)$ - though I'm having trouble spotting how they might be using them.

Edit: Silly mistake, going to leave this question up as a reminder for myself to stay humble.


1 Answer 1


Your misunderstanding is simple. They actually use $\pi(\sqrt{x})\le \sqrt{x}$, which is obviously true and equivalent to $-\pi(\sqrt{x})\ge -\sqrt{x}$. Since the term has a negative sign, this is the correct result.

  • $\begingroup$ That was silly of me. Thanks. $\endgroup$ Commented Jun 19, 2019 at 16:41
  • $\begingroup$ @Alena No problem ;) $\endgroup$
    – F.U.A.S.
    Commented Jun 19, 2019 at 16:45

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