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I was reading Julian Havil’s book Gamma where he talks about a short proof by Paul Erdős on the infinitude of primes.

As I understand it, here are the steps:

(1) Let $N$ be any positive integer and $p_1, p_2, p_3, \dots, p_n$ be the complete set of primes less than or equal to $N$

(2) Each $1 \le x \le N$ can be written as $p_1^{e_1}p_2^{e_2}p_3^{e_3}\ldots p_n^{e_n} \times m^2$ where $e_i \in \left\{0,1\right\}$

(3) So, there are $2^n$ ways of choosing square-free numbers and $m^2 \le N$

(4) Since $m \le \sqrt{N}$, each $2 \le x \le N$ can be chosen in at most $2^n \times \sqrt{N}$ ways.

(5) Thus, $N \le 2^n \times \sqrt{N}$ and $2^n \ge \sqrt{N}$ so that $n \ge \dfrac{1}{2}\log_2 N$

I am confused by step #4. Why does it follow that $x$ can be chosen in at most $2^n \times \sqrt{N}$ ways. I would think that it would be chosen in $2^n \times m^2$ ways. Why is he allowed to replace $m^2$ with $m$ in this case?


Edit: I figured out my misunderstanding.

When I reread the proof in the book this morning, I noticed the following sentence in the paragraph before the proof:

"In 1938 the consummate practitioner Paul Erdos (1913-1996) gave the one that follows, which uses a counting technique and a neat device used by number theorists: that any integer can be written as the product of a square and a square-free integer"

This device is easily proven. Let $u = p_1^{e_1}p_2^{e_2}\ldots p_n^{e_n}$ so that for any $x \le N$, $x = um^2$:

  • Let $p^v | x$ where $p^{v+1} \nmid x$ and $v \ge 1$
  • If $v \equiv 0 \pmod 2$, then $p \nmid u$ and $p^v | m^2$
  • If $v \equiv 1 \pmod 2$, then $p | u$ and $p^{v-1} | m^2$

So that it follows that $m$ is an integer. Now, the full proof works for me.

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  • $\begingroup$ Isn't it true that $m$ is not necessarily an integer. For example, if $x=4$, then $m^2 = 2$ In that case, wouldn't there be $m^2$ choices for $m$? I have a feeling that I am missing something obvious. $\endgroup$ – Larry Freeman Jan 23 '15 at 10:54
  • $\begingroup$ @LarryFreeman $m$ is an integer!! That's where the even powers of $p_i$ go. Apparently, your problem was the step 2. $\endgroup$ – user2345215 Jan 23 '15 at 11:01
  • $\begingroup$ Would it be possible to give the page reference for the passage in Havil since it is not in the index? $\endgroup$ – daniel Jan 23 '15 at 12:31
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    $\begingroup$ Hi Daniel, It is on page 29 of my edition which is in Chapter 3: Sub-Harmonic Series. $\endgroup$ – Larry Freeman Jan 23 '15 at 15:06
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    $\begingroup$ Just a detail, but shouldn't it be $1\leq x\leq N$? Or there is $2^n-1$ ways to chose a square-free number in (3). This also make more sense in (5). $\endgroup$ – Lehs Jan 24 '15 at 11:19
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The number of values $m^2$ can have is given by $m$. If you used $m^2$, it would look as if it could have any value from $1$ to $m^2$, which isn't true.

You "replace" $m^2$ by $m$ in the same sense you replaced $p_1^{e_1}\ldots p_n^{e_n}$ by $2^n$ (and not by $p_1\ldots p_n$).

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  • $\begingroup$ Thanks very much! I am almost getting it. I guess I just need to better understand why you replace $m^2$ by $m$. I appreciate your point. $\endgroup$ – Larry Freeman Jan 23 '15 at 10:56
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    $\begingroup$ @LarryFreeman Think of $x$ as a function $f(e_1,\ldots, e_n, m)$. The number of outputs never greater than the number of inputs. Do you understand now? $\endgroup$ – user2345215 Jan 23 '15 at 10:58
  • $\begingroup$ Thanks! I think that my issue was that I thought $m$ wasn't necessarily an integer. Once I see that it must be an integer, I am clear on your explanation. $\endgroup$ – Larry Freeman Jan 24 '15 at 3:07

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