# Can the product of infinitely many elements from $\mathbb Q$ be irrational? [closed]

I know there are infinite sums of rational values, which are irrational (for example the Basel Problem). But I was wondering, whether the product of infinitely many rational numbers can be irrational. Thank you for your answers.

## closed as off-topic by kjetil b halvorsen, user21820, Namaste, Xander Henderson, user223391 Apr 3 '18 at 2:53

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• Do you know Wallis's product for $\pi$? – NickD Mar 28 '18 at 13:23
• $e=\lim_{n\to \infty}{(1+1/n)^n}$ – Vasya Mar 28 '18 at 13:29
• @Vasya Correct limit of rationals, but not an infinite product. – Ethan Bolker Mar 28 '18 at 13:29
• @DzamoNorton Not a limit of partial products. The number of factors increases each time, but the factors change: $(1 + 1/2)(1+1/2)$, $(1+1/3)(1 + 1/3)(1+ 1/3)$ and so on. – Ethan Bolker Mar 28 '18 at 19:51
• @toliveira: An "infinite product" is not multiplication. It's a limit. Fundamentally, multiplication has two operands; you can inductively extend that to any finite number of operands. But "any finite number" does not include "infinitely many". – R.. Mar 29 '18 at 15:10

Yes, it can.

Consider any sequence $(a_n)$ of non-zero rational numbers which converges to an irrational number. Then define the sequence $b_n$ by $b_1 = a_1$ and $$b_n = \frac{a_n}{a_{n-1}}$$ for $n > 1$.

We then have that $$b_1 b_2 \cdots b_n = a_1 \frac{a_2}{a_1} \frac{a_3}{a_2} \cdots \frac{a_n}{a_{n-1}} = a_n.$$

We thus see that every term of $(b_n)$ is rational, and that the product of the terms of $(b_n)$ is the same as the limit of $a_n$, which is irrational.

• Great answer. Very simple. – Matt Samuel Mar 29 '18 at 9:57
• But it does look like a "cheat" since everything is divided out by itself. So, is there an infinite product which does not contain a collection of $\frac{a_n}{a_n}$ ? both your answer and mohammad Riazi's have this "cheat", while Kumar's does not. – Carl Witthoft Mar 29 '18 at 18:56
• @Carl This isn't a cheat. In fact you can rewrite any infinite product in this way. – Matt Samuel Mar 29 '18 at 19:23
• +1 But I think it needs at least a short explanation (or link to one, maybe the question is already out here too?) that such a sequence "of non-zero rational numbers which converges to an irrational number" exists. It is easier too see than the initial question (I think) but I guess some of those who will fins the initial question interesting may not see this as obvious – Rolazaro Azeveires Mar 29 '18 at 20:08

Yes, every irrational number is an infinite product of rationals.

We can write an infinite sum of rationals as an infinite product of rationals.

\begin{align} a&=a,\\ a+b&=a\times\frac {a+b}{a}\\ a+b+c &= a \times \frac {a+b}{a}\times\frac {a+b+c}{a+b}\\.\\.\\.\\.\end{align}

For example, $$\sqrt 2 =1.414213....=1+.4+.01+.004+.....=$$

$$1\times \frac {1.4}{1}\times \frac {1.41}{1.4}\times\frac {1.414}{1.41}\times .....$$

• Your answer is chosen for a review audit of mine, which I upvoted (+1) because it is simple, generalized and easy to understand. – Trần Thúc Minh Trí Mar 29 '18 at 5:10
• @TrầnThúcMinhTrí Thanks for your attention. – Mohammad Riazi-Kermani Mar 29 '18 at 11:19
• I took the liberty to align equalities at '=' sign. Feel free to rollback if you don't like it. – CiaPan Mar 30 '18 at 7:29
• Thanks, I like it – Mohammad Riazi-Kermani Mar 30 '18 at 11:12

Yes!

$\cfrac{\pi}{2} = \cfrac{2}1 \cfrac 23 \cfrac 43 \cfrac 45 \cfrac 65 \cfrac 67 \cdots$

• This looks cool, but what is the geometric picture to go along with it? I an almost see polygons in there... – Mike Wise Mar 28 '18 at 16:40
• @MikeWise I din't understand which polygon you are talking about. – kayush Mar 28 '18 at 16:42
• General term is $\cfrac{2n}{(2n – 1)}\cfrac{2n}{(2n + 1)}$ – kayush Mar 28 '18 at 16:43
• @MikeWise: I was curious about that as well. Here's a short note on how when you draw rectangles of those areas, they add up to a quarter circle in the limit: math.chalmers.se/~wastlund/monthly.pdf – Eric Lippert Mar 28 '18 at 17:39
• Cool, thanks for that, will print it out and read it on my flight in the morning. :) – Mike Wise Mar 28 '18 at 17:47

Too big to be a comment: it should be noted that the order is more crucial in infinite products than in infinite sums, which is strikingly seen on the example cited many times already:

\begin{align*}\cfrac{\pi}{2}&=\cfrac{2}1 \cfrac 23 \cdot \cfrac 43 \cfrac 45 \cdot\cfrac 65 \cfrac 67\cdot \ldots\\ &= \cfrac{2^2}{2^2-1}\cdot \cfrac{4^2}{4^2-1}\cdot \cfrac{6^2}{6^2-1}\ldots\\ \end{align*} is an infinite product with partials starting at $\frac43$ and increasing towards $\frac\pi 2$ (every factor is greater than $1$), whereas the seemingly identical

\begin{align*}0&=\cfrac{2}3 \cfrac 23 \cdot \cfrac 45\cfrac 45\cdot\cfrac 67 \cfrac 67 \cdot\ldots\end{align*}

starts below $1$ and decreases, towards $0$. All that happened was a shift of denominators one step to the left.

• Products are isomorphic to sum of logs, so any order effect that shows up in one can be generated in the other. By allowing the denominators to be shifted independently of the numerators, you are treating the fraction (a/b) as ab^-1, so when you take the log, it's log(a)-log(b), which is an alternating series, and of course order is more important in alternating series than monotonic ones. It's the alternation generated by treating the numerator and denominator separately, not the product, that makes order important. – Acccumulation Mar 28 '18 at 15:00
• Allowing the denominators to be shifted independently of the numerators can also be interpreted as replacing each term a/b by two terms, a and 1/b. Thus when you take logs, you get two terms, log a and - log b. – Rosie F Mar 28 '18 at 22:10
• These explanations are of course correct, but the point is that it is not obvious to realise that it can matter, easy to make a mistake here. – Arnaud Mortier Mar 29 '18 at 8:32
• I think the claim is not that one shouldn't point out that order matters in an infinite product, but rather that one shouldn't claim that it matters more in infinite products than in infinite sums, since (as @Acccumulation points out) for products of positive real numbers and sums of real numbers it is literally the same phenomenon. – LSpice Mar 29 '18 at 22:38
• @LSpice when I said that it matters more, what I meant is that one should beware of the intuition that a permutation of terms acting only locally does not affect the final result (you need a violent perturbation of terms to give $\sum{(-1)^n\over n}$ a different limit). Of course here what is happening is that the terms are not only reordered, but broken to pieces and the pieces reordered. *This natural way to think of a rational factor as being made of two pieces does not happen with series. * – Arnaud Mortier Mar 29 '18 at 23:44

Consider the Riemann-Zeta Function: $$\sum_{n=1}^{\infty}\frac{1}{n^{s}}=\prod_{p\text{ prime}}\frac{1}{1-p^{-s}}.$$ For $s=2$, the infinite sum on the left is $\pi^{2}/6$, which is irrational. Thus, $\pi^{2}/6$ is an infinite product of rationals.

There is a simple way to obtain any irrational number as an infinite product:

• take any sequence $s_n$ of rational numbers converging to the targeted irrational one (say the approximations of $\pi$ to $n$ decimals);

• form the product of the numbers $f_n:=\dfrac{s_{n+1}}{s_n}$, with $f_0=1$.

$$\pi=\prod_{n=0}^\infty f_n=\frac{31}{10}\cdot\frac{314}{310}\cdot\frac{3141}{3140}\cdot\frac{31415}{31410}\cdot\frac{314159}{314150}\cdots$$

• Ooops, I just noticed that my answer is a duplicate. – Yves Daoust Mar 30 '18 at 15:29