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As far as I know the set of rational numbers is closed under addition. That is, when we add rational numbers we always get another rational number. The sum $\frac{1}{1^{2}}+\frac{1}{2^{2}}+...$ is basically doing that a lot. How come it ends up being $\frac{\pi^{2}}{6}$, an irrational number?

coing2000
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  • One way to define set of real number is use limit of sequence of rational numbers. That mean, sum of a infinite series of rational number could possible converge to any real number. Of course it is possible that the sum is still a rational number. The statement "closed under additional" only apply to finite sum. – Abel Wong Oct 07 '22 at 03:58
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    The sum $3 + 0.1 + 0.04 + 0.001 + 0.0005 + 0.00009 + 0.000002 + \dots$ is also adding rational numbers a lot, but it equals $\pi$. – Misha Lavrov Oct 07 '22 at 04:05

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Saying that "when we add rational numbers we always get another rational number" is a shortcut to saying that when we add finitely many rational numbers we always get another rational number. It does not apply to infinite sums (as your example shows).

Another User
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The simple reason is that the rationals are closed under the addition of finitely many elements.

The rationals form a field, $\mathbb{Q}$, under the usual addition and multiplication. That means that $\mathbb{Q}$ under $+$ is a group. In the axioms of a group, one is that of closure, which is precisely stated as (for generic groups $(G,\ast)$)

$$\forall a,b \in G \text{ we have that } a \ast b \in G$$

This is a statement about (in the context of $\mathbb{Q}$) precisely two rationals summing up to another rational. Of course, induction lets us claim that $\sum_{i=1}^n q_i \in \mathbb{Q}$ whenever $q_i \in \mathbb{Q}$; the argument is fairly trivial.

However, the induction only says that any sum of finitely many elements are in $\mathbb{Q}$. Sure, you may take that "finitely" to be as big as you desire, be it $n=10$ or $n=10^{100}$ or whatever, but it is nonetheless a finite value.

As the example of $\sum_{n=1}^\infty 1/n^2 = \pi^2/6$ shows, this simply need not extend to infinite sums.

PrincessEev
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