# $\pi \not\in \mathbb{Q}$?

I've taken this fact for granted; some thinking tells me that indeed, I cannot express it with fractions. So it's not rational.

But well, if $p,q \in \mathbb{Q}$ then $p+q \in \mathbb{Q}$ since it is closed under addition, yes?

Then, it gave rise to a question. What happens if I "infinitely summed rationals"? Namely, for $p_1,p_2,..., \in \mathbb{Q}$, if I did $\sum p_i$ from $i=1$ to $i=\infty$, what happens? Can this still be in $\mathbb{Q}$? It seems so, to me, since I can iterate the argument that $\mathbb{Q}$ is closed under addition, and thus should not limit myself to only two rationals.

But as always, considering $\infty$ gives rise to weird consequences or exceptions quite often.

So my point is, $\pi =3.14....$ and each digit can be represented as a rational...right? $3,0.1,0.04...$ and so on. So I have $3, \frac{1}{10}, \frac{4}{100}...$.

$\pi$ is a sum of all of this. So, since each digit is in $\mathbb{Q}$...the sum should also be in $\mathbb{Q}$... which can't be true.

I'm sure there's a simple explanation to debunk this, it's more of like a quiz to me it seems. I feel something similar to the classic $1=0.99999$ theory(which, at first glance, seems false). But can someone give me a comprehensive answer?

• No $\mathbb Q$ is not closed under infinite summation. Apr 1 '16 at 15:02
• Iterating the argument just says that you can add any finite number of fractions and get a fraction. Apr 1 '16 at 15:03
• Hi all, so, my question is more of like a counterexample for the claim that infinite sums of rationals are not rational then? Apr 1 '16 at 19:10
• You provided your own counterexample. Every real number is an infinite sum of rationals given by its decimal expansion. Apr 1 '16 at 19:20

$\mathbb{Q}$ is closed under the addition of two elements, and so is closed under the addition of a finite number of elements

But it is not necessarily closed under the addition of a countably infinite terms, even when the partial sums are increasing and bounded above, as shown by your example

In other words, the rationals are not complete, and Cauchy sequences of rationals do not necessarily converge to a rational

It is a common error to use induction to prove something about infinity. Using repeated applications of the rule $a, b \in \mathbb{Q} \Rightarrow a + b \in \mathbb{Q}$ you can only prove that any finite sum of rational numbers is again rational, but nothing about infinite series.

Indeed, it is not true that an infinite sum of rational numbers needs to be a rational number; you're example with $3+.1+.04+\cdots$ proves this. While it is true that any finite sum of rationals is again a rational, this is false for the infinite case.

Remember that an infinite sum is just like an infinite sequence of partial sums. So we can consider infinite sequences of rational numbers and examine what kinds of limits these sequences have. The rationals are dense in $\mathbb{R}$, so every $x\in\mathbb{R}$ is the limit of some sequence of rationals, or, equivalently, for any real number $x$, we can find a sequence of rationals $(q_{n})_{n\in\mathbb{N}}$ such that $\underset{n\geq 1}{\sum}q_{n}=x$.

I hope this helps.

The distinction here is the difference between finite sums and infinite sums. You correctly observe that finite sums of rational numbers are again rational numbers. However, infinite sums of rational numbers need not be rational.

This is because, in some sense, real numbers are what happens when you take rational numbers and add infinity. Real numbers are the limit points of rational numbers --- they lie at the end of sequences of rational numbers. In fact, one way of defining the real numbers is as equivalence classes of Cauchy sequences of rational numbers.

• «and add infinity» does not mean anything, really :-| Apr 1 '16 at 15:23
• @MarianoSuárez-Alvarez Well of course not, that's why I said "in some sense" :P
– Neal
Apr 1 '16 at 15:27