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If you exclude 0 in the denominator of a fraction, then the set of rational numbers (a rational number is anything than be expressed as a fraction with integers in the numerator and the denominator) is closed with respect to addition, subtraction, multiplication, and division. For example, if j, k, m, and n are integers, then j/k + m/n is rational.

There are many irrational numbers, e.g. the square root of an integer not a perfect square or the trigonometric functions, or pi or Euler's constant e. These numbers can be calculated by summing an infinite series. The elements of such a series are rational numbers, fractions.

If an infinite series is composed of fractions and sums, and the set of rational numbers is closed with respect to fractions and sums, then how can the sum of an irrational series be irrational?

I have no doubt in my mind that this is so. I know how to prove that the square root of 2 is irrational, and I know that the Taylor series will converge to the square root of two. Although I don't remember how to prove that trig functions, pi and e are irrational, I know I could dig that up if I needed to. Suffice to write that there exists at least one Taylor series comprised of rational numbers that converges to a provably irrational number.

How does this happen?

It may be that What is meant by "The set of rational numbers is not closed when taking limits"? has the answer and I am unable to understand that. My question might also be a generalization of Prove the series ∑∞n=115n! converges to an irrational number but I am not sure I understand that answer, either.

user1928764
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    The closedness of the rational numbers with respect to addition etc applies to finite sums only. – Martin R Sep 13 '22 at 07:05
  • For any irrational number $x$ and any $\epsilon > 0$ there is rational number $y$ such that $|x-y|<\epsilon$. That's why limit of sequence consisting of rational numbers (partial sums) can be irrational number. – Ivan Kaznacheyeu Sep 13 '22 at 15:18

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An infinite sum of rational numbers is in the end just the limit of a sequence of rational numbers.

The rational numbers are not closed to limits. For example the sequence 1, 1.4, 1.41, 1.414, 1.4142, 1.41423 etc. converges to the square root of two, and not a rational number.

The reason is that the completeness axiom of the real numbers doesn't apply to the rational numbers (completeness axiom: Any bounded set of real numbers has a least upper bound in the real numbers).

gnasher729
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  • I also wanted write decimal representation of $\sqrt{2}$ as example but I thought about sum: $\sqrt{2}=1+4/10+1/100+4/1000+2/10000+...$. – Ivan Kaznacheyeu Sep 13 '22 at 15:20