4

$$e=1+\frac{1}{1!}+\frac{1}{2!}+...$$

The rational numbers are closed under addition that means RHS should be rational number but LHS is an irrational number. So, I think that in RHS there are infinitely many terms that's why this is happening.

More such examples:

$$\frac{\pi^2}{6}=1+\frac{1}{2^2}+\dfrac{1}{3^2}+...$$
$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-...$$

How can you show systematically that such infinite rational terms add up to give an irrational number?

I'm not sure even how to start.

Sorry if this question is too trivial for MSE.

  • 3
    An infinite sum of rational numbers can be rational too. So there is nothing to prove here. – Kavi Rama Murthy Jan 06 '21 at 07:15
  • $1 = \frac 12 + \frac 14 + \frac 18 + \frac{1}{16} + \cdots$ – Martin R Jan 06 '21 at 07:17
  • Yes, sir I know it can be rational too. But how does an irrational number come in the case of e by adding rational terms? –  Jan 06 '21 at 07:18
  • 1
    You are not calculating the actual sum. You are calculating the limit. –  Jan 06 '21 at 07:22
  • Possible duplicate of Sum of rational numbers – Martin R Jan 06 '21 at 07:25
  • Consider $\pi = 3.1415...... = 3 + \frac 1{10} + \frac 4{100} + \frac 1{1000} + \frac 5{10000}+ ....$. – fleablood Jan 06 '21 at 07:58
  • When you study Real Analysis you will learn that the fundamental property and definition of the Real Numbers, is that every real number is a limit of a converging sequence of rational numbers. Each of the finite sums is rational. But the infinite sum is the limit of the finite sums. And that is a limit of a converging sequence of rational numbers. And all real numbers (including and especially the irrational numbers) are limits of a converging sequence of rational numbers. – fleablood Jan 06 '21 at 08:02
  • @HansLundmark No, the answers here are quite different from your linked question. The accepted answer here actually address my problem. –  Jan 06 '21 at 09:17
  • "How can you show systematically that such infinite rational terms add up to give an irrational number?" That's actually fairly difficult and depends on the particular number. We don't even know if $\pi +e$ is rational or irrational. – fleablood Jan 06 '21 at 16:51

3 Answers3

2

The rational numbers are closed under addition

True.

that means RHS should be rational number

False.

The right hand side is not an addition operation. It is a limit operation, i.e.

$$\sum_{n=0}^\infty a_n$$ is defined as $$\sum_{n=0}^\infty a_n = \lim_{N\to\infty} \sum_{n=0}^N a_n.$$

Since rational numbers are not closed under limits, there is no contradiction.

In fact, any real number $x\in\mathbb R$ can be the result of an infinite sum. In fact, that's what decimal expansions are. They are a way of writing any real number as a sum of powers of $10$.

How can you show systematically that such infinite rational terms add up to give an irrational number?

What do you mean by "such" in the sentence above? In general, an infinite sum can have either a rational sum (see $\sum_{n=0}^\infty 2^{-n}$) or an irrational sum (the ones above). In fact, for any real number $x\in\mathbb R$, you can find an infinite sum that sums up to $x$. So, since there is no way to "systematically" prove an arbitrary number is irrational, we can't expect to have a systematic proof that an arbitrary sum is irrational.

5xum
  • 123,496
  • 6
  • 128
  • 204
2

Being closed under addition means that two rational numbers added together is a rational.

And by induction that means any finite number of rational numbers added together is a rational. (We add up the first two and that makes one number, then we add the third and that is one number, than the fourth).

But an infinite sum is .... not actually a sum. The "sum" $\sum_{k=1}^\infty a_k$ is actually defined as a limit $\lim_{n\to \infty} \sum_{k=1}^n a_k$ (assuming that such a limit actually exists)

Now it is true that if all $a_k$ are rational, and if we write $q_n = \sum_{k=1}^n a-k$ then all $q_n$s are rational. But then $\sum_{k=1}^{\infty} a_k = \lim_{n\to \infty} q_n$. So there is no reason at all, that a converging sequence of rational numbers will converge to rational number!

Indeed that is the fundamental property of real analysis! Every real number, irrationals included, is actually a limit of a converging sequence of rational numbers.

And if we think about it that is not a surprise. Consider the decimal expansion of an irrational number. We've been dealing with those for years!

$3.1415..... = 3 + \frac 1{10} + \frac 4{100} + \frac 1{1000} + \frac 5{10000}+.....$

fleablood
  • 124,253
0

A finite sum of rational numbers is rational as $\mathbb Q$ is a field. So the fact that you won't get an irrational number by summing up a finite number of rational numbers is always the case.

An infinite sum of rational numbers can be either rational or irrational.

Your second question is about how to prove that a number is not rational. And this is a difficult topic. It usually goes through proceeding by contradiction.

mathcounterexamples.net
  • 70,018