# Why there are irrational numbers?

I do not quite get it. Why can't we represent all real numbers as a sum of rational numbers? Why do we need irrational numbers?

For example,

• $\pi=3.14159265358\cdots=3+10^{-1}+4*10^{-2}+10^{-3}+5*10^{-4}+\cdots$
• $e=2.71828182846\cdots=2+7*10^{-1}+10^{-2}+8*10^{-3}+2*10^{-4}+\cdots$
• And so on
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For completeness ;) –  Daniel Fischer Jun 24 at 18:46
Those aren't sums, those are limits of sequences. –  Git Gud Jun 24 at 18:47
$\frac12$ is not a sum of integers, no matter how hard you try. –  Asaf Karagila Jun 24 at 18:49
What is the last rational number in one of your sums? –  J. W. Perry Jun 24 at 20:11
That is much more clear. The answer to your question is: any finite sum of rationals is a rational, but that is not the case for the limit of an infinite sum. More generally, get into your head now that a property shared by every member of a set and a property of the limit of that set need not be the same. Every one of your finite sums is rational, but the limit need not be. Accidentally treating set elements and limits as the same thing is the cause of a great many mathematical mistakes. –  Eric Lippert Jun 24 at 22:15

You can represent any real number as some convergent sequence of rational numbers, as you do above. However, irrational numbers are those numbers that cannot be expressed as any finite such sequence. However long of a finite sequence of rational numbers approaching $\pi$ you have, there is a positive real number $\varepsilon$ such that the last term in the sequence is at least $\varepsilon$ away from $\pi$ (In particular, we can take the last term of the sequence, $t$, and take $\varepsilon = \frac{|\pi - t|}{2}$).

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For a long time people believed that the rational numbers were enough to perform any arithmetic operation we needed. It wasn't until 500 BC that the Pythagoreans started to become aware that the rational numbers were not quite sufficient.

For instance, the Pythagoreans tried to compute something that seemed benign. They wanted to find the length of the diagonal of a square with sides of unit length. We know now (by the Pythagorean theorem) that this should be the number we represent as $\sqrt{2}$. The Pythagoreans tried to find a rational number corresponding to this and came to a contradiction. It's said that the Pythagoreans were so upset that they drowned the person who discovered this. I have also heard speculation that this is why Greek mathematics was so focused on geometry, since this is something you can "see" and away from uncomfortable concepts like irrational numbers.

This was the first instance where irrational number started to seem necessary. The next big irrational number to come forward was the number $\pi$. This was in 350 BC, and the first to try to approximate this number was Archimedes. This is where the estimate 22/7 came from. He also had a better estimate than that, it escapes me at the moment. The Bible also estimates this as 3, when it describes a fountain in Solomon's temple. I think the proof of the irrationality of $\pi$ had to wait for Lambert in the 1700s.

Irrational numbers came about to solve problems that rational numbers were not up to the task to do. Early mathematicians resisted the concept, but they have been well accepted for the past millennium at least.

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Here is Archimedes' method of calculating the value of $\pi$. –  Yashbhatt Jun 30 at 15:56
@David thanks for the edit! Can't believe I missed that. –  Joel Jul 1 at 13:49

Some relevant definitions seem to be in order.

Real number: Any number on the continuous real line from $-\infty$ to $\infty$.

Integer: A real number which can be expressed without a fractional component.

Rational number: A number which can be expressed as a ratio of two integers. Note that integers themselves are rational, since we can express any integer $n$ as $\frac{n}{1}$.

Irrational number: A number which cannot be expressed as a ratio of two integers.

Now, in particular, in your question $k\cdot10^{-n}$ is $\textit{not}$ an integer where $k$ and $n$ are positive integers and $k$ is a digit between $1$ and $9$. These are instead just $\textit{rational}$ numbers since $k\cdot10^{-n}=\dfrac{k}{10^n}$, which is a ratio of integers.

Upsettingly for Pythagoras, irrational numbers do exist. Here is the standard proof that $\sqrt{2}$ is irrational.

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Because you can not represent them as p/q with both p and q integers. That is why they are called irrational numbers. Not representable as p/q. 0.33333....is 1/3. root 2 can not be written as p/q where p and q are integers.

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