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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
goblin GONE
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Kira
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    For completeness ;) – Daniel Fischer Jun 24 '14 at 18:46
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    Those aren't sums, those are limits of sequences. – Git Gud Jun 24 '14 at 18:47
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    $\frac12$ is not a sum of integers, no matter how hard you try. – Asaf Karagila Jun 24 '14 at 18:49
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    What is the last rational number in one of your sums? – J. W. Perry Jun 24 '14 at 20:11
  • @J.W.Perry Probably there is no last. Because the sum never terminates – Kira Jun 24 '14 at 20:13
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    @Kira Very good. Case in point. You cannot represent these numbers as sums of rationals as you do not even know what the last [ pick a big number] of rationals would look like. – J. W. Perry Jun 24 '14 at 20:44
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    I don't understand the question. You begin by asking a question "why can't we represent all reals as infinite sums of rationals?" You can do so, therefore the question has a false premise and cannot be answered. Your second question "why do we need irrational numbers?" I simply don't understand. Why does who need irrationals to do what? Try re-stating your questions more precisely. – Eric Lippert Jun 24 '14 at 21:56
  • Your title question is yet a third question: "why are there irrational numbers?" Irrational numbers are defined as those reals which are not a ratio of two integers. There are irrational numbers; asking why there are irrational numbers, I don't know how to answer. Why is there anything at all? – Eric Lippert Jun 24 '14 at 22:00
  • @EricLippert I meant to say "if we can represent all real numbers as a sum, why we do specify that this number is irrational and this one is not". For example, $3.1=3+10^{-1}=\dfrac{31}{10}$. We can do the same for all real numbers. That's my question. And I know I am missing things here that's why I would like to get some helps. Thanks. – Kira Jun 24 '14 at 22:06
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    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 '14 at 22:15

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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 $\varepsilon = \frac{|\pi - t|}{2}$).

qaphla
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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.

David
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Joel
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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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Numbers that cannot be expressed rationally arise naturally as solutions of equations. The solution of $x^2=2$ & of $x(1-x)=1$, & any of an infinitude of specificable polynomial equations, cannot be expressed rationally. It can be proven directly, by a simple recipe, that no rational number can satisfy either of the two equations I have particularly cited; and the same can be proven similarly for the general case.

Moreover there is a class of numbers yet beyond this: the transcendental numbers, any of which can be shown to be not the solution of any polynomial equation comprising a finite number of terms having integer coefficients.

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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.

Yashbhatt
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