How are irrational numbers, fixed points on the number line?

Question

Please, while answering/reading this question, only keep in mind my point of view only.

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The question is, that how come an irrational number on a number line is a fixed point.

To make things more clear, let us consider $\pi$. We see that it is "fixed" on the number line below:

And the value of $\pi$ is incessantly:

So let us take the number $3.14$. It will evidently be fixed on the number line, because it has a finite number of digits. But when we add $1$ at the end, it shifts a little forward. Similarly, when we add $5$ in the end, it again shifts a little forward. So it goes on indefinitely, infinitely and the point keeps on shifting. So how come, an irrational number even exists on a number line. (Because infinity is not defined, it can never be achieved in reality. For example, $\lim\limits_{n\to\infty}\frac1n=0$. But this doesn't mean that the function $y=\frac1n$ will ever become $0$, it will keep on going towards, tending towards it.)
P.S.: Please don't answer as: "Don't look at the problem this way...." etc etc. Just tell me, why doesn't this point of view satisfy the existence of irrational numbers.

Answer 1

This reminds me at the Zeno's paradox, in which Zeno said that dividing infinitely many times the distance between Achille and the turtle of a half he will not be able to move, because of the infinitude of the process (I'm not writing for now the details of the paradox).

The matter is that not all the processes which are infinite are automatically not "known" or divergent in some naive sense (this naive approach is the one that you have shown).

So this argument involves all the infinite processes that somehow try to calculate a quantity with an infinite number of steps, I'm talking about series and improper integrals, which are summations in a domain that is not limited. Not all the convergent functions you know have a finite sum (here sum stands for series or improper integral), the one which is of our interest is the class of function $f_{\alpha}(x):= \frac{1}{\alpha^{x}}$. It's almost easy to prove, once you know the basics of the theory, that this class of functions has a convergent integral (or series, if you are looking at it as a succession) at infinity only when $|\alpha| >1$. This means that with this theory you can evaluate an infinite sum with a rigorous mathematical approach. When we have to evaluate a number with an infinite number of decimals we can write this number as $n=IntegerPart,n_1n_2n_3...n_i...$ that is equal to $n=IntegerPart+ \sum_{i\in N} \frac{n_i}{10^i}$ which is a particular case of function defined above.

In particular, we know this sum, and we are able to find a "fixed" point on the real line for every number. Another example could be the number $0.999...$ which is equal to $1$. This theory has profound links with a lot of naive approaches on infinity and clarify most of misconceptions about "infinity".

Answer 2

You are rewriting the old Zeno paradoxes.

They come from the confusion between a well-defined finite value and an infinite process that defines it.

By the way, this phenomenon is not at all specific to irrational numbers, you could rephrase it for any unlimited decimal number, and even integers!

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Taking your example, $\lim_{n\to\infty}\frac1n=0$, the meaning is: "you can let the LHS expression be as close to $0$ as you want". You don't need to explicitly try all values of $n$, it suffices to prove that there exists one (and those that follow) that works.

The proof concerns an infinity of cases, but the proof itself is finite and all cases are settled in a single go.

Answer 3

You need a definition of the "real numbers" to answer this.It is possible to extend the reals to a larger arithmetic system in which 3,3.1,3.14,... does not have a limit point, so without a proper def'n of R we can get nowhere.Starting with Q, the rationals, we extend it to a larger set R while retaining all the usual arithmetic rules, including the order-properties (in particular if $A<B$ then $A+c<B+c$, and $(A<B\land C>0\to A C<B C$.) And we add another demand : That if S is any non-empty subset of R, and S has an upper bound in R, then S has a LEAST upper bound in R. There is more than one way to define R but all ways lead to isomorphic structures. That is why we speak of THE reals. I suggest you find an axiomatic definition of R and study it. The def'n will be your answer then.$.$