0
$\begingroup$

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

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.

|cite|improve this question
$\endgroup$
  • 5
    $\begingroup$ I don't see where your problem is... Each one of the numbers you take ($3$, $3.1$ etc.) is not $\pi$, but just an approximation. If you want, the number $\pi$ just stays there, it is you that are moving closer and closer to it. $\endgroup$ – Daniel Robert-Nicoud Nov 26 '15 at 13:05
  • 5
    $\begingroup$ You have asked: "how come an irrational number on a number line is a fixed point". But you have not given a mathematical definition of a "fixed point". Perhaps you don't have one, perhaps you just mean "one, fixed position on the number line"? If so, every real number corresponds to one fixed position on the number line, irrespective of whether it has a finite decimal expansion, an infinite decimal expansion, or anything else. If that's not what you mean, you'll have to explain your "point of view" better, or else you will not get any satisfactory answer. $\endgroup$ – Lee Mosher Nov 26 '15 at 13:13
  • 2
    $\begingroup$ Do you have a problem with $1/3 = 0.333333333...$ ? $\endgroup$ – mercio Nov 26 '15 at 13:15
  • 5
    $\begingroup$ Also, the answers you will get will all be a rephrasing of "you're looking at real numbers the wrong way". $\endgroup$ – Daniel Robert-Nicoud Nov 26 '15 at 13:16
  • 2
    $\begingroup$ Every one of those approximations is a different real number: $3$, $3.1$, $3.14$, and so on. Each of them is different from $\pi$. Each of them corresponds to a different point on the number line, and those points are different from the point corresponding to $\pi$. $\endgroup$ – Lee Mosher Nov 26 '15 at 13:19
6
$\begingroup$

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".

|cite|improve this answer
$\endgroup$
3
$\begingroup$

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!

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Yes, but it still doesn't answer my confusion. $\endgroup$ – Aditya Agarwal Nov 26 '15 at 13:08
  • 3
    $\begingroup$ Just start from $1$ on the number line. go Right by $\frac{1}{2}$ units. from there again go right by $\frac{1}{2^2}$ units. Again go right by $\frac{1}{2^3}$ units. Eventually after infinite such moves you will end up in $2$. This process cannot be done practically since there are infinite moves. But still $2$ exists on number line. The same is with $\pi$. $\endgroup$ – Ekaveera Kumar Sharma Nov 26 '15 at 14:20
1
$\begingroup$

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.$.$

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.