I am confused at the very core my understanding of numbers for this particular subject, for some $n \in \mathbb N \backslash {\{1}\}$,the following appears to become an increasingly accurate approximation for any $x \in \mathbb R$, regardless of it's algebraicity or rationality:

$$\lim _{k\rightarrow \infty}\Biggl(n^{\lfloor\ln_n(x)\rfloor-k+1}\Bigl \lfloor xn^{k-\lfloor\ln_n(x)\rfloor-1}\Bigr\rfloor\Biggr)\approx{\{x}\}$$

Take for example, $n=2$ and $x=\pi^2$, if $k \lt N$ for some $N$,the difference between the expression inside the limit on left hand side and the fractional part of $\pi^2$ on the right hand side approaches $0$ as $N$ is made larger, which, by the epsilon test, would mean the limit is indeed true and we can replace the approximation sign with an equality.

But how can this be true? the left hand side is clearly rational, and ${\{\pi^2}\}$ is of course a transcendental number.

So an example of my assertion is that:

$$\lim _{k\rightarrow \infty}\Biggl(2^{\lfloor\frac{\ln(\pi^2)}{\ln(2)}\rfloor-k+1}\Bigl \lfloor {\pi}^{2}2^{k-\lfloor\frac{\ln(\pi^2)}{\ln(2)}\rfloor-1}\Bigr\rfloor\Biggr)=\pi^2-9$$

And I haven't been able to get my version of maple to make a float approximation, (and I am not allowed to know the specifics of the method for which the software makes float approximations anyway, as it is inbuilt code) the rational value that is computed for increasingly large values of $N$ for:

$$\lim _{k\rightarrow N}\Biggl(2^{\lfloor\frac{\ln(\pi^2)}{\ln(2)}\rfloor-k+1}\Bigl \lfloor {\pi}^{2}2^{k-\lfloor\frac{\ln(\pi^2)}{\ln(2)}\rfloor-1}\Bigr\rfloor\Biggr)$$

Does appear to support the assertion, however rigor is missing in a proof of such.

So my question is, how can I reconcile this contradiction? Is this an error in intuitive judgement I have made?

Later I will add the epsilon definition of a limit and assign the values appropriate to this example to confirm this to be true, and try to construct an epsilon-delta proof that will hopefully lead me to a stronger conclusion. I am not very good with these constructions, so help is appreciated greatly.

I guess another way Id like to put my question. can I somehow use the expression above as some kind of rationality test based on an epsilon-delta construction as per the definition of a limit?

This particular case is of special interest to me:

enter image description here

As the above expression is always zero if $N \lt \infty$.

  • 3
    $\begingroup$ I might want to note that infinite likes to muddy the lines a bit. Just because one side has purely rational stuff doesn't mean that, once you throw in infinities, that it maintains that. Consider a famous example: $$\sum_{n=1}^\infty \frac 1{n^2} = \frac{\pi^2}{6}$$ $\endgroup$ Apr 10, 2019 at 9:00
  • $\begingroup$ Sure it's most certainly a subject that's above my calibre mathematically, but I had previously thought that results as you mentioned were more due to something I didn't understand about infinite summation so for this I'm in a state of disbelief because there is no summation $\endgroup$ Apr 10, 2019 at 9:05
  • $\begingroup$ I guess another way Id like to put my question. can I somehow use the expression above as some kind of rationality test based on an epsilon-delta construction as per the definition of a limit? $\endgroup$ Apr 10, 2019 at 9:09

1 Answer 1


No epsilon-delta test can distinguish between rational and irrational numbers. Every interval contains infinitely many of those. It is easy to find infinite sums of rationals that converge to an irrational and infinite sums of irrationals that converge to rationals.

Dedekind cuts, one of the standard ways of constructing the reals from the rationals makes use of this. To construct $\sqrt 2$ you divide the positive rationals into two disjoint sets, those whose square is less than $2$ and those whose square is more than $2$.

  • $\begingroup$ So is there a means for constructing a transcendental number using disjoint subsets as with is done in Dedekind cuts? $\endgroup$ Apr 10, 2019 at 14:22
  • $\begingroup$ I can definitely see why no epsilon-delta test would be adequate for rationality, but is the same true for the algebraicity of a real? $\endgroup$ Apr 10, 2019 at 14:23
  • 2
    $\begingroup$ Yes, you construct all the irrationals the same way. There is nothing special about whether a number is algebraic or not as we do not use any algebra in the construction. The epsilon-delta proof likewise does not care about whether a number is algebraic. The important thing is that the rationals are dense in the reals, which gives you that you can find a sequence of rationals converging to any real you want, rational, algebraic, or transcendental. $\endgroup$ Apr 10, 2019 at 14:32
  • 1
    $\begingroup$ I think that, as written, the claim you are making is too strong. Transcendentals can be approximated by distinct rationals much better than rationals can. This is a quantitative difference that can be described via what is in essence an "$\epsilon$-$\delta$ test", isn't it? $\endgroup$ Apr 11, 2019 at 0:42
  • 1
    $\begingroup$ @AndrésE.Caicedo: some of them can be for a given denominator, but I think most cannot. I don't think we care about that here. OP seemed to have an intuition that the limit of a sequence of rationals must be rationals. $\endgroup$ Apr 11, 2019 at 0:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .