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I can construct a Rational like $3/4$. And then I can construct anther one like $31/41$, and then another like $311/411$.

I can envisage a Rational whose numerator is $31111111...$ and denominator is $41111111...$, where the $1$s continue indefinitely and equally.

Does this number exist, and is it a Rational?

PrincessEev
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Al.
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    There are no "infinite Naturals". – Angina Seng May 03 '19 at 06:37
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    To whoever voted to close this as unclear: The language might not be conventional, but to me it's pretty clear that this question is about a limit, and it's pretty clear what limit that is. It just doesn't use $\lim$ in a formula. – Arthur May 03 '19 at 06:39

2 Answers2

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This number does not exist, simply for the reason that - whatever number it represents - $311111...$ is not a natural number (or an integer or a real number or ... etc.). Same for $411111...$ - and of course this also means their ratio is not a rational.

By definition, if $n \in \Bbb N$, $n$ is finite. There is no "infinite natural number."

But let's push this aside for the moment; we could get into a whole discussion about how this notation isn't well-defined, but I think what you're trying to get at is something closer to this.

Let us define a sequence of rationals $(a_n)$ by the pattern you describe:

$$\frac{3}{4} , \frac{31}{41} , \frac{311}{411} , \frac{3111}{4111} , ...$$

Then we can ask: does $(a_n)$ converge, and if so, what to? That is to say, what is $\lim \limits_{n\to\infty} a_n$, provided it exists?

For convenience, let $a_0$ be the first term of the sequence; then in the numerator/denominator of the $n^{th}$ term, $n$ ones appear. Then as a result, this motivates a formula for the $n^{th}$ term in the sequence:

$$a_n = \frac{3\times 10^{n+1} + \sum \limits_{k=0}^n 10^k}{4\times 10^{n+1} + \sum\limits_{k=0}^n 10^k}$$

Note that the summations are geometric series, with ratio $10$. Then from known formulas about finite geometric series,

$$\sum_{k=0}^n 10^k = \frac{10^{n+1}-1}{9}$$

Then

$$a_n = \frac{3\times 10^{n+1} + \frac{10^{n+1}-1}{9}}{4\times 10^{n+1} + \frac{10^{n+1}-1}{9}}$$

We can multiply through the denominator and numerator by nine to get rid of that pesky complex fraction; a simplification results by combining like terms:

$$a_n = \frac{27\times 10^{n+1} + 10^{n+1} - 9}{36\times 10^{n+1} + 10^{n+1} - 9} = \frac{28\times 10^{n+1} - 9}{37\times 10^{n+1} - 9}$$

Bear in mind - this still gives us a generic term of your sequence of rationals! So finding the limit of the original sequence is equivalent to finding the limit of this!

As $n \to \infty$, it's obvious that the $(-9)$ terms have negligible effect: the growth of $a_n$ is entirely dominated by the powers of ten. As a result of this asymptotic equivalence, we claim

$$\lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{28\times 10^{n+1} - 9}{37\times 10^{n+1} - 9} = \lim_{n\to\infty} \frac{28\times 10^{n+1} }{37\times 10^{n+1} } = \lim_{n\to\infty} \frac{28}{37} = \frac{28}{37}$$

In this sense, the number you're trying to construct in your original approach is $28/37$. Bear in mind that I mean this very literally - "in this sense" - as in your original sense it's simply not well-defined enough to even bother discussing. So in truth most of this post relies on the underlying premise of what I assume you're trying to get at.

PrincessEev
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  • I am glad that you understood my answer, in the end. –  May 03 '19 at 07:36
  • Yes, I will admit it was a happy surprise in the end to find out where your answer came from. I still maintain that the explanation was important though. – PrincessEev May 03 '19 at 07:37
  • My answer is meant to address curious people. What does is take to compute $\dfrac{28000-1}9$ and understand the "why" ??? I am not sure that a ready-made answer is more pedagogical. –  May 03 '19 at 07:48
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$$\lim_{n\to\infty}\frac{\dfrac{28\cdot10^n-1}9}{\dfrac{37\cdot10^n-1}9}=\frac{28}{37}$$ is a rational.