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Is this a mathematically valid method of creating a sequence of rationals that converges to an irrational, or is it a handwaving argument?

I know that I could create a sequence by actually giving a formula.

Let $x$ be an irrational. Choose rationals between $x-1/n$ and $x+1/n$ for all $n$.

I find it a bit suspicious because I'm not specifying clearly what rational I'll be choosing. I can't say choose the smallest/largest.

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  • $\begingroup$ I think you mean "between $x$ and $x-(1/n)$". It's valid as an existence proof. It's not valid as a constructive proof. Do you want to know more about that? or do you want to see a constructive proof? $\endgroup$ – Gerry Myerson Nov 11 '18 at 11:32
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    $\begingroup$ You can use the sequence $\lfloor nx-1\rfloor/n$. $\endgroup$ – Mainkit Nov 11 '18 at 11:35
  • $\begingroup$ @GerryMyerson Yes, I meant that. I don't understand why this sequence valid because a sequence has to be a bijection between Natural Numbers and the terms of the sequence, and I'm not actually pinpoint where each natural number is going. $\endgroup$ – user1752323 Nov 11 '18 at 11:41
  • $\begingroup$ It is, as I wrote, valid as an existence proof – it shows that such a sequence exists. It's not valid as a constructive proof, as it doesn't actually construct the sequence for you. Mathematics is full of non-constructive existence proofs. $\endgroup$ – Gerry Myerson Nov 11 '18 at 11:45
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    $\begingroup$ @GerryMyerson Right you are. Sorry. $\endgroup$ – Thomas Nov 11 '18 at 17:09
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Put $y=x+\frac 1n$.

$\Bbb R$ is Archimedian,

$$\exists q>0 \; : \; q(y-x)>100$$

thus

$$qx<qx+100<qy$$

choose $p$ such that

$$qx<p<qx+100<qy$$

thus $$x<\frac pq<x+\frac 1n$$

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