# Method to create sequence of rationals converging to irrational

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.

• 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? – Gerry Myerson Nov 11 '18 at 11:32
• You can use the sequence $\lfloor nx-1\rfloor/n$. – Mainkit Nov 11 '18 at 11:35
• @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. – user1752323 Nov 11 '18 at 11:41
• 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. – Gerry Myerson Nov 11 '18 at 11:45
• @GerryMyerson Right you are. Sorry. – Thomas Nov 11 '18 at 17:09

Put $$y=x+\frac 1n$$.

$$\Bbb R$$ is Archimedian,

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

thus

$$qx

choose $$p$$ such that

$$qx

thus $$x<\frac pq