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Let $a$ and $b$ be rationals. Prove that if $a<b$ then there exists an irrational $x$ such that $a<x<b$.

Before anyone starts linking other posts, I have gone through the following posts.

However, I think I have a different solution and I would be grateful if anyone could check it.

Note that $$\lim_{x\rightarrow \infty} \frac{\sqrt{2}}{2^x}=\lim_{x\rightarrow \infty} 2^{-x}\sqrt{2}=\sqrt{2}\lim_{x\rightarrow \infty} 2^{-x}=0.$$

And now we state and prove our claim which proves the statement too.

Claim: Between any two rationals $a$ and $b$, with $a<b$, $\exists \alpha$ which is irrational, such that $\alpha=a+\frac{\sqrt{2}}{2^n}$ and $a<\alpha<b$.

Proof: We will prove by contradiction. Suppose for all $n$, we get that $a+\frac{\sqrt{2}}{2^n}>b$. Then consider the set $$S=\{a+\frac{\sqrt{2}}{2^n}, n\in \Bbb{N}\}.$$ Note that $S$ is lower bounded by $b$ and hence by completeness axiom, $\exists $ inf$S$. So we are saying that $$\lim_{n\rightarrow \infty} a+\frac{\sqrt{2}}{2^n}\ge \text{Inf}S\ge b.$$ But as notes above, $$\lim_{n\rightarrow \infty} a+\frac{\sqrt{2}}{2^n}=a\implies a\ge b.$$ Not possible.

Hence $\exists n_0$ such that $$a<a+\frac{\sqrt{2}}{2^{n_0}}<b.$$ And clearly, $a+\frac{\sqrt{2}}{2^{n_0}}$ is irrational.

Sunaina Pati
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  • "Note that $S$ is lower bounded by $b$" $;-;$ Maybe, but that depends on how you argue that point, which you did not elaborate in the post. In any case, it would be easier to write that idea as a forward proof, rather than a proof by contradiction. – dxiv Jun 27 '23 at 05:23
  • @dxiv makes sense. Let me elaborate it. – Sunaina Pati Jun 27 '23 at 05:24
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    That looks fine now. But I still think it would be more direct to argue that $\sqrt{2} / 2^n \to 0$ means there exists an $n_0$ such that $\sqrt{2} / 2^{n_0} \lt b-a \iff a \lt a + \sqrt{2} / 2^{n_0} \lt b$. – dxiv Jun 27 '23 at 05:34
  • To echo @dxiv's post, you just need to show that there exists a sequence of irrationals that converge to zero. – copper.hat Jun 27 '23 at 05:36
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    @dxiv got it! I was thinking of going that way but didn't know how to write it rigorously but got it! – Sunaina Pati Jun 27 '23 at 05:41
  • Please be more specific about what you want to be checked in your proof, compare Best way of asking "check my proof" questions. – Martin R Jun 27 '23 at 06:55

1 Answers1

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We have $a < a + k(b-a) < b$ for every $k$ with $0 < k < 1$. Take any irrational $k$, say $k = \surd\frac12$, and the intermediate number is irrational if $a$ and $b$ are rational.

John Bentin
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gnasher729
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    That is well known (e.g. https://math.stackexchange.com/a/1106178/42969), but here the specific question is “I think I have a different solution and I would be grateful if anyone could check it.” – Martin R Jun 27 '23 at 05:52