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.
- Problem proving that there exists an irrational number between any two real numbers
- Proving that there is an irrational number between any two unequal rational numbers.
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.
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.