Proof there is a irrational number $r$ in $a < b$ so that $a < r < b$

Question

I have a understanding problem of the following proof versus my own attempt. I don't get why you need another intervall in between in which you "place" the shifted version of $\sqrt2$

$\sqrt2$ is irrational (already prooven)

$1 < \sqrt2 < 2~~ \mid ~~-1$

$0 < \sqrt2 -1 < 1$

There are $c$ and $d \in \mathbb{Q}$ with $a < c < d < b$ such that $(c,d) \subseteq (a,b)$.

$0 < \sqrt2 -1 < 1 ~~ \mid \cdot ~~(d-c)$

$0 < (\sqrt2 -1)\cdot(d-c) < d-c$

[$\sqrt2 -1 := x$]

$0 < x\cdot(d-c) < d-c ~~ \mid +~~c$

$c < x\cdot(d-c)+c < d$

$[x\cdot(d-c)+c := r]$

So now we have to proof $r$ is irrational. We can do that by Proof by Contradiction:

Let $r$ be rational. Then $\frac{r-c}{d-c}-1$ must be rational too.

$\frac{x\cdot(d-c)+c-c}{d-c}-1 = x-1 = (\sqrt2 + 1)-1 =\sqrt2 \,\,\,\,\,\,\,\,\,\,\,\,\,\square$

Now my attempt without the $c$ and $d$ part:

$1 < \sqrt2 <2 ~~ \mid - ~~ 1$

$0 < \sqrt2-1 <1 ~~\mid \cdot ~~ (b-a)$

$0 < (\sqrt2-1)\cdot(b-a) <b-a ~~\mid + ~~ a$

$a < (\sqrt2-1)\cdot(b-a)+a <b$

Suppose $(\sqrt2-1)\cdot(b-a) =: r$ is rational, then $\frac{r-a}{b-a}+1$ must be rational too.

$\frac{(\sqrt2-1)\cdot(b-a)+a-a}{b-a}+1 = (\sqrt2-1)+1 = \sqrt2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\square$

Answer 1

If $r$ is rational, $\displaystyle \frac{r-c}{d-c}$ must be rational, because both $c$ and $d$ are rational numbers. And then, you get a contradiction. So $r$ becomes irrational.

However, in your "proof", if $r$ is rational, then $\displaystyle \frac{r-a}{b-a}$ may not be rational when $a$ is rational but $b$ is irrational. In that case, $r-a$ would be rational and $b-a$ would be irrational.

For instance, take $a=-1$ and $b=\sqrt 2$. Then $(\sqrt 2-1) \cdot (b-a) = (\sqrt 2-1)(\sqrt 2 +1) = 1=r$. So, $r$ becomes a rational number. So, you cannot derive a contradiction.

Answer 2

Here's an alternative argument:

If $a\in\mathbb{Q}$, pick $n$ such that $\frac{\sqrt{2}}{n}<(b-a)$. Then $a<a+\frac{\sqrt{2}}{n}<b$ and $a+\frac{\sqrt{2}}{n}$ is irrational.

If $a\not\in\mathbb{Q}$, pick $n$ such that $\frac{1}{n}<(b-a)$. Then $a<a+\frac{1}{n}<b$ and $a+\frac{1}{n}$ is irrational.