I'm working out of John M. Howie's Real Analysis (2001). In section 1.4 (Exercise 1.5), we're asked the following:
Let $x$ and $y$ be real numbers, with $ x < y $. Show that, if $x$ and $y$ are rational, then there exists an irrational number $u$ such that $x < u < y$.
(Assuming) a proof by construction, the book gives the solution as:
Take $u = x + \frac{1}{\sqrt{2}}*(y-x)$ .
I'm not sure how to arrive at this construction. The book gives the Archimedean Property as follows:
$\forall x>0$ in $\mathbb{R}$ there exists $n$ in $\mathbb{N}$ such that $n > x$.
Regards, Brian
Ok, so I think the justification is as follows: Suppose the hypothesis is true. Now $$ x<y $$ $$ 0 <y-x $$ and $$0<\frac{1}{\sqrt{2}}<1$$ so $$ 0<\frac{1}{\sqrt{2}}(y-x)< (y-x).$$ Adding $x$ yields $$x < x+\frac{1}{\sqrt{2}}(y-x)< x+(y-x) $$ which simplifies to $$ x < x+\frac{1}{\sqrt{2}}(y-x) < y$$
so choose $$u = x+\frac{1}{\sqrt{2}}(y-x)$$ which is not rational as already shown.
Thanks for the help!