# Proof of Density of Rational Numbers as given in Howie

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

• Hi Brian, do you understand why the solution is correct? In other words, the book tells us that $u$ is the irrational number we seek. Do you know how to prove that $u$ works? Do you know that $\frac{1}{\sqrt{2}}$ is irrational? – Prototank May 20 at 20:36
• Why do you need the Archimedean axiom here? All you need is to show that $x<u$ (which is clear), $u<y$ (a brief check), and $u\notin Q$ which follows from the fact that $\sqrt 2 \notin Q$. – lulu May 20 at 20:36
• Prototank, I am starting to think that I get why it works... my intuition was saying that starting with $x$, you just need to add a number that is guaranteed to be smaller than the difference between $x$ and $y$. And yes, I know that $\frac{1}{\sqrt{2}}$ is irrational, I just showed that in the last section. – fantasticasm89 May 20 at 20:39

Ok, so I think the justification is as follows: Suppose the hypothesis is true. Now $$x $$0 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.