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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

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  • $\begingroup$ 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? $\endgroup$ – Prototank May 20 at 20:36
  • $\begingroup$ 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 $\sqrt 2 \notin Q$. $\endgroup$ – lulu May 20 at 20:36
  • $\begingroup$ 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. $\endgroup$ – fantasticasm89 May 20 at 20:39
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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!

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