Using a specific Lemma, how to prove that between any two distinct rational numbers there exists an irrational number?

Question

This is probably a really straightforward problem that I'm messing up, but I'm reading a section in a book on irrational numbers and there's a proposition that follows from a lemma, and I'm struggling to see how it follows.

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Lemma 2.1: If m/n and r/s are rational, with r/s $\neq$ 0, then m/n + r/s $\times$ $\sqrt{2}$ is irrational.

Proposition 2.2: Between any two distinct rational numbers there exists an irrational number.

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Proof of Proposition 2.2:

Let the rational numbers be m/n and r/s, where m/n < r/s.

Then, m/n $<$ m/n + $\frac{\sqrt{2}}{2}$ (r/s - m/n) $<$ r/s

(because $\frac{\sqrt{2}}{2} < 1$), and the number in the middle is irrational by the lemma.

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How is it that you can apply the lemma? Since the lemma refers to the number m/n + r/s $\times$ $\sqrt{2}$, and the number in the middle is m/n + $\frac{\sqrt{2}}{2}$ (r/s - m/n). I assume there is some way to manipulate m/n + r/s $\times$ $\sqrt{2}$ into m/n + $\frac{\sqrt{2}}{2}$ (r/s - m/n), but I'm struggling to actually do so. Is there something I'm missing?

Thanks. Proof is from the book "The Foundations of Mathematics" by Stewart and Tall.

Answer 1

Notice that the number $\frac12(r/s-m/n)$ is rational. Call it $r/s$, at the same time trying to forget that a microsecond ago you used letters $r,s$ to denote something else. That is how you get that $m/n+\frac{\sqrt 2}{2}(r/s-m/n)$ is of the form $m/n+\sqrt 2\times r/s$.

To put it another way: your confusion is that you thought $r,s$ had to mean the same thing in both formulas, but the solution effectively introduces "a new $r$" and "a new $s$".

Answer 2

The lemma simply says that if $p$ and $q$ are rational, and $q\ne 0$, then $p+q\sqrt2$ is irrational; it doesn’t matter which rational numbers $p$ and $q$ are (as long as $q\ne 0$). Now take $p=\frac{m}n$ and $q=\frac{r}s-\frac{m}n$; these are rational, and by hypothesis $\frac{m}n<\frac{r}s$, so $q\ne 0$, so the lemma applies to let us conclude that $p+q\sqrt2=\frac{m}n+\sqrt2\left(\frac{r}s-\frac{m}n\right)$ is irrational.

Answer 3

The general idea would be this:

The rational numbers may be supposed to be positive and written as fractions (not necessarily irreducible) with the same denominator, say $\;\frac ad<\frac bd$.

Now, consider any irrational number $0<\alpha < 1$ (e.g. $\:\alpha =\frac1{\sqrt2}, \frac 1\pi, \frac 1{\mathrm e}$. Then the number $$c=\frac{\alpha a+(1-\alpha)b}d$$ belongs to the interval $\Bigl[\frac ad,\frac bd\Bigr]$ and is irrational, which is easy to prove by contrapositive:

If for a coefficient $\alpha$, $\:c=\dfrac bd+\alpha\dfrac{a-b}d$ is rational, then $\alpha$ is rational too.

Note : One can prove more or less the same way, that between two rational numbers, there exists a transcendental number.

Answer 4

I'm gonna rewrite the lemma without the denominators:

If $p,q \in \mathbb{Q}$ with $q\neq 0$ then $p + \sqrt{2}\times q$ is an irrational number.

You want to show that there exists an irrational number between two differents rational numbers $p<q$. The intuitive way to "create" a number between two other numbers given is making their mean: if you call $m=\frac{p+q}{2}$ then you have $p<m<q$

So, following your book, you have $p,q$ and you want to obtain two rationals $a,b$ which verify $p < a + \sqrt{2}\times b < q$. If you need to surpass $p$, the easiest way is taking $a = p$ and adding it something non-negative. So now, you need $b$ making $p + \sqrt{2}\times b < q$, in other words, $b < \frac{q-p}{\sqrt2}$, and you can make this possible if you take $b=\frac{q-p}{2}$, just because $\frac{1}{2} < \frac{1}{\sqrt2}$, or like you put in parenthesis, because $\frac{\sqrt{2}}{2} < 1$