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For every $r,s \in \mathbb{Q}$ with $r<s$, there is an irrational number $u$ for which $r < u < s$.

Can this be proved as follows:

Take $r=\frac{a}{b}$, $s=\frac{c}{d}$ for $a,b,c,d \in \mathbb{Z}$. Then take $u$ to be $\frac{a}{b}+\dfrac{\frac{c}{d}-\frac{a}{b}}{\pi}$. Since $\pi>1$ and since $\frac{c}{d}-\frac{a}{b}$ is always positive, $0<\dfrac{\frac{c}{d}-\frac{a}{b}}{\pi} < \frac{c}{d}-\frac{a}{b}$, i.e. less than the difference of $r$ and $s$, so $r<u=\frac{a}{b}+\dfrac{\frac{c}{d}-\frac{a}{b}}{\pi}<s$.

Lemma 1: The quotient of a (edit: non-zero) rational number and $\pi$ is irrational.

Proof (by contradiction): Assume the quotient of a rational number and π is rational, i.e. $\dfrac{\frac{a}{b}}{\pi}=\dfrac{c}{d}$ for $a,b,c,d \in \mathbb{Z}$. Then $\pi=\dfrac{ad}{bc}$. Since $ad$, $bc$ are both integers, $\pi$ is rational. A contradiction.

Lemma 2: The sum of a rational and an irrational number is irrational.

Proof (by contradiction): Assume the sum of a rational and an irrational number is rational, i.e. $\dfrac{a}{b}+i=\dfrac{c}{d}$ for $a,b,c,d \in \mathbb{Z}$ and some irrational number $i$. Then $i=\dfrac{cb-ad}{db}$. Since $db$ and $(cb-ad)$ are both integers, $i$ is rational. A contradiction.

We show that $u$ is irrational. Since $\pi$ is irrational, $\dfrac{\frac{c}{d}-\frac{a}{b}}{\pi}$ and hence $\frac{a}{b}+\dfrac{\frac{c}{d}-\frac{a}{b}}{\pi}$ are irrational by Lemma 1 and 2 respectively. $\blacksquare$

Max
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  • If I will prove it, I'll go this way. We have $\frac{r}{\sqrt{2}}<\frac{s}{\sqrt{2}}$ and so we can find (using the fact $\Bbb Q$ is dense in $\Bbb R$) a rational number $q$ such that $$\frac{r}{\sqrt{2}}<q<\frac{s}{\sqrt{2}}$$ which implies that $$r<q\sqrt{2}<s.$$. We take $u=q\sqrt{2}$. Its clear that $u$ is an irrational number. – Juniven Acapulco Feb 19 '17 at 13:24
  • Cool. Thanks for the suggestion. Haven't learned about density yet, but evidently it makes things easier. – Max Feb 19 '17 at 13:35
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    You still have to prove that $\pi$ is irrational. ;-) Using $\sqrt{2}$ (which is irrational and greater than $1$) is easier. – egreg Feb 19 '17 at 14:38
  • egreg, oh yes, you're right. That way it's more watertight. – Max Feb 19 '17 at 14:41

1 Answers1

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Nice approach! Good use of the results you have at hand, plus Lemmas. However, any irrational greater than $1$ will suffice, so you may wish to choose one you can prove to be irrational.

One critique of your first Lemma: The quotient of a non-zero rational number and $\pi$ (or any irrational number $\alpha$) is irrational. Do you see where that assumption is necessary in your proof of the Lemma?

Cameron Buie
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  • Thank you for the critique. Yes, I see why it's necessary. Otherwise the quotient would be 0, which is rational. – Max Feb 19 '17 at 16:17