# Proof that for any interval (a,b) with a<b in the real numbers contains both rational and irrational numbers?

Background: We are assuming that the elements of $\mathbb{R}\setminus\mathbb{Q}$ are irrational number. If $x$ is irrational and $r$ is rational then $y=x+r$ is irrational. Also, if $r\neq 0$ then $rx$ is irrational as well. Likewise, if a number is irrational then its reciprocal is irrational as well.

Theorem: Every interval $(a,b)$, no matter how small, contains both rational and irrational numbers.

Proof. First we can see that the interval $(0,1)$ contains both rational and irrational, just select the numbers $1/2$ and $1/\sqrt{2}$. For the general interval $(a,b)$, think of $a$ and $b$ as cuts, that is $a=A\mid A^{'}$ and $b=B\mid B^{'}$ , such that $a<b$ . The fact that $a<b$ implies that $B\setminus A\neq\emptyset$ , moreover since $B$ has no greatest value if we select an a rational element $r\in B\setminus A$ then we can find another rational element $s\in B\setminus A$ such that $r<s$ . Thus $a\leqslant r<s\leqslant b$ . The transformation $$T:t\mapsto r+(s-r)t$$

sends the interval $(0,1)$ to the interval $(r,s)$ . Since $r,s,$ and $s-r$ are all rational, the transformation $T$ sends rationals to rationals and irrationals to irrationals. That is, $(r,s)$ contains and irrationals, and so does the larger interval $(a,b)$.$\space\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \square$

My Confusion: My first confusion is what does the transformation $T$ map from and to? If $t\in \mathbb R$ then for any value of $t$ we should have $r\leqslant r+(s-r)t \leqslant s$ right? However, if $t<0$, $r<0$, and $s<0$ then the preceding inequality shouldn't hold (unless I am missing something); and if the preceding inequality doesn't hold then what good is the transformation $T$ for showing that there is an irrational in the interval $[r,s]$?

• @Billy: I would post that as an answer. – Ross Millikan Jul 16 '13 at 2:34
• Okay, will do... – Billy Jul 16 '13 at 2:36
• @Billy OK, I see it now thanks. I kept trying to think of $t \in \mathbb R$. Thanks! – JimmyJackson Jul 16 '13 at 2:37

As the proof says, we actually define $T$ only on the open unit interval $(0,1)$, which it maps onto $(r,s)$. If we were to use the same formula to define $T$ on all of $\Bbb R$, we’d find that $T$ maps $\Bbb R$ bijectively onto $\Bbb R$.

• $T(0)=r$, $T(1)=s$.
• $T$ maps $(0,1)$ onto $(r,s)$ in order-preserving fashion.
• $T$ maps $(\leftarrow,0)$ onto $(\leftarrow,r)$ in order-preserving fashion: as $t$ moves leftward from $0$, $T(t)$ moves leftward from $r$.
• $T$ maps $(1,\to)$ onto $(s,\to)$ in order-preserving fashion: as $t$ moves rightward from $1$, $T(t)$ moves rightward from $s$.

But we don’t actually care about the first, third, and fourth of these points.

• With such a transformation, shouldn't there be infinitely many primes between the interval $(r,s)$? Since there are infinitely many positive primes, all of which have irrational square-roots, thus there should be infinitely many irrational numbers of the form $1/\sqrt{p}$ where $p\in P$ all of which will be on the interval (1,0). Hence, we can map infinitely many irrational numbers to the interval (r,s). So, could we restate the theorem to include this fact? – JimmyJackson Jul 16 '13 at 3:09
• You can, but it’s unnecessary. Once you’ve shown that every interval $(a,b)$ contains member of some set $S$, it’s not hard to show directly that every interval contains infinitely many members of $S$. Start with $(a,b)$; there’s some $s_0\in S\cap(a,b)$. Then there’s an $s_1\in(a,s_0)$, an $s_2\in(a,s_1)$, and so on. Taking $S$ to be the rationals, you get infinitely many rationals in every open interval; taking it to be the irrationals, you get infinitely many irrationals in every open interval. – Brian M. Scott Jul 16 '13 at 3:13
• It's amazing to think that there are infinitely many rational and irrational numbers between two points on the real number line. Thanks! – JimmyJackson Jul 16 '13 at 3:24
• @Jimmy: You’re welcome! – Brian M. Scott Jul 16 '13 at 3:25

Another method is to use the Archimedean axiom for the reals:

For any two positive reals $x$ and $y$, there is a positive integer $n$ such that $nx > y$.

We want to find a rational number $m/n$ such that $a < \frac{m}{n} < b$.

To do this, choose $n$ such that $n(b-a) > 1$, so that $\frac1{n} < b-a$.

Let $j = \lfloor na \rfloor$. Then $j \le na < j+1$ so $\frac{j}{n} \le a < \frac{j+1}{n} =\frac{j}{n}+\frac{1}{n} <a+(b-a) = b$.

Therefore the rational $\frac{j+1}{n}$ is between $a$ and $b$.

To find an irrational between $a$ and $b$, do the same as before, but choose $n$ such that $n(b-a) > 2$ and $j = \lfloor na \rfloor$. By the same reasoning, both $\frac{j+1}{n}$ and $\frac{j+2}{n}$ are between $a$ and $b$. Then $\frac{j+\sqrt{2}}{n}$ is an irrational that is between $a$ and $b$ since $a < \frac{j+1}{n} < \frac{j+\sqrt{2}}{n} < \frac{j+2}{n} < b$.

• What does the symbol ⌊na⌋ mean? – DrHAL Feb 11 '16 at 1:59
• Integer part or floor. $\lfloor x \rfloor$ is the largest integer $\le x$. – marty cohen Feb 11 '16 at 3:01

T maps from (0,1) to $(r,s)$, as the proof says! The point is that $a < r < r + (s-r)\frac{1}{\sqrt{2}} < r + (s-r)1 = s < b$, and the third number in that inequality is irrational (and lies in $(a,b)$). This only works because $0 < \frac{1}{\sqrt{2}} < 1$ — it doesn't work for all $t\in\mathbb{R}$.