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I was studying for my analysis mid-term paper and was going over the properties of real numbers. I was wondering how to prove the following statement: (Not a textbook problem, it just popped into my head.)

Given rational numbers $p$ and $q$ such that $p < q$, show that there exists an irrational number $r$ such that $p < r < q$.

I know some ways of proving it, like picking a known irrational and shifting it into the open interval $(p,q)$. I was wondering whether there is a way to prove it without referencing to any previously known irrationals. Specifically I am trying to construct a sequence of rational numbers which converges to a irrational in the interval $(p,q)$. Is there any way to do that?

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If $p>q$, then there are no numbers $r$ such that $p<r<q$. –  Chris Eagle Nov 13 '11 at 21:12
@chris edited.Thanks for pointing that out. –  Thiagarajan Nov 13 '11 at 21:15
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4 Answers

up vote 2 down vote accepted

Since there exists two rational numbers with finite decimal writing $x$ and $y$ such that $p < x < y < q$, consider $x = a_n a_{n-1} \dots a_0 . a_{-1} \dots a_{-m}$ and $y = b_N b_{N-1} \dots b_0.b_{-1} \dots b_{-M}$, the decimal expansions of $x$ and $y$. It is a theorem that a number of the form $z= c_{\ell} c_{\ell-1} \dots c_0 . c_{-1} c_{-2} \dots$ is an irrational if and only if its decimal expansion is infinite and non-periodic. Thus choose $x < z < y$ with $z$ an arbitrary non-periodic infinite decimal expansion (this is easy enough to do, just take $z$ with the right decimals for the first few and then choose whatever you like afterwards =D... one way to do this is just add decimals to the writing of $x$, i.e. $$ x = a_n a_{n-1} \dots a_0. a_{-1} \dots a_{-m} \quad \Longrightarrow \quad z = a_n a_{n-1} \dots a_0 . a_{-1} \dots a_{-m} c_{-m-1} \dots c_{-j} \dots. $$ and choose the decimals such that $z < y$, for instance by putting many zeros in $z$ until you are sure that whatever number you put, $z < y$) and you will get irrationals between $p$ and $q$.

Perhaps you might also be interested in this : http://en.wikipedia.org/wiki/Continued_fractions it explains the concept of continued fractions, which is another nice way of looking at real numbers.

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Ok thanks.That helped.So to construct a arbitrary irrational,I would basically need access to some kind of random process which generate those digits in my decimal expansion.Would it be right to assume,in the absence of such process I can't show existence of irrationals in my interval.As in,there is no deterministic way to,say,come with a convergent sequence which converges to my irrational,or construct a set such that my irrational is its supremum? –  Thiagarajan Nov 13 '11 at 21:34
Yes =) But there is one way to generate arbitrary processes such as the ones you want. For instance, let the decimals of $z$ fixed up to a certain point so that we have choice for whatever decimal we want, and then let $j$ be the last fixed in advance, so that we have arbitrary choice for $c_{j+k}$ with $k > 0$. Take any non-periodic sequence on $\mathbb N$, and define $c_{j+k} = 1$ if $k$ is in the sequence and $0$ if not. You do not need "randomness", you just need non-periodicity. This gives you a $z$. If you want to know if there exists such a sequence, consider $2^n$, the prime numbers.. –  Patrick Da Silva Nov 13 '11 at 21:38
There are lots of examples. –  Patrick Da Silva Nov 13 '11 at 21:40
Thank you :).That cleared up lot of difficulties I was having.I spent quite some time trying to construct convergent sequences but couldn't find a way to show that the limit is irrational.The comment about non-periodicity made me realize that varying structure(non-periodicity) is quite different from no structure(randomness).Thank you. –  Thiagarajan Nov 13 '11 at 21:49
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Look at all the translates $(p,q) + \frac{q-p}{2} \mathbb{Z}$. They cover all of $\mathbb{R}$. Since $(q-p)/2$ is rational, if all of $(p,q)$ were rational then there would be no irrational numbers.

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:D.Thanks.Liked the proof very much.Abstract and elegant. –  Thiagarajan Nov 13 '11 at 22:19
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The main idea is that all that matters here is what happens between $0$ and $1$, so that if we find an irrational between $0$ and $1$, we can do a combination of scaling and translation (by rationals, in both cases) to generate an irrational between $0$ and $1$, and these two operations preserve irrationality:

If you assume that, e.g., $2^{1/2}$ is irrational, and that both the sum of an irrational plus a rational is irrational (otherwise, the sum of two rationals is irrational), and that the ratio of an irrational by an integer is irrational, then $2^{1/2}-1$ is irrational, and $0<2^{1/2}-1<1$ . By the Archimedean principle, there is an integer $n$ with $n(x-y)>1$, so that there is an integer $z$ with $nx<z<ny$. Then $nx< nx+ 2^{1/2}-1<ny$, and now, dividing thru by $n$, we get: $x<x+ \frac{2^{1/2}-1}{n}<y$ is an irrational between $x$ and $y$.

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For some reason, the last part of the answer I included is not appearing . Hope the punchline is clear. –  C.O.Jones Nov 13 '11 at 22:37
Yes,it is.Thanks.I was looking for a proof such that I can avoid choosing a irrational number(like the square root of 2) and generating the required irrational as a function of it.Turns out there isn't. –  Thiagarajan Nov 13 '11 at 22:47
I couldn't think of another one. A good thing about this proof, I think, is that it can be generalized to other arguments, including the proof of the existence of a rational between two rationals, etc. –  C.O.Jones Nov 13 '11 at 23:00
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I want to mention how the notions of countable and uncountable can be used to give a solution. Because $\mathbb Q$ is countable, so is $(p,q)\cap \mathbb Q$. But $(p,q)$ is uncountable, because $$\displaystyle{f(x)=\frac{x-\frac{p+q}{2}}{(x-p)(q-x)}}$$ is a bijective map from $(p,q)$ to $\mathbb R$, and $\mathbb R$ is uncountable, or because $g(x)=\frac{x-p}{q-p}$ is a bijective map from $(p,q)$ to $(0,1)$, and $(0,1)$ is uncountable. Therefore $(p,q)\cap \mathbb Q\neq(p,q)$, meaning that there are irrational numbers in $(p,q)$.

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