Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $a$ and $b$ be reals with $a<b$. Show that there are infinitely many rationals $x$ such that $a<x<b$.

My plan of action was to assume that $x$ is the smallest such rational and find another rational in the interval $(a, x)$, but I am struggling to make it work. A hint will be much preferred to a full solution.

share|improve this question
    
Just take the average of $a$ and $x$. It's not hard to show that's greater than $a$ and smaller than $x$. –  crf Sep 20 '12 at 0:30
3  
@crf The average of $a$ and $x$ need not be rational. –  Jon Sep 20 '12 at 0:32
1  
! sorry. Somehow I missed the whole thing about them being rationals. –  crf Sep 20 '12 at 0:35

6 Answers 6

This PDF may help, but I improved the picture :

enter image description here

Whence we observe by inspection : Because $\color{#1FB4BF}{1/n} < \color{#E431D2}{(y - x)}$ for all $x, y \in \mathbb{R}$,
thus $ \; k/n < x \iff k/n + \color{#1FB4BF}{1/n}< x + \color{#E431D2}{(y - x)}$.
In summary, $k/n + \color{#1FB4BF}{1/n} = \dfrac{k+1}{n}$ $\in (x,y)$ is the rational number desired.

share|improve this answer

Note that the real numbers are an Archimedean field, so for any real number $r$ we have some integer $n>r$. This means that for any real number $\epsilon>0$, we have some $n>1/\epsilon$, so $1/n<\epsilon$. Furthermore, the rationals are dense in the reals, so we can find some rational $x$ such that $a<x<b$.

Let $n$ be such that $1/n<b-x$. Then $x+\frac{1}{n},x+\frac{1}{n+1},\ldots$ is an infinite set of rationals between $a$ and $b$.

share|improve this answer

There is no smallest such rational. But your basic strategy will work if we can show there is at least one such rational. We sketch a proof of the fact that there are at least two. There is some detail that needs to be filled in.

Let $\epsilon=\dfrac{1}{b-a}$. Then by something that has undoubtedly already been proved, there is a positive integer $N$ such that $\dfrac{1}{N}\lt \epsilon/2$.

There is a largest integer $m$ such that $\dfrac{m}{N}\lt a$. Argue that $$a\lt \frac{m+1}{N}\lt \frac{m+2}{N}\lt b.$$

share|improve this answer

Maybe one can try something like this. Using the following

Lemma 1. For every real number $x$ there is exactly one integer $N$ such that $N \le x < N + 1$. (This integer $N$ is called the integer part of $x$, and is sometimes denoted $N = \lfloor x \rfloor$.)

Lemma 2. For any positive real number $x > 0$ there exists a positive integer $N$ such that $0 < 1/N < x$.

We now show

Proposition 3. Given any two real numbers $x < y$, we can find a rational number $q$ such that $x < q < y$.

By hypothesis, we have $y -x$ is positive. By Lemma 2, exists a positive integer $N$ such that $0 < 1/N < y - x$. Since $xN$ is a real number, by Lemma 1, there exists a integer $n$ such that $n - 1 \le xN < n$, i.e., $n/N - 1/N \le x$ and $x < n/N$. Thus $x < n/N \le x + 1/N$. Since $1/N \le y - x$, i.e., $x + 1/N < y$, we have $x < n/N < y$. Thus $n/N$ is rational, the claim follows.

share|improve this answer
    
Did you forget to state the lemmas? –  egreg Sep 26 at 21:11
    
Sorry. I pressed a wrong key. –  Cristhian Gz Sep 26 at 21:42
    
To prove Lemma 1, you can see math.stackexchange.com/questions/509711/…. To prove Lemma 2, use the Archimedeam property. –  Cristhian Gz Sep 27 at 3:51

Say $a\ne b$ and we want to show that infinitely many rationals are between $a$ and $b$. Then $|a-b|>0$. Is there an integer $n$ so big that $1/n < |a-b|$? If not, then $|a-b|>0$ is a lower bound of the set $\{1/n:n\in\{1,2,3,\ldots\}\}$, which therefore has an infimum $c$ that is positive and therefore has a reciprocal $1/c>0$, and $c=\sup\{1,2,3,\ldots\}$. Since $c>0$ is the smallest number greater than every positive integer, $c/2$ is smaller than sum positive integer $n$, so $c$ is smaller than $2n$, but $2n$ is a positive integer, so we have a contradiction. Conclusion: for some positive integer $n$, we have $1/n<|a-b|$. From there it's not hard to show that for every denominator $m\ge n$, some rationals with denominator $m$ are between $a$ and $b$.

share|improve this answer

If you already know (or can prove) that there is at least one rational between any two real numbers, then you can do this for $a < b$:

There is a rational number $x$ such that $a < x < \frac{a+b}{2}.$

There is a rational number $y$ such that $\frac{a+b}{2} < y < b.$

Now $a < x < y < b,$ with $x$ and $y$ rational.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.