I'm trying to prove there is infinite rational numbers between any two rational number

Question

I understand how to explain but can't put it down on paper.

$\displaystyle \frac{a}{b}$ and $\displaystyle \frac{c}{d}$ are rational numbers. For there to be any rational between two numbers I assume $\displaystyle \frac{a}{b} < \frac{c}{d}$.

I let $\displaystyle x = \frac{a}{b}$ and $\displaystyle y = \frac{c}{d}$ so $x < y$. the number in the middle of $x$ and $y$ is definitely a rational number so $\displaystyle x < \frac{x+y}{2} < y$. I know that there is another middle between $x$ and $\displaystyle \frac{x+y}{2}$ and it keeps on going. How would I write it as a proof?

Answer 1

One natural way to complete your proof is by contradiction.

You noticed that, between any two rational numbers, there is a third.

Now, pick any two rational numbers $x < y$, and assume that there are a finite number of (say, only $n$) rational numbers between them. Call these numbers $a_0 < a_1 < \cdots < a_{n-1}$. Now, look between $x$ and $a_0$. We know there is some rational number, call it $z$ between $x$ and $a_0$, which then must also be between $x$ and $y$. But this cannot be true, because $z$ is less than $a_0$, so cannot equal any of the $a_i$, which were assumed to be every rational between $x$ and $y$.

Answer 2

I wanted to add to Cameran Buie's answer with the actual proof by induction, since it seems that the question was "How do I word the proof?" I'm going overboard here, being fully precise; in any normal situation, much of this reasoning can be brushed over and still get the point across.

First, a precise wording of what the original post has so far.

Let $x=a/b$ and $y=c/d$ be two distinct rational numbers. We may as well assume $x < y$, since we can easily swap the letter assignment otherwise. We may also assume $b,d$ are positive since they are nonzero and we can multiply be $\frac{-1}{-1}$ to swap signs. We want to show that there are infinitely many other rational numbers in between these.

We will start by showing that the average of $x$ and $y$ is in between $x$ and $y$. (This can be done in many ways. I'll do it the most tedious but direct way here.) The average is $$ q=\frac{x+y}{2} = \frac{\frac ab + \frac cd}{2} = \frac{ad+bc}{2bd} $$ which is a rational number because $x,y$ are rational imply $b,d$ are nonzero imply $2bd$ is nonzero.

To show that $x\lt q\lt y$, we can show the left is true: $$ \begin{aligned} x &\lt y && \text{given} \\ \frac ab &\lt \frac bd && \text{substitution} \\ ad &\lt bc &&\text{by multiplying by $bd$}\\ &&&\text{(where $b,d$ are positive)} \\ 2ad&\lt ad+bc && \text{by adding $ad$}\\ \frac ab&\lt \frac{ad+bc}{2bd} && \text{by dividing by $2bd$} \\ &&&\text{(where $b,d$ are positive)} \\ x&\lt q && \text{substitution} \end{aligned} $$

The right side, $q\lt y$, is shown to be true in a very similar fashion. (To figure out these steps, work them backwards, starting from what you want to show and simplifying to what you know.)

So, we now know that between any distinct rationals $x,y$, there is a rational $q$ between them.

Now, a precise wording of the proof by induction.

Let $x,y$ be given distinct rationals as before, and we will show that there are infinitely many distinct rationals between them.

In the base step, we show that there is $1$ distinct rational between $x,y$. We proved that for $q$ above.

In the induction step, we assume that we have shown there are $n$ distinct rationals between $x,y$ for some positive integer $n$, and will show that there are $n+1$ distinct rationals between $x,y$. Let $y_0$ be the minimum of all the $n$ distinct rationals between $x,y$. By the above proof, there must be a rational $q$ between $x,y_0$, and since it is less than the smallest of the others between $x,y$, it must be a new distinct rational between $x,y$. Thus there are $n+1$ distinct rationals between $x,y$.

By induction, there are infinitely many distinct rationals between $x,y$. (The induction is saying we know there is one of them, and for every positive-integer number of them there is one more, so we must have every-positive-integer number of them, which is the same as "infinitely many".)

A note on proof by contradiction.

For some people, proof by contradiction is reserved as a last resort. Sure, a proof by contradiction can be shorter, but if you can avoid it, such as here with the proof by induction, then you usually gain more insight into how to actually construct your answer.

In this case, that's shown by how, if you actually had two rationals, say $0$ and $1$, then the proof by induction tells you exactly what infinite sequence of rationals you are finding, say $1/2^n$ for each positive integer $n$. The proof by contradiction tells you there's infinitely many, but not how to find them.

This is all a matter of personal preference. Whichever you find most intuitive should stay close to your heart; whichever you find most convincing to others should be what you write down.

Answer 3

You've got the idea. Let $a_1=\frac{x+y}2,$ and $a_{n+1}=\frac{x+a_n}2.$ A quick induction proof will show that each $a_n$ is rational, that $x<a_1<y$ and that $x<a_{n+1}<a_n$ for all $n.$ Thus, you have explicitly determined infinitely-many rational numbers between $x$ and $y.$

More explicitly, you can show by induction that $$a_n=x+\frac{y-x}{2^n}$$ for all $n,$ if you want a closed (non-recursive) form.

Answer 4

Slight adaption to Steven Stadnicki linear method (see his comment).

Let $p$ and $q$ be any two rational numbers with $p \lt q$. Now

$\tag 1 p + 1/n \lt q \text{ iff } 1/n \lt q - p \text{ iff } n \gt \frac{1}{q-p}$.

By Archimedean property, we can find an $n_0$ so that (1) is true, and of course it will also be true for all integers greater than $n_0$.

Answer 5

For rationals $r_1$ and $r_2$ you can always find the rational $(r_1 + r_2) / 2$ between them. As you can repeat this at will, you can construct infinitely many rationals between them (essentially the set $r_1 + (r_2 - r_1) \cdot 2^{-n - 1}$).