# Rational number

• Oct 22nd 2009, 04:18 PM
spikedpunch
Rational number
I need help proving that between any two real numbers there exists a rational number.
• Oct 22nd 2009, 08:09 PM
tonio
Quote:

Originally Posted by spikedpunch
I need help proving that between any two real numbers there exists a rational number.

$\displaystyle r,s \in \mathbb{Q} \Longrightarrow x:=\frac{r+s}{2} \in \mathbb{Q}\,,\,\,so...$

Tonio
• Oct 23rd 2009, 06:36 AM
HallsofIvy
Tonio's proof is that between any two rational numbers, there exist an irrational number. If x is NOT rational, It's a little more complicated.

Suppose that r and s are the given real numbers, with r< s. Again, look at t= (r+s)/2. That is a real number between r and s but may not be rational. Let $\displaystyle \epsilon= t-r= (r+s)/2- 2r/2= (s- r)/2$. There exist an increasing sequence of rational number converging to any real number (The sequence got by truncating the decimal expansion of the number at the $\displaystyle n^{th}$ decimal place is such a sequence.) so there exist an increasing sequence of rational numbers, $\displaystyle \{a_n\}$ converging to t. Since it converges to t, there exist N such that if n> N $\displaystyle |a_n- t|< \epsilon$. $\displaystyle a_n$ for n> N is then larger than t- (s- r)/2= (s+r)/2- (s- r)/2= r and is less than t< s because it is an increasing sequence.
• Oct 23rd 2009, 09:07 AM
tonio
Quote:

Originally Posted by HallsofIvy
Tonio's proof is that between any two rational numbers, there exist an irrational number. If x is NOT rational, It's a little more complicated.

$\displaystyle \color{red}\mbox{No. Tonio's proof is that between the two rational numbers r,s there exists}$ $\displaystyle \color{red}\mbox{ another RATIONAL number which I denoted by x}$

$\displaystyle \color{red}\mbox{This in fact is the easy part. To prove there exists an irrational}$ $\displaystyle \color{red}\mbox{between those two is harder}$

$\displaystyle \color{red}Tonio$

Suppose that r and s are the given real numbers, with r< s. Again, look at t= (r+s)/2. That is a real number between r and s but may not be rational. Let $\displaystyle \epsilon= t-r= (r+s)/2- 2r/2= (s- r)/2$. There exist an increasing sequence of rational number converging to any real number (The sequence got by truncating the decimal expansion of the number at the $\displaystyle n^{th}$ decimal place is such a sequence.) so there exist an increasing sequence of rational numbers, $\displaystyle \{a_n\}$ converging to t. Since it converges to t, there exist N such that if n> N $\displaystyle |a_n- t|< \epsilon$. $\displaystyle a_n$ for n> N is then larger than t- (s- r)/2= (s+r)/2- (s- r)/2= r and is less than t< s because it is an increasing sequence.

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• Oct 23rd 2009, 11:12 PM
proscientia
All tonio has shown is that between two rational numbers there exists a rational number; this does not imply that between two real numbers there exists a rational number.

More precisely, we want to show that between two distinct real numbers there exists a rational number. Let $\displaystyle x,y\in\mathbb R$ with $\displaystyle x<y.$ Then $\displaystyle 0<y-x$ $\displaystyle \implies$ $\displaystyle \exists\,n\in\mathbb N$ with $\displaystyle \frac1n<y-x$ by the Archimedean principle. $\displaystyle \therefore\ 1<ny-nx.$ Let $\displaystyle m=\lfloor nx\rfloor$ (so $\displaystyle m$ is the largest integer such that $\displaystyle m\leqslant nx).$ Then $\displaystyle nx<m+1$ and also $\displaystyle m+1<ny$ (otherwise $\displaystyle m+1\geqslant ny$ $\displaystyle \implies$ $\displaystyle 1\geqslant ny-m\geqslant ny-nx).$ Hence

$\displaystyle x\ <\ \frac{m+1}n\ <\ y$

and $\displaystyle \frac{m+1}n\in\mathbb Q.$