Rational number

**spikedpunch** · October 22nd 2009, 04:18 PM

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

**tonio** · October 22nd 2009, 08:09 PM

Originally Posted by spikedpunch

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

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

Tonio

**HallsofIvy** · October 23rd 2009, 06:36 AM

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 $\epsilon= t-r= (r+s)/2- 2r/2= (s- r)/2$ . 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 $n^{th}$ decimal place is such a sequence.) so there exist an increasing sequence of rational numbers, $\{a_n\}$ converging to t. Since it converges to t, there exist N such that if n> N $|a_n- t|< \epsilon$ . $a_n$ 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.

**tonio** · October 23rd 2009, 09:07 AM

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.

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

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

$\color{red}Tonio$

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 $\epsilon= t-r= (r+s)/2- 2r/2= (s- r)/2$ . 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 $n^{th}$ decimal place is such a sequence.) so there exist an increasing sequence of rational numbers, $\{a_n\}$ converging to t. Since it converges to t, there exist N such that if n> N $|a_n- t|< \epsilon$ . $a_n$ 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.

.

**proscientia** · October 23rd 2009, 11:12 PM

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 $x,y\in\mathbb R$ with $x<y.$ Then $0<y-x$ $\implies$ $\exists\,n\in\mathbb N$ with $\frac1n<y-x$ by the Archimedean principle. $\therefore\ 1<ny-nx.$ Let $m=\lfloor nx\rfloor$ (so $m$ is the largest integer such that $m\leqslant nx).$ Then $nx<m+1$ and also $m+1<ny$ (otherwise $m+1\geqslant ny$ $\implies$ $1\geqslant ny-m\geqslant ny-nx).$ Hence

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

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

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