How to find a rational number between two irrational numbers?

Question

Find the rational number between $\sqrt 2$ and $\sqrt3$. I try to solve by using some methods in my book but can not understand steps.

Answer 1

The rational number? Rational numbers are a dense subset of $\mathbb{R}$, there are infinitely many of them between $\sqrt{2}$ and $\sqrt{3}$. For instance, since the difference between $\sqrt{2}$ and $\sqrt{3}$ is between $\frac{3}{10}$ and $\frac{1}{3}$, there is for sure an integer number between $4\sqrt{2}$ and $4\sqrt{3}$: $6$, for instance. By rescaling, we get: $$ \sqrt{2} < \frac{6}{4}=\color{red}{\frac{3}{2}} < \sqrt{3}.$$

In general, we may assume to have two distinct irrational numbers $a,b$ with their difference, multiplied by some $n\in\mathbb{N}$, being greater than one. Then there is a natural number $m$ in the range $[na,nb]$, and by rescaling we get a rational number $\frac{m}{n}$ between $a$ and $b$.

Another approach may be to consider a convergent of the continued fraction of $\frac{a+b}{2}$.
Since the convergents $\frac{p}{q}$ of some $\alpha\not\in\mathbb{Q}$ fulfill $$\left|\alpha-\frac{p}{q}\right|\leq\frac{1}{q^2},$$ as soon as $q$ is big enough we have $\frac{p}{q}\in (a,b)$. With this approach we get, for instance, that $\color{red}{\large\frac{11}{7}}$
is very close to the midpoint of $(\sqrt{2},\sqrt{3})$.

Answer 2

A quick and inelegant approach is to use the (beginnings of the) decimal expansions $\sqrt2=1.41\dots$ and $\sqrt3=1.73\dots$. Any terminating decimal between these two will solve your problem, for example $1.5$ or $1.6$ or $1.7$. You could also use any periodically repeating decimal between the two, like $1.6666666\dots$.

Once you've obtained an answer in this way, you can check it by squaring it and seeing that its square is between $2$ and $3$. For example, $(1.5)^2=(3/2)^2=9/4$ and $(1.666666\dots)^2=(5/3)^2=25/9$.

Answer 3

As $\sqrt2 = \sqrt{200/100}$ and $\sqrt3= \sqrt{300/100}$, we need to find a rational number $x$ such that $$\frac1{10}\sqrt{200}< x<\frac1{10} \sqrt{300}$$

Choose any perfect square such as $225$ or $256$ in between 200 and 300. Then $x=\sqrt{225}/10 = 15/10=5/3$ and similarly $16/10=8/5$ would be the numbers with the desired property.

Answer 4

An alternative approach:

$\sqrt2$ and $\sqrt 3$ are non but solutions to the equations $x^2-2$ and $x^2-3$.

Consider their graphs and notice that any positive rational solution to the polynomial $x^2-a$, for $a\in (2,3)$, will satisfy your requirement.

We know the solutions to $x^2-a$ are given by $\pm\sqrt a$.

Then consider only the positive and write $a=\frac{m^2}{n^2} \in(2,3)$

From here it is easy to find examples like $\frac{3^2}{2^2}$.

Answer 5

We look for some rational number $r = p / q$, where $p, q \in \mathbb{N}$: $$ \sqrt{2} < \frac{p}{q} < \sqrt{3} \iff \\ 2 < \frac{p^2}{q^2} < 3 \iff \\ 2 q^2 < p^2 < 3 q^2 $$ E.g. $q=10$ will lead to $$ 200 < p^2 < 300 $$ which is fulfilled for $p \in \{ 15, 16, 17 \}$.

Answer 6

The meaning of 'density' you mention is the following:

A set $S$ of real numbers is dense in $\mathbb{R}$ if between any two real numbers there exists another real number from $S$.

In your case you wish to replace $S$ with the set of rational numbers $\mathbb{Q}$ (which remember are still real numbers, being a subset of $\mathbb{R}$) and these are dense in $\mathbb{R}$.

The Archimedean Property of $\mathbb{R}$ (APR) states that for each positive $a$, $b\in\mathbb{R}$ there exists a natural number $n\in\mathbb{R}$ such that $na>b$.

In general let $a$, $b\in\mathbb{R}$, with $c<d$, and $c>0$, then by the APR there is an $m\in\mathbb{N}$ such that $1/m<d-c<d$. Invoking the APR again there exists an $n\in\mathbb{N}$ such that $n/m\ge d$. There will be a smallest element, $e$ say, for which $e/m\ge d$, and this implies by minimality of $e$ that $(e-1)/m< d$. We then have using our inequalities,
$$c=d-(d-c)< (e/m)-(1/m)=(e-1)/m<d$$ and so $(e-1)/m$ is a rational number between two irrational numbers.

In your problem $c=\sqrt{2}\approx 1.414$, $d=\sqrt{3}\approx 1.732$, and $\sqrt{3}-\sqrt{2}\approx 0.318$. Hence for a constructive example let $m=5$, then $1/5=0.2<0.318<1.732$. Now we need $e$ such that $e/5>1.732$, and $(e-1)/5<1.732$: we find this to be $e=9$ since $9/5=1.8>1.732$ and $8/5=1.6<1.732$. Putting it all together we obtain: $$\sqrt{2}<\frac{8}{5}<\sqrt{3}.$$

You may now tailor this method to find any requested rational that lies between two given irrationals.