Find two rationals, one greater and one smaller than a given irrational number. Given an irrational number  0 < i < 1. Find two rational numbers a and b such that 0 < a < i < b < 1.
 A: Since $0<i<1,$ then there exists some least positive integer $n$ such that $$\frac1n<i<1-\frac1n.$$ To see why, apply the Archimedean property to $$\frac1{\min(i,1-i)}.$$
A: Find the first non-zero digit in the decimal explasion, set everything after to 0 this gives a. Now find the first non 9 digit set add one to this digit and set everything after to 0 this gives b.
A: It is easy to do. Note that $\Bbb Q$ is dense in $\Bbb R$. Therefore $(0,i) \cap \Bbb Q \not=\emptyset$ and $(i,1)\cap \Bbb Q \not= \emptyset$ since $(0,i)$ and $(i,1)$ are two open sets. Then there exist $a$ and $b$ in $\Bbb Q$ such that $0<a<i<b<1$.
A: Take $a_0$ = 0/1 and $b_0$ = 1/1.  Those enclose your number $i$.  Now for each next step $n+1$, take the "mediant" fraction of $a_n$ and $b_n$ gained by adding their nominators and denominators and making a new fraction from the sum.
Call it $m_n$.  Now if $m_n$ > i, let $a_{n+1}$ = $a_n$ < i < $b_{n+1}$ = $m_n$, otherwise let $a_{n+1}$ = $m_n$ < i < $b_{n+1}$ = $b_n$.
Repeat until you are satisfied with the result.  For i=$\sqrt{1/2}$, you get the enclosing fractions $\frac01<i<\frac11$, $\frac12<i<\frac11$, $\frac23<i<\frac11$, $\frac23<i<\frac34$, $\frac23<i<\frac57$, $\frac7{10}<i<\frac57$, $\frac{12}{17}<i<\frac57$ and so on.  Yes, the fractions are always automagically in smallest terms.
A: Choose an $N$ large enough such that $\frac{1}{N}<i$.  Certainly this is possible since $lim(\frac{1}{N}) = 0$.
Now let x = $min(\frac{1}{N}, 1-\frac{1}{N})$.  And so:
$$0<x<i<1-x<1$$
