Why the set of rational numbers is not an order complete field as it is the subset of real numbers which is an order complete field? So far as I am concerned an order complete field is that one which have supremum and infimum belonging to that set. And for the proof of the above mentioned question I already have the proofs using some k^2 greater and lesser than 2 which I can't decode.
Please make it as simple as possible.
 A: Let x be an irrational number , then take two rational sequences an and bn (rational sequences means that every term of that sequence is a rational number) such that both of them converge to x and an is less than or equal to bn for every natural n.
Then it's easy to show that sup(an)=inf(bn)=x but yet x is not a rational number therefore the set of rational numbers is not an order complete field.
A: Re : "some k^2 greater or less than 2". The idea is to get a  sequence $x_1,x_2,x_3,...$ of positive rationals where each $x_n^2>2$ and such that no positive rational is less than every member of $\{x_n^2-2:n\in \Bbb N\}.$ 
Assume that that $x=\inf_{n\in \Bbb N}x_n$ exists and belongs to $\Bbb Q.$ Note  that $0< x$ (because $1$ is a lower bound for $\{x_n\}_n$ ) and that $x<x_n$ for all $n.$ 
For any $q\in \Bbb Q^+$ there exists $n$ such that $x^2-2< x_n^2-2<q.$ So $x^2-2$ is less then any positive rational. 
Now if $x^2<2$ let $x'=(x+2/x)/2$ and $x''=2/x.$ Then $x<x''\in \Bbb Q$ and $(x'')^2<2$ so $x''<x_n$ for every $n.$ But then $x$ would not be the $greatest$ lower bound for $\{x_n\}_n$. Therefore $x^2\geq 2 .$ 
So either $x^2-2=0$ (contradicting $x\in \Bbb Q$) or $x^2-2$ is a positive rational that's less than every positive rational  (which is absurd). 
Therefore the assumption that some $x\in \Bbb Q$ is equal to $\inf_nx_n$ is paradoxical. So the non-empty subset $\{x_n\}_n$ of $\Bbb Q$ has a lower bound in $\Bbb Q$ but no greatest lower bound in $\Bbb Q,$ so $\Bbb Q$ is not order-complete.
Note : We can obtain $\{x_n\}_n$ by taking any $x_1\in \Bbb Q^+$ such that $x_1^2>2$ and letting $x_{n+1}=(x_n+2/x_n)/2.$ (Reference topic: Heron's method for approximate square roots: Heron (a.k.a. Hero) of Alexandria, circa 100 A.D.) 
A: Another proof is the following. We will use that

*

*$\mathbb{Q}^{+}\ni x\mapsto x^2$ is a monotone injection: $y^2-x^2=(y-x)(y+x)$

*If $0<x,y$ are 'close', then $x^2,y^2$ are close too, the idea which motivates the proof

Let $A=\{x\in \mathbb{Q}^+:2\leq x^2\}$. Since $1^2=1<2$, $1$ is a lower bound of $A$, therefore, if $\mathbb{Q}$ is complete, $c=\inf A$ exists in $\mathbb{Q}$.
We show that
$$1<c<2.\qquad  (*)$$
Since $1$ is a lower bound of $A$, we have to have $1<c$. On the other hand, if $2\leq x$, then $2^2=4\leq x^2$, but then, $2<(5/3)^2<4$, so $5/3\in A$ and $(5/3)<2\leq c$, which contradicts to the definition of $c$.
Now, we examine the value of $c^2$: $c^2$ cannot be equal to two, due to the fact that $x^2-2=0$ cannot be solved in $\mathbb{Q}$. Therefore, $2<c^2$. We would like to find $\Delta\in\mathbb{Q}^+$ suct that $2<(c-\Delta)^2<c^2$. Such a $\Delta$, if exists leads to a contradiction to the definition of $c$. We have two separate cases

*

*Assume $2c<c^2-2$. Note that $0<c-1$ due to $(*)$ and $c-1<c$, so $(c-1)^2<2$. But
$$(c-1)^2-2=(c^2-2)-2c +1>1,$$
and so $(c-1)^2>3>2$, a contradiction to $(c-1)^2<2$.


*Assume $2c\geq c^2-2$. Then, since $\mathbb{Q}$ is Archimedian, there exists $n\in\mathbb{N}$, $n>0$ such that $n^{-1}2c\leq c^2-2$. Of course, $0<c-1\leq c-n^{-1}<c$, therefore $(c-n^{-1})^2<2$. But
$$(c-n^{-1})^2-2=(c^2-2)-2n^{-1}c+n^{-2}>n^{-2}>0,$$
resulting in $2<(c-n^{-1})^2$ a contradiction to $(c-n^{-1})^2<2$.
