Question regarding partially ordered sets I have encountered few questions while reading the book 'Modern Algebra'.
Let $\mathbb Q$ be the set of rational numbers.
Let

$B = \{ x : x\in\mathbb Q,\sqrt2 < x < \sqrt3 \}$.

How it can be shown that -

*

*$B$ has infinite number of upper and lower bounds.


*$\inf B$ and $\sup B$ do not exist .
Why $\sqrt2$ and $\sqrt3$ cannot be taken as lower aqnd upper bounds?
I am not able to understand that how can infimum and supremum can exist at the first place as the relation defined is not a binary relation.
A detailed proof would be helpful.
 A: 
Why $\sqrt2$ and $\sqrt 3$ can not be taken as lower and upper bounds?

Because the exercise is implicitly asking about an inf and sup that are in $\mathbb Q$. The exercise is perhaps not stating this very clearly, but remember that infimum and supremum is always about how a set fits into a certain ordered superset.

I am not able to understand that how can infimum and supremum can exist at the first place as the relation defined is not a binary relation.

The order relation that is being considered is the ordinary comparison of rational numbers. $x\le y$ is certainly a binary relation. (This is the default assumption when you speak about order properties of sets of numbers).
A: *

*Every (rational) integer $n=2,3,...$ is an upper bound, since $x < {\sqrt3} < n$ 
(since $3<2^2=4$).
Similarly every (rational) integer $n=1,0,-1,-2,-3,...$ is a lower bound, since $n\le1<\sqrt 2<x$. 

*$\sup B$ DNE. Take any upper bound $r$ for $B$. 
Then $r$ is a rational number hence $r\not=\sqrt3$. 
If $r<\sqrt3$ we get a contradiction since there are rational 
numbers $q$ with $r<q<\sqrt3$ showing that $r$ is not an upper bound. If $r>\sqrt3$ we get a contradiction since there are rational 
numbers $q$ with $\sqrt3<q<r$ showing that $r$ is an upper bound, but not the least one, i.e. not $\sup B$). 

*This part of the question was meanwhile edited, and is now answered in 2. 
Here are some details showing for example that 
if $r>\sqrt3$ then there are rational 
numbers $q$ with $\sqrt3<q<r$. 
Note that $\displaystyle\lim_{m\to\infty}(r-\frac1m)=r$, hence for all large enough $m$ we have $\displaystyle\sqrt3<r-\frac1m<r$ and we can take $\displaystyle q=r-\frac1m$ for some $m$. 
There ought to be a different proof, involving only arithmetic and inequalities between rational or integer numbers, starting with $3<r^2$, and looking for a positive rational $q$ with $3<q^2<r^2$, but this seems to require some extra work to come up and write down the details. 
And ... just in case, $\sqrt2$ and $\sqrt3$ are not rational numbers, that is why they cannot be taken as $\inf$ and $\sup$. They are a lower and an upper bound, and they are $\inf$ and $\sup$ if we are allowed to work with the reals, but if we only consider rational numbers, $\sqrt2$ and $\sqrt3$ do not belong and cannot be used. There are well-known proofs (since antiquity) that $\sqrt2$ is not a rational number (and there ought to be some of these on MSE). 
The relation $<$ is binary, whether you consider it on the reals or on the rational numbers. First it is defined on natural numbers, it could then be extended to positive rational numbers, and to all rational numbers, and to all real numbers. For the positive rational numbers, if $a,b,c,d$ are positive integers, define $\frac ab < \frac cd$ if 
$a\cdot d<c\cdot b$. (The extension, from here to all rational numbers is easy, e.g. use things like each negative rational is smaller than each positive rational.)
Once $<$ is defined for all rational numbers, one can talk about lower and upped bounds. There are sets that have a $\sup$ that is a rational number, e.g. $C = \{ x : x\in\mathbb Q,\sqrt2 < x < \sqrt4 \}$, in this case $\sup C =2$. The relation is binary, you compare two numbers and tell which is smaller, e.g. $5<6$ is the same as saying that the pair $(5,6)$ belongs to the $<$ relation. 
