Why would the supremum of a bounded set in $\mathbb Q$ actually not exist? It is not hard to show that the supremum of a bounded set $A=\{x\in\mathbb R\mid x^2\leq 2\}$ equals $\sqrt 2$, yet the reason I am confused is that why would it not exist in $\mathbb Q$ (the supremum, not the $\sqrt 2$).
It is true that $\sqrt 2$ is not in $\mathbb Q$ and we need $\mathbb R$ to fill the gap, but why can't we look for another number greater than $\sqrt 2$ that is rational and has the minimum difference with $\sqrt 2$, letting it be assumed that $\mathbb R$ is not found by humanity to save the day?
I understand that my question may be silly in many ways, so it would be nice that someone explains the intuition as to why we say that the supremum does not exist instead of finding another value in $\mathbb Q$ to be our hero.
Thanks!
 A: To be fully precise: when $L$ is a linearly ordered set and $X\subseteq L$, by "$X$ has a supremum in $L$" I mean "There is some $a\in L$ such that $a>_Lx$ for all $x\in X$ and for all $b<_La$ there is some $x\in X$ with $b\le_Lx$." In particular, changing $L$ can change whether $X$ has a supremum or what that supremum is.

You're right that the fact that $\sqrt{2}$ is irrational does not immediately imply that $A=\{x\in\mathbb{Q}: x^2\le 2\}$ has no supremum in $\mathbb{Q}$. For example, $\{x\in\mathbb{Z}: x^2\le 2\}$ obviously does have a supremum in $\mathbb{Z}$ (namely $1$).
However, it isn't hard to show that $A$ in fact does not have a supremum in $\mathbb{Q}$ as follows:


*

*Suppose $q=\sup(A)$.

*Then since $\sqrt{2}\not\in\mathbb{Q}$ we have either $q^2<2$ or $q^2>2$. Assume the former (the argument for the latter is the same).

*We now just show that there is a positive rational $p$ such that $(q+p)^2<2$. This isn't too hard: let $p$ be some positive rational such that $p<q$ and $q^2+3pq<2$, and note that $$(q+p)^2=q^2+2pq+p^2<q^2+3pq<2.$$

*But then $q+p\in A$ and $q+p>q$, which is impossible by assumption on $q$.
A: A set in $\mathbb{Q}$ being bounded (above) just means that there is some rational number larger than every number in the set. But that doesn't guarantee there is a minimum element of that set of upper bounds. The natural numbers are well-ordered, so actually every bounded subset of the natural numbers does have a supremum in $\mathbb{N}$. The real numbers more or less restores this least-upper-bound property to $\mathbb{Q}$ without going back to being well-ordered.
Another thing to keep in mind is that not every bounded subset of $\mathbb{Q}$ is missing a supremum. The set $(0,1)\subset\mathbb{Q}$ clearly has supremum 1, for example. It's just that some subsets, like the set $A$ you mentioned, don't have a set of upper bounds with a minimal element. 
If you just want to convince yourself that the set $A$ has no supremum in $\mathbb{Q}$, just try assuming that there is a least upper bound $x$ and then find a smaller rational number $y$ that has $y^2>2$. If you can do that, you've shown why the supremum doesn't exist: there are upper bounds, but there isn't a smallest one.
