Does $f(x) = \sqrt{2} x - \lfloor\sqrt{2} x \rfloor,x\in \Bbb N$ attain a minimum? Let $x \ge 1$ be an integer, does there exist an minimum value for  
$f(x) = \sqrt{2} x - \lfloor\sqrt{2} x \rfloor$ ?
I think the answer is no. But I really don't know how to prove it, hope someone help me out, thanks!
 A: let $p_0 = 1, q_0 =1$ and 
$$
p_{n+1} = p_n^2 + 2 q_n^2 \\
q_{n+1} = 2p_nq_n
$$
so
$$
p_{n+1} -q_{n+1}\sqrt 2 = (p_n - q_n\sqrt 2)^2
$$
this gives
$$
p_{n+1}\sqrt 2 - q_{n+1} = \sqrt 2(\sqrt 2 - 1)^{2^{n+1}}
$$
which can be made arbitrarily small for large enough $n$
A: I want to extend on the answer of David Holden. This will show that the more general function
$$f_r(x)=rx-\lfloor rx\rfloor$$
will not attain a minimum for irrational numbers $r$ (here $\lfloor x\rfloor$ is the usual way to write down $floor(x)$). In your case you are dealing with $f_{\sqrt{2}}$.

Maybe you have heard about the irrationality measure $\mu(r)$, a way to measure how good or bad a number $r$ can be approximated by rational numbers. Rational numbers have $\mu(r)=1$. If $\mu(r)>1$ then $r$ is irrational. If I am not mistaken then $\mu(\sqrt 2)=2$. From the definition behind the link you can see that $\mu(r)>1$ means that there is some small $\epsilon>0$ so that
$$(*)\qquad\left|r-\frac pq\right|<\frac 1{q^{1+\epsilon}}$$
for infinitely many numbers $p,q\in\Bbb N$. More precisely: there are sequences $p_n,q_n$ with $p_n,q_n\to\infty$ so that $(*)$ holds. Multiply by $q_n$ to find
$$|r q_n-p_n|<\frac 1{q_n^{\epsilon}}$$
Because of the pidgeonhole principle we have $rq_n>p_n$ or $rq_n<p_n$ infinitely often (equality can only hold for rational numbers). Lets say we can choose an infinite subsequence with $rq_n>p_n$ (actually, we can always do this: the sequences $p_n,q_n$ emerge from the continued fraction expansion of $r$, and as you can read here every second term satisfies above condition). Then
$$rq_n-p_n<\frac1{q_n^{\epsilon}}.$$
For sufficiently large $n$, the right side is $<1$, hence  $p_n=\lfloor rq_n\rfloor$. We then have
$$f_r(q_n)=rq_n-\lfloor rq_n\rfloor=rq_n-p_n<\frac1{q_n^\epsilon}\to 0.$$
But as discussed, $f_r(x)=0$ can only happen for rational numbers. So this indicates that the minimum is not attained (for irrational numbers $r$).
