How to prove that $[2^n \sqrt{2}],[2^{n+1} \sqrt{2}],\ldots, [2^{2n} \sqrt{2}]$ contains at least one even numbers for every integer $n\ge 1$? I tried as follows: 
If not, denote $x=\{2^{n-1}\sqrt{2}\}$, then 
  $$1-\frac{1}{2^{n+1}}<x<1.$$
Denote $y=[2^{n-1}\sqrt{2}+1]$ and assume $|\sqrt{2}-p/q|<1/q^2$, then
  $$\frac{2^{n-1}p}{q}-\frac{2^{n-1}}{q^2}<y<\frac{2^{n-1}p}{q}+\frac{2^{n-1}}{q^2}+\frac{1}{2^{n+1}}.$$
But it doesn't work.
 A: As you observed, $2^{n-1}\sqrt2$ would have to be pretty close to an integer $y$ (in your notation), so that $|2^{n-1}\sqrt2-y|<\frac1{2^{n+1}}$ and with $q:=2^{n-1}$,
$$\tag1\left|\sqrt 2-\frac{y}{q}\right|<\frac14\cdot\frac1{q^2}. $$
In particular, $$\tag21<\frac yq<2.$$
We know that $\sqrt2$ is a root of the irreducible polynomial $f(X)=X^2-2$, hence by the IVT,
$$\tag3f(\tfrac yq)=f(\tfrac yq)-f(\sqrt 2)=(\tfrac yq-\sqrt2)f'(\xi)$$
for some $\xi$ between $\sqrt2$ and $\frac yq$; from (2), we see that $1<\xi<2$. 
As $f(\tfrac yq)$ is a nonzero(!) integer multiple of $\frac 1{q^2}$ and as $|f'(\xi)|=|2\xi|<4$, we have
$$\left|\sqrt 2-\frac yq\right|=\frac{|f(\tfrac yq)|}{|f'(\xi)|}>\frac{\frac1{q^2}}4 $$
contradicting $(1)$.
A: By a basic trick of Diophantine approximation we have, for all positive integers $p,q$, the inequality
$$
\left|\frac pq-\sqrt2\right|\cdot\left|\frac pq+\sqrt2\right|=\left|\frac{p^2}{q^2}-2\right|\ge\frac1{q^2}.
$$
In particular, if $3/2>p/q>\sqrt2$, we get the estimate
$$
\left|\frac pq-\sqrt2\right|\ge\frac1{3q^2}.
$$
Therefore, with $M=[2^{n-1}\sqrt2]$ we get, using $x=\{2^{n-1}\sqrt2\}$,
$$
1-x=2^{n-1}(\frac{M+1}{2^{n-1}}-\sqrt2)\ge\frac1{3\cdot2^{n-1}}>\frac1{2^{n+1}}.
$$
Therefore $x<1-2^{-(n+1)}$, so your contrapositive case cannot happen.
