At least one even number among $\{ \lfloor 2^{n}\sqrt{2} \rfloor, \lfloor 2^{n+1}\sqrt{2} \rfloor,..., \lfloor 2^{2n}\sqrt{2} \rfloor \}$ 
For any positive integer $n$, prove that the set
$$\{ \lfloor 2^{n}\sqrt{2} \rfloor, \lfloor 2^{n+1}\sqrt{2} \rfloor,..., \lfloor 2^{2n}\sqrt{2} \rfloor \}$$
contains at least one even number.

I tried to prove this question by contradiction, assuming that each element is odd. There exist the positive integers $k_1, k_2, ..., k_{n+1}$ such that
$$2k_1-1<2^n\sqrt{2}<2k_1$$
$$2k_2-1<2^{n+1}\sqrt{2}<2k_2$$
$$...$$
$$2k_{n+1}-1<2^{2n}\sqrt{2}<2k_{n+1}$$
But i can't find a contradiction among these inequalities.

If $\lfloor x\rfloor$ and $\lfloor {2x}\rfloor$ are both odd, then $\lfloor {2x}\rfloor=2\lfloor x\rfloor+1$.  Use this fact to prove that if all the floor functions are odd, then $\lfloor{2^n\sqrt2}\rfloor$ is one less than a multiple of $2^{n+1}$.  
Then show that for $n\ge 1$, in fact $2^{n+1}>1+2^n\sqrt2$.  Thereby the assumption that all the floor functions are odd will meet the same fate as most of my college basketball tournament brackets.  
For further review:  If we allow $n=0$ then we do have all the floor functions (one value) odd.  Where does the above proof break down if we allow $n=0$?
