Irrational number not ocurring in the period of rational numbers Write each rational number from $(0,1]$ as a fraction $a/b$ with $\gcd(a,b)=1$, and cover $a/b $ with the interval 
$$
\left[\frac ab-\frac 1{4b^2}, \frac ab + \frac 1{4b^2}\right].
$$
Prove that the number $\frac 1{\sqrt{2}}$ is not covered. What I did was the following:-
 We define the given period as P. Assume that $\frac 1{\sqrt{2}}$ is not present in P. Hence set $\frac ab=k=\frac 1{\sqrt{2}}+x$. Therefore we have to prove that 
$$
\begin{align}x\gt\frac 1{4b^2}&\implies x \gt \frac {k^2}{4a^2}\\
&\implies4a^2x\gt\left(\frac 1{\sqrt{2}}+x\right)^2\\
&\implies4a^2x\gt x^2+x\sqrt{2}+\frac 12\\
&\implies x^2-x(4a^2-\sqrt{2})+\frac 12\lt 0\end{align}
$$ 
Hence 
$$
\begin{align} D\lt 0 & \implies (4a^2-\sqrt{2})^2-4\cdot\frac 12 \lt 0 \\ 
& \implies(4a^2-\sqrt{2})^2-\sqrt{2}^2\lt 0 \\
 & \implies(4a^2-2\sqrt{2})*4a^2\lt 0\\
&\implies4a^2\lt2\sqrt{2}\\
&\implies a^2\lt \frac 1{\sqrt{2}}\end{align}
$$ 
This is impossible since $a$ is a natural number???? So, what do I do?

I got a good solution from one of my teachers...$$\left|\frac ab-\frac 1{\sqrt2}\right|\left(\frac ab+\frac 1{\sqrt2}\right)=\left|\frac {a^2}{b^2}-\frac 12\right|=\frac {|2a^2-b^2|}{2b^2}\gt\frac 1{2b^2}$$ Also we know that $\frac ab+\frac 1{\sqrt2}\lt2=>\left|\frac ab-\frac 1{\sqrt2}\right|\gt\frac 1{4b^2}$. Hence proved. 
