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?
 A: There are three problems in your attempt, in increasing order of severity:


*

*Although you have a good idea for the solution (use the discriminant of an 
appropriate polynomial) you are missing a coherent strategy and as a result 
you are abusing techniques (proof by contradiction) and their execution (you 
are using one sided implications where only equivalences could give you what 
you want).  In general, it is important to read what you write and see if 
makes logical sense before posting your question.

*Your attempt at a contradiction begins with $x<\frac{1}{4b^2}$.  Such an 
attempt cannot work because there is nothing to contradict; this inequality 
holds for $\frac{1}{\sqrt{2}}$.  What does not hold is the completely 
different statement $$-\frac{1}{4b^2}\leq x\leq \frac{1}{4b^2};$$ thus it should not 
come as a surprise when you get stuck later on trying to show a certain 
polynomial is positive: it is not positive.

*The most serious error, which cannot be mended even if you fix the previous 
ones, is that you are defining $x$ as a quantity that depends on $a/b$.  This 
$x$ is a single number for each $a/b$, giving rise to a single polynomial, 
different for each $a/b$, for which you find that when evaluated at your $x$, 
the result is positive.  This does not mean the polynomial cannot be negative 
at other values of $x$, so we get no information about the discriminant.  In 
fact, you yourself show that the discriminant of those polynomials cannot be 
negative, so the polynomial cannot be positive.  You need something else.


What goes wrong in your approach is that you are making the rational numbers 
$a/b$ the highlight of the problem, relegating $\frac{1}{\sqrt{2}}$ to a side 
role somewhere in the inequalities.  That is not right.  $\frac{1}{\sqrt{2}}$ 
is the star of this show and the rational numbers around it are the chorus; 
and like any self-respecting diva, it doesn't want the chorus to stick too close.
I will give you two hints in case you want to tackle the problem again, and 
then sketch a solution based on Lemma I.2.E of W.M.Schmidt's book Diophantine Approximation.
Hints:


*

*Find the unique polynomial of minimal degree with integer coefficients and 
positive leading term that has $r=\frac{1}{\sqrt{2}}$ as a root.  Call it $P(x)=a_2x^2+a_1x+a_0$.

*Take any rational number $\frac{a}{b}$ in $(0,1)$ and evaluate $|P(x)|$ at that 
number.  Use the properties of $P$ to derive a contradiction from the claim 
$$|\frac{a}{b}-\frac{1}{\sqrt{2}}|\leq \frac{1}{4b^2};$$ the properties that you 
will need are the fact that the coefficents of $P$ are integers and the 
factorization of a polynomial using its roots..
===========
Now here is the solution:
First, $$P(x) = 2x^2-1 = 2(x-\frac{1}{\sqrt{2}})(x+\frac{1}{\sqrt{2}})$$ is 
the minimal polynomial of $r$ over the integers.
Fix $a/b$ in the unit interval that satisfies $$|\frac{a}{b}-\frac{1}{\sqrt{2}}|\leq\frac{1}{4b^2}$$ and estimate $|P(\frac{a}{b})|$.  First of all 
$$|P(\frac{a}{b})|\geq \frac{1}{b^2}.$$  Can you see that?  Here is where you use the 
fact that the coefficients are integers:
$$|P(\frac{a}{b})| = \frac{|2a^2-b^2|}{b^2}\geq \frac{1}{b^2}$$ because the numerator 
is a posiitve integer (why never zero?).
Secondly, and this is the trickier part, $$|P(\frac{a}{b})| = 2|\frac{a}{b}-r||\frac{a}{b}+r| \leq \frac{1}{4b^2}2|\frac{a}{b}+r|;$$
in the last inequality we used the hypothesis about the approximation of 
$r=1/\sqrt{2}$ by rationals.  Now we get rid of the pesky $a$ using our hypothesis again and the triangle inequality:
$$2|\frac{a}{b}+r| = 2|\frac{a}{b}-r+2r|\leq 2|\frac{a}{b}-r| +2\cdot 2r\leq \frac{1}{2b^2} + 2\sqrt{2};$$  now plug this in the last term of the inequality above to get
$$|P(\frac{a}{b})| \leq \frac{1}{8b^4}+ \frac{1}{\sqrt{2}b^2}.$$  Together with 
$$|P(\frac{a}{b}))|\geq \frac{1}{b^2}$$ we get a contradiction if $b\geq 2$.  Finish 
by checking that $b=a=1$ also do not satisfy the inequality by hand.
A: 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. 
