proving $|x - 1| < {1\over4} \Rightarrow |2x - 1| \geq {1\over 2}$ I tried solving the above, consider that: ($x \in R)$,
I know it's not a complicated problem to solve though I struggle getting on with this question,  
What I've done far is:  
$|x-1|<{1\over4} \Longleftrightarrow -{1\over4} < x - 1 < {1\over4} \Longleftrightarrow {1\over2} < 2x-1 < {3\over2}$  
but from here I don't really know how to proceed with proving the question,
what am I doing wrong / should do from here?
or what should I consider when approaching such question?  
edit:
Thanks for the answers, it seems really obvious and clear now.
 A: Your proof is perfect, you simply need to note that $2x-1$ is now positive, so it is equal to its absolute value. Here are two other ways to get there. The first one is a technique which generalizes to normed vector spaces (I also do it below), while your method does not. So it is not bad to get used to it to prepare the future.
1) Reverse triangular inequality yields
$$
|y+z|\geq |y|-|z|
$$
Now try $y=1$ and $z=2x-2=2(x-1)$.
2) Observe that
$$
|2x-1|>\frac{1}{2}\quad\Leftrightarrow\quad\lvert x-\frac{1}{2}\rvert >\frac{1}{4}
$$
Now if $x$ lies within distance $1/4$ from $1$, how close can it be to $1/2$?
3) More generally in a normed vector space, if $\|x_0\|\geq 1$
$$
\|x-x_0\|<\frac{1}{4}\quad\Rightarrow\quad \|2x-x_0\|>\frac{1}{2}.
$$
Indeed, $\|2x-x_0\|=\|2(x-x_0)+x_0 \|\geq \|x_0 \|-2\|x-x_0\|>1-\frac{1}{2}=\frac{1}{2}$.
A: Your proof is complete, indeed overcomplete, since it contains some unnecessary inequalities. 
As you wrote, $|x-1|\lt \frac{1}{4}$ implies that $-\frac{1}{4}\lt x-1$, which implies that $\frac{3}{4}\lt x$.
It follows that $2x-1\gt \frac{1}{2}$, and therefore $|2x-1|\gt \frac{1}{2}$, and, more weakly, $|2x-1|\ge \frac{1}{2}$.   
