How to prove that at least two of the numbers $x,y,z$ are within $\frac{1}{2}$ unit from one another? 
Let $x,y,z$ be real numbers with $0<x<y<z<1$. Prove that at least two of the numbers $x,y,z$ are within $\frac{1}{2}$ unit from one another.

I guess I'll have to prove this by contradiction. However what does it mean that "at least two of the numbers $x,y,z$ are within $\frac{1}{2}$ unit from one another."?
 A: How about pigeon hole principle?  Split the interval $(0,1)$ in half.  If one is $\frac{1}{2}$, you are done since it is close enough to everything.  So now you have three numbers to stick into two boxes.  Two have to go into the same box and are therefore within $\frac{1}{2}$ of each other.
A: Exactly, Assume that this is not true and so  $y-x > \frac{1}{2}$ then this means that $y > x + \frac{1}{2}$ but since we know that $x > 0$ then this implies that $y > \frac{1}{2}$
Also assume that $z-y > \frac{1}{2}$ and so $z > y + \frac{1}{2}$ . However, since we showed that $y > \frac{1}{2}$ so $z > \frac{1}{2} + \frac{1}{2}$ and so  $z > 1$ which contradicts the fact that $z <1$
A: $0<x<y\wedge y-x>\frac12\implies y>\frac12$
$y<z<1\wedge z-y>\frac12\implies y<\frac12$
Contradiction.
A: Well, since $0<x,$ then $-x<0,$ so $$z-x<z-0=z<1.\tag{$\star$}$$
Proceeding by contradiction, then, is fairly straightforward. Suppose that each of $|x-y|,|y-z|,|x-z|\ge\frac12.$ Since $x<y<z,$ then $|x-y|=y-x$ and $|y-z|=z-y,$ whence by assumption, $$\frac12\le y-x\tag{1}$$ and $$\frac12\le z-y.\tag{2}$$ But then by $(1)$ and $(2)$ we have $$1=\frac12+\frac12\le y-x+z-y=z-x,$$ which by $(\star)$ yields $1<1,$ and we have a contradiction.
