Is $\rho(x,y)=(x-y)^2$, with $x,y\in \mathbb{R}^1$, a metric space on $\mathbb{R}^1$? Obviously it has to satisfy the following:
1) For all $x,y\in X$, $0\le d(x,y)$. (positivity) 
2) For all $x,y\in X$, $d(x,y)=d(y,x)$. (symmetry)
3) For all $x,y,z\in X$, $d(x,y)\le d(x,z)+d(z,y)$. (triangle-inequality)
This is a homework problem and I'm not sure where to even start. I'm new to the concept of metric spaces and would appreciate any help/direction. 
If x=y, then $\rho(x,y)=\rho(x,x)=(x-x)^2=0$. I'm assuming that will suffice for (1).
 A: $d(x,y)=(x-y)^2$ 
checking for 3rd axiom:
$$(x-y)^2+(y-z)^2 =x^2 -2xy + y^2 +y^2 -2yz +z^2\\=x^2 -2xz +z^2 +2xz -2xy + y^2 +y^2 -2yz\\=(x-z)^2+2x(z-y)-2y(z-y)\\=(x-z)^2+2(z-y)(x-y)  $$
It means 
$$(x-z)^2=(x-y)^2+(y-z)^2-2(z-y)(x-y)$$
Now to satisfy the third axiom $2(z-y)(x-y)\ge0$ which is not possible say for 
$x<y<z$.
A: Let's look at 1) as an example. You want to prove/disprove that
$$\rho(x,y) = (x-y)^2 \geq 0$$
for any pair of real numbers $x$and $y$. Since the square of something is always non-negative, we can see that this holds.
But for $\rho$ to be a metric, 2) and 3) must hold as well! Try proving 2) by noting that
$$(x-y)^2 = (-y + x)^2 = ((-1)(y - x))^2 = (-1)^2(y-x)^2 = (y-x)^2$$
For 3), try plugging in the specific values that David Mitra mentioned in the comments of your question. Do you see that this is a counter-example? What is your final conclusion about $\rho$?
EDIT: There is a fourth axiom which you forgot to mention: That $d(x,y) = 0$ happens if and only if $x=y$.
A: Let's check the triangle-inequality : let $x=0$, $y=1$, $z=2$, then $\rho(x,y)=1$, $\rho(y,z)=1$, and $\rho(x,z)=4$. From "triangle-inequality" we should have
$$\rho(x,z) \leq \rho(x,y) + \rho(y,z)$$
i.e.
$$4 \leq 1 + 1$$
and that's not true, therefore $\rho$ is not a metric.
A: $$ρ(x,z)=(x−z)^2=(x−y+y-z)^2=(x−y)^2 +(y−z)2 +2(z−y)(x−y)= ρ(x,y)+ρ(y,z)+2(z−y)(x−y),$$
$$ρ(x,z)= ρ(x,y)+ρ(y,z)+2(z−y)(x−y)$$
So  triangular inequality is not satisfied in general.
