Show that $d(x,y)=|x-2y|$ defines a metric on $\mathbb R$

Question

Normally I can solve these problems but THIS one irritates me.

$d(x,y)=|x-2y|$ on the real numbers.

I use the axioms (from my course where I'm a student, see Image: http://puu.sh/BfjWu/7f75958f49.png).

M1: $d(x,y)=0\Leftrightarrow x=y$

M2: $d(x,y)=d(y,x)$ (symmetry)

M3: $d(x,y) \leq d(x,z)+d(z,y)$ (triangle inequality)

So can anyone tell me why M2 is wrong? I know M1 fails because we can let $x=1$ and $y=1$ then $1\neq 2$

I know M3 fails because I can let $x=5,z=2,y=1$ then $d(5,1)=|5-2\cdot 1|=3>1+0=|5-2\cdot 2|+|2-2\cdot 1|=d(5,2)+d(2,1)$

But I can't figure out why M2 fails... First I tried this:

$d(x,y)=|x-2y|=|2y-x|=d(y,x)$

Then M2 holds, but I think it's wrong. In the other hand, I see M2 fails if:

$d(x,y)=|x-2y| \neq |y-2x|=d(y,x)$

Can anyone help me to clear it up?

Answer 1

Yeah! Your last argument is the correct one.

You have to show that |x-2y|=|y-2x|.

Which obviously fails because x=1 &y=0 as given in comments does not satisfy it